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1.

Prove that the distance of the roots of the equation `|sintheta_1|z^3+|sintheta_2|z^2+|sintheta_3|z+|sintheta_4|=3fromz=0`is greater than `2//3.`

Answer» We know that `| sin theta_(k)| lt 1`. Given
`|sin theta_(1)|z^(3) + |sin theta_(2)| z^(2) +|sin theta_(3)| z+ |sin theta_(4)| = 3`
`or |3|=||sin theta_(1)|z^(3)+|sin theta_(2)| z^(2) +|sin theta_(3)| z+|sin theta_(4)| = 3`
`lt 1|z^(3)+z^(2) + z + 1`
`lt|z|^(3) +|z|^(2) +|z| +1`
`lt 1+|z| +|z|^(2)+ |z|^(3) +|z|^(4) +....oo`
`or 3 lt (1)/(1-|z|)`
`or 3-3|z| lt 1`
`or 2lt 3|z|`
`or |z| gt (2)/(3)`
Discussion
2.

` theta in [0,2pi]` and `z_(1)`, `z_(2)`, `z_(3)` are three complex numbers such that they are collinear and `(1+|sin theta|)z_(1)+(|cos theta|-1)z_(2)-sqrt(2)z_(3)=0`. If at least one of the complex numbers `z_(1)`, `z_(2)`, `z_(3)` is nonzero, then number of possible values of `theta` isA. InfiniteB. `4`C. `2`D. `8`

Answer» Correct Answer - B
`(b)` If `z_(1)`, `z_(2)`, `z_(3)` are collinear and `az_(1)+bz_(2)+cz_(3)=0`. Then
`a+b+c=0`
Hence `1+|sintheta|+|costheta|-1-sqrt(2)=0`
`implies |sintheta|+|costheta|=sqrt(2)`
`impliestheta=(pi)/(4)`, `(3pi)/(4)`, `(5pi)/(4)`, `(7pi)/(4)`
Discussion
3.

Find the locus of point `z`if `z ,i ,a n di z ,`are collinear.

Answer» Given that points z, i and iz are colliner.
`therefore arg ((z-i)/(iz -i)) = 0, pi`
So,`(z-i)/(iz -i) `is purely real.
`rArr (z-i)/(iz -i) = (vbarz +i)/(-ibarz +i)`
`rArr (z-1)/(z-1) = (barz +i)/(-ibarz + 1)`
`rArr 2zbarz + iz - ibarz- barz - z = 0`
`2(x^(2) +y^(2))-2x -2y =0`
`rArr x^(2) + y^(2) - x-y =0`
Discussion
4.

Let `O`, `A`, `B` be three collinear points such that `OA.OB=1`. If `O` and `B` represent the complex numbers `O` and `z`, then `A` representsA. `(1)/(barz)`B. `(1)/(z)`C. `barz`D. `z^(2)`

Answer» Correct Answer - A
`(a)` Let `A` represents `z_(1)`.
Since `OA.OB=1 :. |z_(1)-0|.|z-0|=1`
`implies |z_(2)|=(1)/(|z|)`
Also, `arg((z_(1)-0)/(z-0))=0impliesarg((z_(1))/(z))=0`
`impliesargz_(1)=argz`
If `theta` is the argument of `z`, then
`z=|z|e^(itheta)`
`:. z_(1)=(1)/(|z|)e^(etheta)=(1)/(|z|^(2))|z|e^(itheta)-(z)/(zbarz)=(1)/(barz)`
`:. A` is `(1)/(baraz)`
Discussion
5.

If `A(z_(1))`, `B(z_(2))`, `C(z_(3))` are vertices of a triangle such that `z_(3)=(z_(2)-iz_(1))/(1-i)` and `|z_(1)|=3`, `|z_(2)|=4` and `|z_(2)+iz_(1)|=|z_(1)|+|z_(2)|`, then area of triangle `ABC` isA. `(5)/(2)`B. `0`C. `(25)/(2)`D. `(25)/(4)`

Answer» Correct Answer - D
`(d)` `|z_(2)+iz_(1)|=|z_(1)|+|z_(2)|impliesz_(2)`, `iz_(1)`, `o` are collinear
`:.arg(iz_(1))=argz_(2)`
`impliesargi+argz_(1)=argz_(2)`
`impliesargz_(2)-argz_(1)=(pi)/(2)`
`z_(3)=(z_(2)-iz_(1))/(1-i)`
`implies(1-i)z_(3)=z_(2)-iz_(1)`
`impliesz_(3)-z_(2)=i(z_(3)-z_(1))`
`:.(z_(3)-z_(2))/(z_(3)-z_(1))=i`
`impliesarg((z_(3)-z_(2))/(z_(3)-z_(1)))=(pi)/(2)` and `|z_(3)-z_(2)|=|z_(3)-z_(1)|`
`:.AC=BC` and `AB^(2)=AC^(2)+BC^(2)`
`impliesAC=(5)/(sqrt(2))`
Required area `=(1)/(2)xx(5)/(sqrt(2))xx(5)/(sqrt(2))=(25)/(4)`sq. units
Discussion
6.

If the imaginery part of `(z-3)/(e^(itheta))+(e^(itheta))/(z-3)` is zero, then `z` can lie onA. a circle with unit radiusB. a circle with radius `3` unitsC. a straight line through the point `(3,0)`D. a parabola with the vertex `(3,0)`

Answer» Correct Answer - A::C
`(a,c)` Let `z-3=re^(iphi)`
`:. (z-3)/(e^(itheta))+(e^(itheta))/(z-3)=re^(i(phi-theta))+(1)/(r )e^(i(theta-phi))`
Imaginargy part of the above `=r sin (phi-theta)-(1)/(r )sin(phi-theta)`
Given that `r sin(phi-theta)-(1)/(r )sin(phi-theta)=0`
`impliesr-(1)/(r )=0` or `sin(phi-theta)=0`
`impliesr-(1)/(r )=0impliesr=1`
`implies|z-3|=1`
`impliesz` lies on a circle with unit radius.
`sin(phi-theta)=0impliesphi=theta`
`:.z-3=re^(itheta)`
`z-3=rcostheta`, `y=r sin theta`
`impliesx-3=rcostheta`, `y=rsin theta`
`(x-3)/(costheta)=(y)/(sintheta)=r`
This represents a straight line through `(3,0)`.
Discussion
7.

Consider the region `R` in the Argand plane described by the complex number. `Z` satisfying the inequalities `|Z-2| le |Z-4|`, `|Z-3| le |Z+3|`, `|Z-i| le |Z-3i|`, `|Z+i| le |Z+3i|` Answer the followin questions : Minimum of `|Z_(1)-Z_(2)|` given that `Z_(1)`, `Z_(2)` are any two complex numbers lying in the region `R` isA. `0`B. `5`C. `sqrt(13)`D. `3`

Answer» Correct Answer - A
`(a)` `|Z_(1)-Z_(2)|_(min)=0`, occurs when `z_(1)` and `z_(2)` coincide.
Discussion
8.

Consider the region `R` in the Argand plane described by the complex number. `Z` satisfying the inequalities `|Z-2| le |Z-4|`, `|Z-3| le |Z+3|`, `|Z-i| le |Z-3i|`, `|Z+i| le |Z+3i|` Answer the followin questions : The maximum value of `|Z|` for any `Z` in `R` isA. `5`B. `14`C. `sqrt(13)`D. `12`

Answer» Correct Answer - A
`(a)` `|Z_(1)-Z_(2)|_(max)=`Length of diagonal
`=sqrt(3^(2)+4^(2))=5`
Discussion
9.

The roots of the equation `z^(4) + az^(3) + (12 + 9i)z^(2) + bz = 0` (where a and b are complex numbers) are the vertices of a square. Then The value of `|a-b|` isA. `5sqrt(5)`B. `sqrt(130)`C. 12D. `sqrt(175)`

Answer» Correct Answer - B
Given equation is `z(z^(3) + az^(3) +(12 + 9i)z + b) =0`
So , either z =0
or `z^(3) + az^(2) + (12 + 9i) z + b = 0`
So, one of the vertice of the square is origin .
Therefore, other three verticles are `z_(1),iz_(1)` and `(z_(1) + iz_(1))` which are the roots equation (1).
`therefore z_(1)(iz_(1)) + (iz_(1))(z_(1) + iz_(1)) + z_(1)(z_(1) + iz_(1)) = 12 +9i`
`rArr_(1)^(2)(i + i+ i^(2)+1+i)= 12+9i`
`rArr 3iz_(1)^(2) = 12+9i`
`rArr z_(1)^(2) = 3 - 4i`
` rArr z_(1) = sqrt(3-4 i) = pm (2-i)`
`therefore z_(1) = 2 - i or -2 +i or -2 + i ` (both value will give the same result )
Also, `-a= z_(1) + iz_(1) + z_(1)(1+i) = 2z_(1)(1+i)`
`= 2(2-i) (1+i) = 6 + 2i`
` therefore a= - 6- 2i`
Futher, `(z_(1))(iz_(1))(1+i) z_(1) = b`
`therefore - b = z_(1)^(3)(-1+i)`
or `b = (2-i)^(3) (1-i) = -9-13i`
` b = -9-13i`
`|z - b|=|3+11i|= sqrt(130)`
`|z_(1)| = sqrt(5)`
Therefore, area of square is 5sq. units.
Discussion
10.

If `z = (3+ 7i) (a +ib)` where a,`b in in Z- {0}`, is purely imaginary , then the minimum value of `|z|` isA. 74B. 45C. 58D. 65

Answer» Correct Answer - C
`z=(3a-7b)+i(3b+7a)`
For z to be purely imaginary, we have
`3a=7b`
`rArra =7 or b=3` (for least value of `|z|`)
`rArr |z| =|3+7i|`
Discussion
11.

Let `z=x+i y`be a complex number where `xa n dy`are integers. Then, the area of the rectangle whose vertices are theroots of the equation `z z ^3+ z z^3=350`is48 (b)32 (c) 40(d) 80A. 48B. 32C. 40D. 80

Answer» Correct Answer - A
`zbarz(barz^(2) + z^(2))= 350`
Putting z +iy, we have
`(x^(2) + y^(2)) (x^(2) -y^(2)) = 175`
`(x^(2) + y^(2)) (x^(2) -y^(2)) = 5 xx 5xx7`
`x^(2)+ y^(2) = 25`
`and x^(2) -y^(2) = 7`
(as other combinations given non-intergral values of x and y )
`therefore x = pm 4, y = pm 3 (x,y in I)`
Hence, area is `8 xx 6 = 48` sq . units
Discussion
12.

If `a`, `b` are complex numbers and one of the roots of the equation `x^(2)+ax+b=0` is purely real whereas the other is purely imaginery, and `a^(2)-bara^(2)=kb`, then `k` isA. `2`B. `4`C. `6`D. `8`

Answer» Correct Answer - B
`(b)` Let us consider `alpha` as the real and `ibeta` as the imaginergy root. Then
`alpha+ibeta=-a`
`implies alpha-ibeta=-bara`
`implies2alpha=-(a+bara)` and `2ibeta=-(a-bara)`
`implies 4ialphabeta=a^(2)-bara^(2)`
`implies a^(2)-bara^(2)=4b`
Discussion
13.

The roots of the equation `t^3+3a t^2+3b t+c=0a r ez_1, z_2, z_3`which represent the vertices of an equilateral triangle. Then`a^2=3b`b. `b^2=a`c. `a^2=b`d. `b^2=3a`A. `a^(2) = 3b`B. `b^(2) = a`C. `a^(2) = a`D. `b^(2) = 3a`

Answer» Correct Answer - C
`S_(1)=Sigmaz_(1)=-3a,S_(2)=Sigmaz_(1)z_(2)=3b`
`Sigmaz_(1)^(2)=Sigmaz_(1)z_(2)`
`rArr (Sigmaz_(1))^(2)-2Sigmaz_(1)z_(2)=Sigmaz_(1)z_(2)`
`rArr (Sigmaz_(1))^(2)=3Sigmaz_(1)z_(2)`
`rArr (-3a)^(2)=3(3b)`
`rArr a^(2)=b`
Discussion
14.

Let `arg(z_(k))=((2k+1)pi)/(n)` where `k=1,2,………n`. If `arg(z_(1),z_(2),z_(3),………….z_(n))=pi`, then `n` must be of form `(m in z)`A. `4m`B. `2m-1`C. `2m`D. None of these

Answer» Correct Answer - B
`(b)` `arg(z_(1),z_(2),z_(3)……….z_(n))=pi`
`impliesarg(z_(1))+arg(z_(2))+….+arg(z_(n))=pi+-2mpi`, `m in I`
`implies (pi)/(n)[3+5+7+….+(2n+1)]=pi+-2mpi`
`implies(pi)/(n)[(n)/(2)[6+2(n-1)]]=pi+-2mpi`
`implies3+n-1=1+-2m`
`impliesn=-1=1+-2m`
Discussion
15.

If `|z_(1)| = sqrt(2), |z_(2)| = sqrt(3) and |z_(1) + z_(2)| = sqrt((5-2sqrt(3)))` then arg `((z_(1))/(z_(2)))` (not neccessarily principal)A. `(3pi)/(4)`B. `(2pi)/(3)`C. `(5pi)/(4)`D. `(5)/(2)`

Answer» Correct Answer - A::C
`|z_(1)+z_(2)|^(2) = |z_(1)|^(2) + |z_(2)|^(2)+ z_(1)barz_(2)+ barz_(1)z_(2)`
`rArr 5-2sqrt(3) = 2+ 3 + 2|z_(1)||z_(2)|cos(theta_(1)-theta_(2))`,
where `arg(z_(1))= theta_(1) and arg(z_(2))= theta_(2)`
`rArr -2sqrt(3) = 2 + 3+ 2|z_(1)||z_(2)|cos(theta_(1)-theta_(2))`
`rArr cos(theta_(1)-theta_(2))= -(1)/(sqrt(2))`
`rArr theta_(1) - theta_(2) = (3pi)/(4),(5pi)/(4)`
`rArr arg(z_(1))-arg(z_(2)) = arg((z_(1))/(z_(2))) = (3pi)/(4),(5pi)/(4)`
Discussion
16.

The roots of the equation `x^(4)-2x^(2)+4=0` are the vertices of `a` :A. square inscribed in a circle of radius `2`B. rectangle inscribed in a circle of radius `2`C. square inscribed in a circle of radius `sqrt(2)`D. rectangle inscribed in a circle of radius `sqrt(2)`

Answer» Correct Answer - D
`(d)` `x^(4)-2x^(2)+4=0`
`impliesx^(2)=1+-sqrt(3)i`
`=2e^(+-(pi)/(3))`
`implies x=+-sqrt(2)e^(+-i(pi)/(6))`
These points lie on a circle of radius `sqrt(2)`, as the diagonals intersect at an angle `pi//3`,
`:.` These are the vertices of an inscribed rectangle.
Discussion
17.

If `z=(3+7i)(a+ib)`, where `a`, `b in Z-{0}`, is purely imaginery, then minimum value of `|z|^(2)` isA. `74`B. `45`C. `65`D. `58`

Answer» Correct Answer - D
`(d)` `z=(3a-7b)+i(3b+7a)`
For purely imaginary, `3a=7b`
`impliesa=7` or `b=3` (for least value of `|z|`)
`implies |z|=|3+7i|=sqrt(58)`
Discussion
18.

If `z_(1),z_(2),z_(3)………….z_(n)` are in `G.P` with first term as unity such that `z_(1)+z_(2)+z_(3)+…+z_(n)=0`. Now if `z_(1),z_(2),z_(3)……..z_(n)` represents the vertices of `n`-polygon, then the distance between incentre and circumcentre of the polygon isA. `0`B. `|z_(1)|`C. `2|z_(1)|`D. none of these

Answer» Correct Answer - A
`(a)` Let vertices be `1, alpha,alpha^(2),……….,alpha^(n-1)`
Given `1+alpha+alpha^(2)+…….+alpha^(n-1)=0impliesalpha^(n)-1=0`
`impliesz_(1),z_(2),z_(3),……….,z_(n)` are roots of `alpha^(n)=1`
which form regular polygon. So distance is zero.
Discussion
19.

`z_(1)`, `z_(2)` are two distinct points in complex plane such that `2|z_(1)|=3|z_(2)|` and `z in C` be any point `z=(2z_(1))/(3z_(2))+(3z_(2))/(2z_(1))` such thatA. `-1 le Re z le 1`B. `-2 le Re z le 2`C. `-3 le Re z le 3`D. None of these

Answer» Correct Answer - B
`(b)` `z=(2z_(1))/(3z_(2))+(3z_(2))/(2z_(1))`
`=(2)/(3)(|z_(1)|)/(|z_(2)|)e^(i(theta_(1)-theta_(2)))+(3)/(2)(|z_(2)|)/(|z_(1)|)e^(i(theta_(2)-theta_(1)))`
`=e^(i(theta_(1)-theta_(2)))+e^(i(theta_(2)-theta_(1)))=2cos(theta_(1)-theta_(2))=2cosalpha`
Discussion
20.

If `z_(1)`, `z_(2)` are complex numbers such that `Re(z_(1))=|z_(1)-2|`, `Re(z_(2))=|z_(2)-2|` and `arg(z_(1)-z_(2))=pi//3` , then `Im(z_(1)+z_(2))=`A. `2//sqrt(3)`B. `4//sqrt(3)`C. `2//sqrt(3)`D. `sqrt(3)`

Answer» Correct Answer - B
`(b)` Let `z_(1)=x_(1)+iy_(1)` and `z_(2)=x_(2)+iy_(2)`
Given `Re(z_(1))=|z_(1)-2|`, `Re(z_(2))=|z_(2)-2|`
`:.y_(1)^(2)-4x_(1)+4=0` and `y_(2)^(2)-4x_(2)+4=0`
So that `(y_(1)-y_(2))/(x_(1)-x_(2))=(4)/(y_(1)+y_(2))` ……………`(i)`
Given `arg(z_(1)-z_(2)=pi//3`
`implies (y_(1)-y_(2))/(x_(1)-x_(2))=sqrt(3)` .........`(ii)`
From `(i)` and `(ii) implies y_(1)+y_(2)=(4)/(sqrt(3))`
Discussion
21.

Write the following complex number in polar form :(i) `-3 sqrt(2) + 3 sqrt(2) i`(ii) `1+i`(iii) `(1+7i)/(2-i)^2`

Answer» (i) Let `z = - 3sqrt(2) + 3sqrt(2)i` Then,
`|z| = sqrt(-3sqrt(2)^(2) + (3sqrt(2))^(2)) =6`
Let `tan alpha |(Im(z))/(Re(z))| = 1 rArr alpha = (pi)/(4)`
Since the point reqresenting z lies in the second quadrant, the agrument of z is given by
`theta = pi - alpha - ((pi)/(4)) = ((3pi)/(4))`
So, the polar form of `z - 3sqrt(2)+3sqrt(2)`i si
`z=|z|(cos theta + i sin theta) = 6 (cos.(3pi)/(4) + i sin .(3pi)/(4))`
(ii) Let `z = 1+i.` Then `|z| = sqrt(1^(2)+1^(2)) = sqrt(2)`. Let
`tan alpha = |(Im(z))/(Re(z))| `
Then, `tan alpha|(1)/(1)| =1 or alpha = (pi)/(4)`
Since the point (1,1) representing z lies in the first quadrant, the argument of z is given by `theta = aloha = pi//4`. So, the polar form of `z =1 + i` is
`z =|z| (cos theta + isin theta) = (cos.(pi)/(4) +isin.(pi)/(4))`
(iii) Let `z = -1-i`. Then `|z| = sqrt((-1)^(2)+ (-1)^(2)) = sqrt(2)`
Let `tan alpha = |(Im(z))/(Re(z))|`
Then , `tan alpha = |(-1)/(-1)| = 1 or alpha = (pi)/(4)`
Since the point (-1,-1) representing z lies in the third quadrant, the argument of z is given by
`theta = -(pi -alpha) = - (pi-(pi)/(4))= (-3pi)/(4)`
So, the polar form of z = - 1- is
`z = |z| (cos theta + isin theta) = sqrt(2) {cos ((-3pi)/(4))+isin((-3pi)/(4)) }`
(iv) Let `z = 1 - i` . Then `|z|= sqrt(1+(-1)^(2)) = sqrt(2)`. Let
`tan alpha =|(Im(z))/(Re(z))|`
Then, `tan alpha =|(-1)/(1)| =1 or alpha = (pi)/(4)`
since the point (1,-1) lies in the fourth quadrant, the argument of z is given by `theta = alpha = - pi//4`. So the polar form of z = 1 - i si
`z =|z|(cos theta + isin theta) = sqrt(2) {cos((-pi)/(4)) +isin((-pi)/(4))} = sqrt(2) (cos.(pi)/(4) -isin.(pi)/(4))`
(v) Let `z = (1+7i) //[(2-i)^(2)]`. Then
`z = (1+7i)/(4-4i+^(2)) = (1+7i)/(3-4i) = ((1+7i)/(3-4))((3+4i)/(3+4i))=(-25+25i)/(25) = -1+i`
`therefore |z| = sqrt((-1)^(2) + (1)^(2) ) = sqrt(2)`
Let `alpha` be the acutue angle given by
`tan alpha =|(Im(z))/(Re(z))| = |(-1)/(1)| = 1`
Then `alpha = pi//4`. Since the point (-1,1) represeting z lies in the second quadrant, we have `theta = arg(z) = pi - alpha = pi - pi//4 = 3pi//4`
Hence, z in the polar form is given by
`z = sqrt(2) (cos. (3pi)/(4) + isin.(3pi)/(4))`
Discussion
22.

Evaluate : (i) `i^(135)` (ii) `i^((1)/(47)` (iii) `(-sqrt(-1))^(4n +3) , n in N` (iv) `sqrt(-25) + 3sqrt(-4) + 2 sqrt(-9)`

Answer» (i) 135 leaves reaminder as 3 when it is divided by 4. Therefore,
`i^(135) = i^(3) = - i`
(ii) `i^((1)/(47)) = (1)/(i^(44)i^(3)) = (1)/(-i^(2))=i`
(iii) `(-sqrt(-1))^(4n + 3) = (-i)^(4n + 3)`
` =(-i)^(4n) (-i)^(3)`
`= {(-i)^(4n) } (-i)^(3)`
`= 1 xx (-i)^(3) = - i^(3) = -i^(3) = - (-i)=i`
(iv) `sqrt(-25) + 3sqrt(-4) + 2 sqrt(-9) = 5i + 6 i + 6i = 17i`
Discussion
23.

If `z_1=a + ib and z_2 = c + id` are complex numbers such that `|z_1|=|z_2|=1 and Re(z_1 bar z_2)=0` , then the pair ofcomplex nunmbers `omega=a+ic and omega_2=b+id` satisfiesA. `|omega_(1)|=1`B. `|omega_(2)|=1`C. `Re(omega_(1)baromega_(2)) = 0`D. `Im(omega_(1)baromega_(2))=0`

Answer» Correct Answer - A::B::C
`|Z_(1)|=|Z_(2)| = 1`
`rArr a^(2) + b^(2) =c^(2) + a^(2) = 1" "(1)`
and `Re(z_(1)barz_(2)) =0`
`rArr Re{(a+ib)(c+id)} = 0`
`rArrac + bd = 0" "(2)`
Now from Eqs. (1) and (2) we get
`a^(2) + b^(2) =1`
` rArr a^(2) + (a^(2)c^(2))/(d^(2)) = 1`
`rArr a^(2) =d^(2)" "(3)`
Also, `c^(2)+d^(2) =1`
`rArr c^(2) +(a^(2)+c^(2))/(b^(2))=1`
`rArr b^(2) = c^(2)`
`|omega_(1)| = sqrt(a^(2) + b^(2)) = sqrt(a^(2) +b^(2)) = 1` [From (1) and (4)]
`and |omega_(2)|=sqrt(b^(2) + d^(2))= sqrt(c^(2) + d^(2))=1`[From (1) and (4)]
Further `Re(omega_(1)baromega_(2)) = Re{(a+ ic)(b-id)}`
`ab+ cd`
`=ab -(ac^(2))/(b)" "["From (2)"]`
`= (ab^(2) - ac^(2))/(b) =0" "["Form (4)"]`
Also, `Im(omega_(1)baromega_(2)) = bc-ad`
`= bc-a(-(ac)/(b)) = ((a^(2) +b^(2))c)/(b) = (c)/(b) = pm1ne 0`
`therefore |oemga_(1)|= 1, |omega_(2)|=1 and Re(omega_(1)baromega_(2))=0`
Discussion
24.

Find the minimum value of `|z-1`if `||z-3|-|z+1||=2.`

Answer» `2=||z-3|-|z+1||le|(z-3)+(z-1)|`
`implies|2z-2|ge2`
`implies|z-1|ge1`
Thus, minimum value of `|z-1|` is 1.
Discussion
25.

If `z_1, z_2, z_3, z_4`are the affixes of four point in the Argand plane, `z`is the affix of a point such that `|z-z_1|=|z-z_2|=|z-z_3|=|z-z_4|`, then prove that `z_1, z_2, z_3, z_4`are concyclic.

Answer» We have,
`|z-z_(1)|=|z-z_(2)|=|z-z_(3)|=|z-z_(4)|`
Therefore, the point having affix z is equidistant from the four points having affixes `z_(1),z_(2),z_(3),z_(4).` Thus, z is the affix of either the center of a circle or the point of intersection of diagonals of a rectangle. Therefore, `z_(1),z_(2),z_(3),z_(4)` are concyclic.
Discussion
26.

`|{:("6i " "-3i " "1" ),("4 " " 3i" " -1"),("20 " "3 " " i"):}|`=x+iy thenA. x=3,y=1B. x=1,y=1C. x=0,y=3D. x=0,y=0

Answer» Correct Answer - D
Given`|{:("6i " "-3i " "1" ),("4 " " 3i" " -1"),("20 " "3 " " i"):}|=x+iy `
`rArr -3i|{:("6i " "-3i " "1" ),("4 " " 3i" " -1"),("20 " "3 " " i"):}|=x+iy`
`rArr " " x+ iy=0 [ because C_2 and C_3` are identical ]
`rArr " "x=0, y=0`
Discussion
27.

A value of for which `(2+3isintheta)/(1-2isintheta)`purely imaginary, is :(1) `pi/3`(2) `pi/6`(3) `sin^(-1)((sqrt(3))/4)`(4) `sin^(-1)(1/(sqrt(3)))`A. `(pi)/(6)`B. `sin^(-1)((Sqrt(3))/(4))`C. `sin^(-1)((1)/(sqrt(3))`D. `(pi)/(3)`

Answer» Correct Answer - C
`(2+ 3isin theta)/(1-2 sin theta) = (2+3i sin theta)/( 1-2i sin theta) xx(1+ 2i sin theta)/(1+ 2isin theta)`
`= ((2- 6 sin^(2) theta)+i( 7 sin theta))/(1 + 4 sin^(2) theta)`
For purely imaginary number
`sin^(2) theta = (1)/(3) or sin theta = (1)/(sqrt(3))`
`therefore theta= sin^(-1).(1)/(sqrt(3))`
Discussion
28.

Find the value of `theta`if `(3+2isintheta)//(1-2isintheta)`is purely real or purely imaginary.A. `(pi)/3`B. `(pi)/6`C. `sin""^(-1)((sqrt(3))/4)`D. `sin""^(-1)(1/sqrt(3))`

Answer» Correct Answer - D
Let `z=(2+3 I sin theta )/(1-2 I sin theta ) ` is purely imaginary . Then we have Re (z)=0
Now , consider `z =(2+3 I sin theta)/(1-2i sin theta)`
`=((2+3 is sin theta )(1+2i sin theta))/((1-2i sin theta)(1+2i sin theta))`
`(2+4i sin theta + 3i sin theta +6i^2 sin ^2 theta)/(1^2-(2i sin theta)^2`
`rArr (2+ 7i sin theta - 6 sin ^2 theta)/(1+4 sin ^2 theta)`
`=(2-6 sin ^2 theta )/(1+4 sin ^ 2 theta)+i(7 sin theta)/(1+4 sin^2 theta)`
`therefore (Re(z)=0)`
`therefore (2-6 sin^2 theta)/(1+ 4sin ^2 theta)=0 rArr 2 =6 sin^2 theta`
`rArr sin^2 theta =1/3`
`rArr sin theta = pm 1/(sqrt(3))`
`rArr theta sin^(-1)(pm 1/(sqrt3)= pm sin^(-1)(1/(sqrt 3)))`
Discussion
29.

The complex number `sin(x)+icos(2x)` and `cos(x)-isin(2x)` are conjugate to each other forA. `x=npi`B. x=0C. `x=(n+1/2)pi`D. no value of x

Answer» Correct Answer - D
Since (sin x+ I cos2 x )=cos x - I sin 2x
`rArr ` sin x- icos 2x =cos x -I sin 2x
`rArr` sin x= cos x and cos 2 x = sin 2x
`rArr ` tan x=1 and tan 2x=1
`rArr x=pi//4 and pi//8 ` which is not possible at same time .
Hence no solution exists.
Discussion
30.

If `cos alpha+cos beta+cos gamma=0=sin alpha+sin beta+sin gamma`, then `(sin3alpha+sin3beta+sin3gamma)/(sin(alpha+beta+gamma))` is equal toA. `1`B. `-1`C. `3`D. `-3`

Answer» Correct Answer - A
`(a)` `a=e^(ialpha)`, `b=e^(ibeta)`, `c=e^(igamma)`
Clearly `a+b+c=0`
`impliesa+b+c=0=(1)/(a)+(1)/(b)+(1)/(c )`
`impliesa^(3)+b^(3)+x^(3)=3abc`
`impliessin3alpha+sin3beta+sin3gamma=sin(alpha+beta+gamma)`
Discussion
31.

Consider the complex number `z = (1 - isin theta)//(1+ icos theta)`. The value of `theta` for which z is purely real areA. `npi-(pi)/(4), n in I`B. `pin +(pi)/(4) , n in I`C. `npi, n in I`D. None of these

Answer» Correct Answer - A
`z=(1-i sin theta)/(1+ i cos theta) = ((1- isin theta)( 1- icos theta))/((1+ i cos theat)(1-i cos theta))`
`=((1-sin theta cos theta) - icos theta + sintheta)/((1+ cos^(2)theta))`
If z is purely real , then
`cos theta + sin theta=0`
or `tan theta = -1`
`rArr = npi - (pi)/(4), n in I`
If z is purely imaginary, `1- sin theta cos theta = 0 or sin theta coe theta = 1`, which is not possible
`|z|= |(1- isin theta)/(1+ i cos theta)| = (sqrt(1+ sin^(2) theta))/(sqrt(1+ cos^(2) theta))`
If `|z|=1`,then
`cos^(2) theta = sin theta^(2) theta`
`rArr tan^(2) theta = 1`
`rArr theta = npipm(pi)/(4), ninI`
We have,
`agr(z) = tan^(-1) ((-(cos theta + sin theta))/((1-sin theta cos theta)))`
Now `arg(z) = pi//4`
`rArr (-(costheta + sintheta))/((1-sin theta cos theta)) = 1`
`rArr cos^(2) theta+ sin^(2) + 2 sin theta cos theta = 1 + sin^(2) theta cos^(2) theta - 2 sin theta cos theta`
`rArr 1+ 4 sin theta cos theta = 1 + sin ^(2) theta cos^(2) theta`
`rArr sin^(2) theta cos^(2) theta - 4 sin theta cos theta =0`
`rArr sin theta cos theta (sin theta cos theta -4) = 0`
`rArr sin theta cos theta = 0 " "(because sin theta cos theta = 4" is not possible ")`
`rArr = (2n + 1) pi or theta ( 4n - 1) pi//2, n in I " "(because - cos theta - sin theta gt 0)`
Discussion
32.

Consider the complex number `z = (1 - isin theta)//(1+ icos theta)`. The value of `theta` for which z is unimodular give byA. `npi pm (pi)/(6), n in I`B. `npi pm (pi)/(3), n in I`C. `npi pm (pi)/(4), n in I`D. no real values of `theta`

Answer» Correct Answer - C
`z=(1-i sin theta)/(1+ i cos theta) = ((1- isin theta)( 1- icos theta))/((1+ i cos theat)(1-i cos theta))`
`=((1-sin theta cos theta) - icos theta + sintheta)/((1+ cos^(2)theta))`
If z is purely real , then
`cos theta + sin theta=0`
or `tan theta = -1`
`rArr = npi - (pi)/(4), n in I`
If z is purely imaginary, `1- sin theta cos theta = 0 or sin theta coe theta = 1`, which is not possible
`|z|= |(1- isin theta)/(1+ i cos theta)| = (sqrt(1+ sin^(2) theta))/(sqrt(1+ cos^(2) theta))`
If `|z|=1`,then
`cos^(2) theta = sin theta^(2) theta`
`rArr tan^(2) theta = 1`
`rArr theta = npipm(pi)/(4), ninI`
We have,
`agr(z) = tan^(-1) ((-(cos theta + sin theta))/((1-sin theta cos theta)))`
Now `arg(z) = pi//4`
`rArr (-(costheta + sintheta))/((1-sin theta cos theta)) = 1`
`rArr cos^(2) theta+ sin^(2) + 2 sin theta cos theta = 1 + sin^(2) theta cos^(2) theta - 2 sin theta cos theta`
`rArr 1+ 4 sin theta cos theta = 1 + sin ^(2) theta cos^(2) theta`
`rArr sin^(2) theta cos^(2) theta - 4 sin theta cos theta =0`
`rArr sin theta cos theta (sin theta cos theta -4) = 0`
`rArr sin theta cos theta = 0 " "(because sin theta cos theta = 4" is not possible ")`
`rArr = (2n + 1) pi or theta ( 4n - 1) pi//2, n in I " "(because - cos theta - sin theta gt 0)`
Discussion
33.

Consider the complex number `z = (1 - isin theta)//(1+ icos theta)`. If agrument of z is `pi//4`, thenA. `theta = npi, n in I` onlyB. `theta= (2n + 1), n in I `onlyC. both `theta= npi and theta = (2n + 1)(pi)/(2), n in I`D. none of these

Answer» Correct Answer - D
`z=(1-i sin theta)/(1+ i cos theta) = ((1- isin theta)( 1- icos theta))/((1+ i cos theat)(1-i cos theta))`
`=((1-sin theta cos theta) - icos theta + sintheta)/((1+ cos^(2)theta))`
If z is purely real , then
`cos theta + sin theta=0`
or `tan theta = -1`
`rArr = npi - (pi)/(4), n in I`
If z is purely imaginary, `1- sin theta cos theta = 0 or sin theta coe theta = 1`, which is not possible
`|z|= |(1- isin theta)/(1+ i cos theta)| = (sqrt(1+ sin^(2) theta))/(sqrt(1+ cos^(2) theta))`
If `|z|=1`,then
`cos^(2) theta = sin theta^(2) theta`
`rArr tan^(2) theta = 1`
`rArr theta = npipm(pi)/(4), ninI`
We have,
`agr(z) = tan^(-1) ((-(cos theta + sin theta))/((1-sin theta cos theta)))`
Now `arg(z) = pi//4`
`rArr (-(costheta + sintheta))/((1-sin theta cos theta)) = 1`
`rArr cos^(2) theta+ sin^(2) + 2 sin theta cos theta = 1 + sin^(2) theta cos^(2) theta - 2 sin theta cos theta`
`rArr 1+ 4 sin theta cos theta = 1 + sin ^(2) theta cos^(2) theta`
`rArr sin^(2) theta cos^(2) theta - 4 sin theta cos theta =0`
`rArr sin theta cos theta (sin theta cos theta -4) = 0`
`rArr sin theta cos theta = 0 " "(because sin theta cos theta = 4" is not possible ")`
`rArr = (2n + 1) pi or theta ( 4n - 1) pi//2, n in I " "(because - cos theta - sin theta gt 0)`
Discussion
34.

Consider the complex number `z = (1 - isin theta)//(1+ icos theta)`. The value of `theta` for which z is purely imaginary areA. `npi-(pi)/(4), n in I`B. `pin +(pi)/(4) , n in I`C. `npi, n in I`D. no real values of `theta`

Answer» Correct Answer - D
`z=(1-i sin theta)/(1+ i cos theta) = ((1- isin theta)( 1- icos theta))/((1+ i cos theat)(1-i cos theta))`
`=((1-sin theta cos theta) - icos theta + sintheta)/((1+ cos^(2)theta))`
If z is purely real , then
`cos theta + sin theta=0`
or `tan theta = -1`
`rArr = npi - (pi)/(4), n in I`
If z is purely imaginary, `1- sin theta cos theta = 0 or sin theta coe theta = 1`, which is not possible
`|z|= |(1- isin theta)/(1+ i cos theta)| = (sqrt(1+ sin^(2) theta))/(sqrt(1+ cos^(2) theta))`
If `|z|=1`,then
`cos^(2) theta = sin theta^(2) theta`
`rArr tan^(2) theta = 1`
`rArr theta = npipm(pi)/(4), ninI`
We have,
`agr(z) = tan^(-1) ((-(cos theta + sin theta))/((1-sin theta cos theta)))`
Now `arg(z) = pi//4`
`rArr (-(costheta + sintheta))/((1-sin theta cos theta)) = 1`
`rArr cos^(2) theta+ sin^(2) + 2 sin theta cos theta = 1 + sin^(2) theta cos^(2) theta - 2 sin theta cos theta`
`rArr 1+ 4 sin theta cos theta = 1 + sin ^(2) theta cos^(2) theta`
`rArr sin^(2) theta cos^(2) theta - 4 sin theta cos theta =0`
`rArr sin theta cos theta (sin theta cos theta -4) = 0`
`rArr sin theta cos theta = 0 " "(because sin theta cos theta = 4" is not possible ")`
`rArr = (2n + 1) pi or theta ( 4n - 1) pi//2, n in I " "(because - cos theta - sin theta gt 0)`
Discussion
35.

The complex number `sin(x)+icos(2x)` and `cos(x)-isin(2x)` are conjugate to each other forA. `x = npi, n in Z`B. `x= 0`C. `x = (n+1//2)pi, n in Z`D. no value of x

Answer» Correct Answer - D
Let `z_(1) = sin x + i cos 2x = cos x - i sin 2x`
Then `barz_(1) = z_(2)`
`rArr sin x - i cos 2x = cos x - i isn 2x `
`rArr sin x = cos x and cos 2x = sin 2x`
`rArr tan x = 1 and tan 2x = 1`
`rArr x = (pi)/(4) and x = (pi)/(8)`
This is not possible.
Hence, there is no value of x.
Discussion
36.

Express the following in `a + ib` form: (i) `((cos theta+ isin theta)/(sin theta + icos theta))^(4)` (ii)` ((cos 2theta -isin2theta)^(4)(cos4theta +isin4theta)^(-5))/((cos 3theta +isin3theta)^(-2) (cos 3theta -isin 3theta)^(-9))` (iii) ` ((sinpi//8 +icos pi//8)^(8))/((sinpi//8-icospi//8)^(8))`

Answer» (i) `((costheta+isintheta)/(sintheta+icostheta))^(4)=((costheta+isintheta)^(4))/(i^(4)(costheta-isintheta)^(4))`
`=((costheta+isintheta)^(4))/((costheta+isintheta)^(-4))`
`=(cos8theta+isintheta)^(8)`
`=cos8theta+isin8theta`
(ii) `((cos2theta-isin2theta)^(4)(cos4theta+isin4theta)^(-5))/((cos3theta+isin3theta)^(-2)(cos3theta-isin3theta)^(-9))`
`=([(costheta+isintheta)^(-2)]^(4)[(costheta+isintheta)^(4)]^(-5))/([(costheta+isintheta)^(3)]^(-2)[(costheta+isintheta)^(-3)]^(-9))`
`=((costheta+isintheta)^(-8)(costheta+isintheta)^(-20))/((costheta+isintheta)^(-6)(costheta+isintheta)^(27))`
`=(costheta+isintheta)^(-8-20+6-27)`
`=(costheta+isintheta)^(-49)`
`=cos49theta-isin49theta`
(iii) `((sinpi//8+icospi//8)^(8))/((sinpi//8-icospi//8)^(8))`
`(i^(8)(cospi//8-isinpi//8)^(8))/((-i)^(8)(cospi//8+isinpi//8)^(8))=(cospi-isinpi)/(cospi+isinpi)=1`
Discussion
37.

If `z_1,z_2,z_3` are three points lying on the circle `|z|=2` then the minimum value of the expression `|z_1|z_2|^2+|z_2+z_3|^2+|z_3+z_1+^2=`

Answer» Correct Answer - 12
`|z|=|z_(2)| =|z_(3)| = 2`
Now, `|z_(1) + z_(2) + z_(3)|^(2) ge 0`
`rArr (z_(1) + z_(2) +z_(3))( barz_(1)+barz_(2) + barz_(3)) ge 0`
`rArr sum|z_(1)|^(2) +sum(z_(1)barz_(2) + barz_(1)z_(2)) ge 0`
`2sum|z_(1)|^(2)+sum(z_(1)barz_(2)+barz_(1)z_(2)) ge sum|z_(1)|^(2)`
`rArr |z_(1) +z_(2)|^(2)+|z_(2)+z_(3)|^(2)+|z_(3) + z_(1)|^(2) ge 12`
Discussion
38.

If `|z_1|=|z_2|=|z_3|=1` then value of `|z_1-z_3|^2+|z_3-z_1|^2+|z_1-z_2|^2` cannot exceedA. `6`B. `9`C. `12`D. none of these

Answer» Correct Answer - B
`(b)` Let `y=|z_(1)-z_(2)|^(2)+|z_(2)-z_(3)|^(2)+|z_(3)-z_(1)|^(2)`
`=(z_(1)-z_(2))(barz_(1)-barz_(2))+(z_(2)-z_(3))(barz_(2)-barz_(3))+(z_(3)-z_(1))(barz_(3)-barz_(1))`
`=6-(z_(1)barz_(2)+z_(2)barz_(1)+z_(2)barz_(3)+barz_(2)z_(3)+z_(3)barz_(1)+z_(1)barz_(3))`...........`(i)`
Now we know
`|z_(1)+z_(2)+z_(3)|^(2) ge 0`
`implies 3+(z_(1)barz_(2)+z_(2)barz_(1)+z_(1)barz_(3)+z_(3)barz_(1)+z_(2)barz_(3)+barz_(2)z_(3)) ge 0`.........`(ii)`
From `(i)` and `(ii)`, `y le 9`
Discussion
39.

If `|z_1-1|lt=,|z_2-2|lt=2,|z_(33)|lt=3,`then find the greatest value of `|z_1+z_2+z_3|dot`

Answer» `|z_(1)+z_(2)+z_(3)|=|(z_(1)-1)+(z_(2)-2)+(z_(3)-3)+6|`
`le|z_(1)-1|+|z_(2)-2|+|z_(3)-3|+6`
`ge1+2+3+6=12`
Hence, the greatest value is 12.
Discussion
40.

If `|z_(1)|=|z_(2)|=|z_(3)|=1 and z_1+z_2+z_3=0`then the area of the triangle whose vertices are `z_1, z_2, z_3`isA. `3sqrt(3//4)`B. `sqrt(3//4)`C. 1D. 2

Answer» Correct Answer - A
`|z_(1)|=|z_(2)|=|z_(3)|=1`
Hence, the cirumcenter of triangle is origin. Also, centroid `(z_(1)+z_(2)+z_(3))//3=0`, which coincides with the circumcentrer. So the triangle is equilateral . Since radius is 1, length of side is `a=sqrt(3)`. Therefore, the area of the triangle is `(sqrt3//4)a^(2)=(3sqrt3//4)`.
Discussion
41.

If the complex numbers `z_1,z_2 " and " z_3` represent the vertices of an equilateral triangle such that `|z_1|=|z_2|=|z_3|," then " z_1+z_2+z_3=0`

Answer» Correct Answer - True
Since `z_1,z_2,z_3` are vertices of equilateral triangle and `|z_1|=|z_2|=|z_3|`
`rArr z_1,z_2,z_3` lie on a circle with centre at origin
`rArr` Circumentre = centroid
`rArr " "0=(z_1+z_2+z_3)/(3)`
`therefore z_1+z_2+z_3=0`
Discussion
42.

`A(z_1)`,`B(z_2)` and `C(z_3)` are the vertices of triangle ABC inscribed in the circle |z|=2,internal angle bisector of angle A meets the circumcircle again at `D(z_4)`.Point D is:A. H.M of `z_(2)` and `z_(3)`B. A.M of `z_(2)` and `z_(3)`C. G.M of `z_(2)` and `z_(3)`D. none of these

Answer» Correct Answer - C
Discussion
43.

Let `z_1, z_2, z_3`be three complex numbers and `a ,b ,c`be real numbers not all zero, such that `a+b+c=0a n da z_1+b z_2+c z_3=0.`Show that `z_1, z_2,z_3`are collinear.

Answer» Given
`a+ b+ c = 0" "(1)`
and ` az_(1) + bz_(2) + cz_(3) = 0" "(2)`
Since a, b c are not all zero, from (2), we have
`az_(1) bz_(2) -(a+b) z_(3) = 0 " "["From"(1), c = - (a+b)]`
or ` az_(1) + bz_(2) = (a+b)z_(3)`
or `z_(3) =(az_(1) + bz_(2))/(a+b) " "(3)`
From (3), it follows tha `z_(3)` divides the line segment joining `z_(1)` and `z_(2)` internally in the ratio b:a
Hence `z_(1), Z_(2) and z_(3)` are coliner.
If a and b are of same sign , then division is in fact internal,
and if a and b are of opposte sign, then division is external in the ratio `|b|:|a|`
Discussion
44.

If `|2z-1|=|z-2|a n dz_1, z_2, z_3`are complex numbers such that `|z_1- (alpha)|< alpha,|z_2-beta||z|`d. `>2|z|`A. `lt|z|`B. `lt2|z|`C. `gt|z|`D. ` gt2|z|`

Answer» Correct Answer - B
`|2z-1|=|z-2|`
or `|2z-1|^(2)=|z-2|^(2)`
`or (2z-1)(2barz-1)=(z-2)(barz-2)`
`or 4zbarz-2barz-2z+1=zbarz-2barz-2z+4`
`or 3|z|^(2)=3`
`or |z|=1`
Again
`|z_(1)+z_(2)|=|z_(1)-alpha+z_(2)-beta+alpha+beta|`
`le|z_(1)-alpha|+|z_(2)-beta|+|alpha+beta|`
`ltalpha+beta+|alpha+beta|`
`=2|alpha+beta|" " [therefore alpha,betagt0]`
`therefore |(z_(1)+z_(2))/(alpha+beta)|lt2`
`or |(z_(1)+z_(2))/(alpha+beta)|lt2|z|`
Discussion
45.

Least positive argument ofthe 4th root ofthe complex number `2-isqrt(12)` isA. `pi//6`B. `5pi//12`C. `7pi//12`D. `11pi//12`

Answer» Correct Answer - B
`(b)` `z^(4)=2(1-sqrt(3)i)=4((1)/(2)-(sqrt(3))/(2)i)`
`=4[cos(-(pi)/(3))+isin(-(pi)/(3))]`
`z=sqrt(2)[cos"(2mpi-(pi//3))/(4)+isin"(2mpi-(pi//3))/(4)]`
For `m=1`, `z=sqrt(2)[cos((5pi)/(12))+isin((5pi)/(12))]`
Discussion
46.

Given that ` 1 + 2|z|^(2) = |z^(2) + 1|^(2) + 2 | z + 1 | ^(2)`, then the value of `|z(z + 1 )|` is ______.

Answer» Correct Answer - 1
`1+2|z|^(2) = |z^(2) +1|^(2)+ 2|z+1|^(2)`
`rArr 1+2barzz = (barz^(2)+1)+2(barz +1)`
`rArr (zbarz)^(2)+ 2(z +barz) +z^(2) + barz^(2) + 2=0`
`rArr (zbarz -1)^(2) + (z+barz +1)^(2) = 0`
`rArr zbarz =1 rArr barz = (1)/(2)`
and `z+barz+1=0`
or `z+(1)/(2)+1=0`
`rArr z^(2) + z =1`
`rArr |z^(2)+z| =1`
Discussion
47.

If `|z-4/z|=2`, then the maximum value of`|Z|`is equal to(1) `sqrt(3)+""1`(2) `sqrt(5)+""1`(3) 2(4) `2""+sqrt(2)`A. `sqrt3 + 1 `B. ` sqrt5 + 1 `C. `2`D. `2 + sqrt2`

Answer» Correct Answer - B
`|Z|= |(Z-(4)/(Z))+(4)/(Z)|le|Z-(4)/(Z)|+(4)/(|Z|) le 2 + (4)/(|Z|)`
`rArr |Z|^(2) - 2 |Z| - 4 le 0`
`rArr [|z| - sqrt(5) + 1)][(Z| - 1(-sqrt(5))]le 0`
`rArr 0 le |Z| le sqrt(5) + 1`
Discussion
48.

The number of complex numbers z such that `|z""""1|""=""|z""+""1|""=""|z""""i|`equals(1) 1 (2) 2 (3) `oo`(4) 0A. `oo`B. `0`C. `1`D. `2`

Answer» Correct Answer - C
Let`z= x +iy`. So
`|z-1| = |z+1| rArr Re(z) =0 rArrx le 0`
`|z -1| = |z + i| rArr x = y`
` |z+1| = |z-i| rArr y = - x`
Only (0,0) will satisfy all conditions.
Therefore, the number of complex numbers(z) is 1.
Discussion
49.

Let `alpha,beta`be real and z be a complex number. If `z^2+alphaz""+beta=""0`has two distinct roots on theline Re `z""=""1`, then it is necessary that :(1) `b"" in (0,""1)`(2) `b"" in (-1,""0)`(3) `|b|""=""1`(4) `b"" in (1,oo)`A. ` beta in ( 1 , oo ) `B. ` beta in ( 0 , 1 )`C. ` beta in (-1, 0 ) `D. `|beta | = 1 `

Answer» Correct Answer - A
Let the roots of the given equation be 1+ ip and 1-ip, where `p in R`
`rArr beta `= product of roots
`= (1+ip)( 1-ip) = 1 p^(2) gt 1 AA p in R`
` rArr beta in (1, oo)`
Discussion
50.

If `(3pi)/(2) lt alpha lt 2 pi`, find the modulus and argument of `(1 - cos 2 alpha) + i sin 2 alpha `.

Answer» `z = (1-cos2 alpha)+ i sin 2alpha`
`= 2sin^(2) alpha + i (2sin alpha cos alpha)`
`=2 sin alpha(sin alpha + i cos alpha)`
`=-2 sin alpha (-sin alpha - i cos alpha)`
`=-2 sin alpha (cos((3pi)/(2)-alpha)+ sin ((3pi)/(2)-alpha))`
Thus, `|z| = - s sin alpha`
Alos, `(3pi)/(2) - alpha in ((-pi)/(2),0)`
Therefore, z lies in the fourth quadrant.
`therefore arg(z) = (3pi)/(2) =- alpha`
Discussion