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151.

If `|z-5i|=|z+5i`, then the locus of `zdot`

Answer» Let `z = x+iy`
Then,
`|x+iy-5i| = |x+iy+5i|`
`=>|x+(y-5)i| = |x+(y+5)i|`
`=>x^2+(y-5)^2 = x^2+(y+5)^2`
`=> -10y = 10y`
`=> -20y = 0`
`=> y = 0`
So, imaginary part of `z` is `0`.
We can write it as,
`(z-barz)/2 = 0`
`z - barz = 0`, which is the required locus.
Discussion
152.

If `a ,b ,c`are nonzero complex numbers of equal moduli and satisfy `a z^2+b z+c=0,`hen prove that `(sqrt(5)-1)//2lt=|z|lt=(sqrt(5)+1)//2.`

Answer» `|a| = |b| = |c| = r `
Again ` az^(2) + bz = - c `
` rArr |c| = |-az^(2) - bz| le |a||z^(2)| + |b| |z| `
` rArr r le r |z|^(2) + r |z| `
` rArr |z|^(2) + |z| - 1 ge 0 `
` rArr |z| ge (sqrt5 - 1 )/( 2 ) " " (1)`
Also from ` a z ^(2) = - bz - c,`
`|z|^(2) - |z| - 1 le 0 `
` rArr 0 lt |z| le (sqrt5 + 1 )/(2) " " ` (2)
From (1) and (2) ,
` (sqrt5 - 1 )/(2) le |z| le (sqrt 5 + 1 )/( 2 )`
Discussion
153.

The region represented by the inequality |2z-3i|

Answer» Correct Answer - B
`(b)` `|2z-3i|^(2) lt |3z-2i|^(2)`
`:. 4x^(2)+(2y-3)^(2) lt 9x^(2)+(3y-2)^(2)`
`:.x^(2)+y^(2) gt 1`
`:. |z|^(2) gt 1`
Discussion
154.

What is the locus of `z ,`if amplitude of `(z-2-3i)`is `pi/4?`

Answer» As `z = x+iy`
`:. z-2-3i = x+iy-2-3i =(x-2)+i(y-3)`
As `pi/4` is the amplitude of ` (x-2)+i(y-3)`,
`:. tan (pi/4) = (y-3)/(x-2)`
`=>1 = (y-3)/(x-2)`
`=>y-3 = x-2`
`=> x - y +1 = 0`, which is the locus of `z`.
`:.` Locus of `z` represents a straight line.
Discussion
155.

If `|z-1|lt=2a n d|omegaz-1-omega^2|=a``(w h e r eomegai sac u b erootofu n i t y)`, then complete set of values of `a`is`0lt=alt=2`b. `1/2lt=alt=(sqrt(3))/2`c. `(sqrt(3))/2-1/2lt=alt=1/2+(sqrt(3))/2`d. `0lt=alt=4`A. `0 le ale 2`B. `(1)/(2) le a le (sqrt(3))/(2)`C. `(sqrt(3))/(2)-(1)/(2)lea le (1)/(2) +(sqrt(3))/(2)`D. `0le ale 4`

Answer» Correct Answer - D
`|omegaz-1-omega^(2)|=a`
`rArr |z+1|=arArr|z-1+2|=a`
`rArr |z-1|+2ge arArr 0 le a le 4`
Discussion
156.

Let `Z in C` with Im (z) = 10 and it satisfies `=(2z-n)/(2z+n)=2i-1` For some natural number n. thenA. n=20 and Re (z) =-10B. n=40 and Re (z) =10C. n=40 and Re(z)D. n=20 and Re (z)=10

Answer» Correct Answer - C
Let z=x+10 I, as Im(z) = 10 ( given )
Since z satisfies
`(2z-n)/(2z +n)=2i -1 , n ne N`
`therefore (2x+20 i-n)=(2i -1)(2x+20 I +n)`
`rArr (2x-n)+20 i=(-2 -n- 40 )+(4x+2n- 20)i`
On comparing real imaginary parts ,we get
`2x-n= -2 x-n 40 and 20 = 4x+2n-20`
`rArr 4x = -40 and 20 = 4x 2n = 40 `
`rArr x=-10 and - 40 +2n = 40 rArr n = 40`
So , n = 40 and x= Re (z) = -10
Discussion
157.

If `((a+i)^2)/((2a-i))=p+i q ,`show that: `p^2+q^2=((a^2+1)^2)/((4a^2+1))`.

Answer» `(a+i)^2/(2a-i) = p+iq->(1)`
`:. bar((a+i)^2/bar(2a-i)) = p-iq`
`=>(a-i)^2/(2a+i) = p-iq->(2)`
Multiplying (1) and (2),
`((a+i)^2(a-i)^2)/((2a-i)(2a+i)) = (p+iq)(p-iq)`
`=>((a+i)(a-i))^2/(4a^2 - i^2) = p^2-i^2q^2`
`=>(a^2-i^2)^2/(4a^2+1) = p^2+q^2 ...[As i^2 = -1]`
`=>(a^2+1)^2/(4a^2+1) = p^2+q^2`
Discussion
158.

Find the real values of x and y for which`(1+i)y^2+(6+i)=(2+i)x `

Answer» `(1+i)y^2+(6+i) = (2+i)x`
`=>y^2+iy^2 + 6+i = 2x+ix`
`=>(y^2+6)+i(y^2+1) = 2x + ix`
Comparing real and imaginary parts,
`:. y^2+6 = 2x->(1)`
`y^2+1 = x`
`=>y^2 = x -1 ->(2)`
Putting value of `y^2` in (1),
`=> x-1+6 = 2x => x = 5`
`:. y^2 = 5-1 = 4`
`=> y = +-2`
`:. x = 5, y = +-2.`
Discussion
159.

If `a+i b=(c+i)/(c-i)`, where `c`is real, prove that:`a^2+b^2=1a n d b/a=(2c)/(c^2-1)dot`

Answer» `a+ib = (c+i)/(c-i)`
`=>a+ib = (c+i)/(c-i)**(c+i)/(c+i) = (c^2+i^2+2ci)/(c^2-i^2)`
`=>a+ib = (c^2-1+2ci)/(c^2+1)`
`=>a+ib = (c^2-1)/(c^2+1)+i(2c)/(c^2+1)`
Comparing real and imaginary parts,
`:. a = (c^2-1)/(c^2+1) and b = (2c)/(c^2+1)`
(i) `L.H.S. = a^2+b^2 = [(c^2-1)/(c^2+1)]^2+ [(2c)/(c^2+1)]^2`
`=(c^4+1-2c^2+4c^2)/(c^4+1+2c^2)`
`=(c^4+1+2c^2)/(c^4+1+2c^2)`
`=1 = R.H.S.`

(ii) `L.H.S. = b/a = ((2c)/(c^2+1))/((c^2-1)/(c^2+1))`
`=(2c)/(c^2-1) = R.H.S.`
Discussion
160.

Find the real part of `(1-i)^(-i)dot`

Answer» Let `z= (1-i)^(-i)`. Taking log on both sides, we have
log `z = -I log_(e) (1-i)`
`= -I log_(e)(sqrt(2)(cos.(pi)/(4)- i sin.(pi)/(4)))`
`= - log_(e) (sqrt(2)e^((-ipi//4)))`
`=-i[(1)/(2)log_(e)2 + log_(e) ^(-ipi//4)]`
`= -i[(1)/(2)log_(e),2-(pi)/(4)]`
`= -(i)/(2)log_(e)2-(pi)/(4)`
`rArr z=e^(-pi//4)e^(-i(log2)2)`
`rArr " "Re(Z) = e^(-pi//4) cos((1)/(2) log2)`
Discussion
161.

If `(sqrt(8)+i)^(50)=3^(49)(a+i b)`, then find the value of `a^2+b^2dot`

Answer» Given that
`(sqrt(8) +i)^(50) = 3^(49) (a+ib)`
Taking modulus and squaring both sides, we get
`(8+1)^(50) = ^(98) (a^(2) + b^(2))`
`or 9^(50) = 3^(98) (a^(2) + b^(2))`
` 3^(100) = 3^(98) (a^(2) + b^(2))`
` or (a^(2) + b^(2)) =9`
Discussion
162.

Show that `(x^2+y^2)^4=(x^4-6x^2y^2+y^4)^2+(4x^3y-4x y^3)^2dot`

Answer» `(x^(2) + y^(2)) = |x+ iy|^(8)`
`=|(x+iy)^(2)|^(4)`
`=|(x^(2)-y^(2))+ 2ixy)|^(4)`
`=|[(x^(2) -y^(2))+ 2ixy)]^(2)|^(2)`
`=|(x^(2) -y^(2))^(2) +(2ixy)^(2) + 2(x^(2)-y^(2))(2ixy)|^(2)`
`=|x^(4) -y^(2) -2x^(2)y^(2) - 4x^(2)y^(2) + i(4x^(3)y - 4xy^(3))|^(2)`
`|x^(4) + y^(4) -6x^(2)y^(2) + i(4x^(3)y-4xy^(3))|^(2)`
`=|x^(4) + y^(4) -6x^(2)y^(2) + i(4x^(3)y- 4xy^(3))|^(2)`
`=(x^(4) -6x^(2)y^(2) +y^(4))^(2) + (4x^(3)y - 4xy^(3))^(2)`
Discussion
163.

If `(1+i)(1+2i)(1+3i)1+m)=(x+i y)`, then show that `2xx5xx10xxxx(1+n^2)=x^2+y^2dot`

Answer» We have ,
`(1+i) (1+2i)(1+3i).....(1+ni) = x+iy`
`rArr |(1+i) (1+2i)...(1+ni)|=|x = iy|`
`rArr |1+ i||1+2i|....|1+ni|=|x+iy|" "[because |z_(1)z_(2)...z_(n)|=|Z_(1)||z_(2)| .....|z_(n)|]`
`rArr sqrt(1+1)sqrt(1+4).....sqrt(1+n^(2)) = sqrt(x^(2)+y^(2))`
`rArr 2xx5xx10...(1+n^(2))= (x^(2) + y^(2))`[On squaring both side]
Discussion
164.

If z is a complexnumber of unit modulus and argument q, then `a r g((1+z)/(1+ bar z))`equal(1) `pi/2-theta`(2) `theta`(3) `pi-theta`(4) `-theta`A. `- theta `B. `pi/2- theta`C. ` theta`D. `pi - theta`

Answer» Correct Answer - C
Given |z| =1 arg z= `theta` :. Z = `e^itheta`
`therefore barz =e^(-itheta)rArr bar z = 1/z`
`therefore arg (1+z)/(1+z)=arg (1+z/1+1/z)=arg (z)= theta`
Discussion
165.

Covert the complex number `z = 1 + cos (8pi)/(5) + i. sin (8pi)/(5)` in polar form. Find its modulus and argument.

Answer» `z = 1 + cos (8pi)/(5) + I sin (8pi)/(5)`
`= 2cos^(2)(4pi)/(5) + cos .(4pi)/(5)`
`= 2cos.(4pi)/(5)(cos.(4pi)/(5) + isin .(4pi)/(5))`
`= - 2cos.(pi)/(5)(-cos.(pi)/(5) + isin .(pi)/(5))`
`= 2 cos.(pi)/(5)(cos.(pi)/(5)- isin .(pi)/(5))`
`= 2cos.(pi)/(5) (cos(-(pi)/(5)) + isin (-(pi)/(5)))`
Thus, `|z| = 2 cos .(pi)/(5) and arg(z) = (pi)/(5)`
Discussion
166.

Find the modulus, argument, and the principal argument of the complex numbers.(i) `(tan1-i)^2`

Answer» `z = (tan1 - i)^(2) = (tan^(2) 1-1)- (2 tan1)i`
`|z|= sqrt((tan^(2)1-1)^(2) + 4 tan^(2)1)=sqrt((tan^(2)1+1)^(2))= sec^(2)1`
Since `tan^(2) 1- 1 lt 0 and -2 tan 1 lt 0,` so z lies in the third quadrant.
`rArr " " arg(z) = - pi + tan^(-1)|(2 tan1)/(1-tan^(2))| = - pi + tan^(-1)|tan2| = 2pi`
Discussion
167.

Find the modulus, argument, and the principal argument of the complexnumbers.`(t a n1-i)^2``(i-1)/(i(1-cos(2pi)/5)+s inn(2pi)/5)`

Answer» Correct Answer - `"Modulus"=(1)/(sqrt(2))"cosec"(pi)/(5),"argument"=(11pi)/(20)`
`z=(i-1)/(i(1-cos.(2pi)/(5))+sin.(2pi)/(5))`
`=(i-1)/(i2sin^(2).(pi)/(5) + 2sin.(pi)/(5)cos.(2pi)/(5))`
`= (i-1)/((2sin.(pi)/(5))(cos.(pi)/(5) +isin.(pi)/(5)))`
`|z|=(|i-1|)/((2sin.(pi)/(5))|(cos.(pi)/(5) +isin.(pi)/(5))|)`
`=(sqrt(2))/((2sin.(pi)/(5))|(cos.(pi)/(5)+isin.(pi)/(5))|)`
`=(1)/(sqrt(2)) cosec.(pi)/(5)`
` argz= arg[(i-1)/((2sin .(pi)/(5))(cos .(pi)/(5)+isin.(pi)/(5)))]`
`=arg(-1+i) - arg(2sin.(pi)/(5))-arg(cos.(pi)/(5) + isin.(pi)/(5))`
`=(3pi)/(4) -0-(pi)/(5)=(11pi)/(20)`
Discussion
168.

If complex number `z(z!=2)`satisfies the equation `z^2=4z+|z|^2+(16)/(|z|^3)`,then the value of `|z|^4`is______.

Answer» Correct Answer - 4
We know that
`|z + iomega| le |z| +|iomega| =|z| +|i| |omega|=1`
Given `|z+iomgea| = 2 rArr|z| =|omega| =1`.
Discussion
169.

Modulus of nonzero complex number z satifying `barz + z =0` and `|z|^(2)-4iz=z^(2)` is _____.

Answer» Correct Answer - 2
`barz + z = 0`
`rArr barz = - z`
Now `|z|^(2) -4zi = z^(2)" "["from (1)"]`
`rArr -z^(2) - 4zi = z^(2)`
`rArr 2z = - 4i`
`rArr z = -2i`
`rArr |z| = 2`
Discussion
170.

Find the complex number `z`satisfying `R e(z^2 0=0,|z|=sqrt(3.)`

Answer» `z = x +iy`
or `z^(2) = x^(2) -y + 2ixy `
` rArr Re(z^(2)) = x^(2) -y^(2)`
Also, `|z| = sqrt(x^(2) + y^(2))`
`rArr x^(2) -y^(2) = 0, x^(2) + y^(2) = 3`
`rArr x^(2) = y^(2) = (3)/(2)`
`rArr x =pm sqrt((3)/(2)), y = pm sqrt((3)/(2))`
`rArr z = pm sqrt((3)/(2)) pm sqrt((3)/(2))i`
Thus, there are four complex numbers.
Discussion
171.

If `|z-i R e(z)|=|z-I m(z)|`, then prove that `z`, lies on the bisectors of the quadrants.

Answer» `z = x +iy`
`rArr Re (z) = x, Im (z) = y`
`|z - iRe(z)|=|z- Im (z)|`
`rArr |x + iy -ix|=|x + iy -y|`
`rArr x^(2) +(x-y)^(2) = (x-y)^(2) + y^(2)`
`rArr x^(2) = y^(2)`
` rArr |x| = |y|`
Hence, z lies on the bisectors the quadrants.
Discussion
172.

If `|z-2-i|=|z|sin(pi/4-a r g z)|` , where `i=sqrt(-1)` ,then locus of z, isA. a pair of straing linesB. circleC. parabolaD. ellispe

Answer» Correct Answer - C
We have,
`|(x-2)+i(y-1)|=|z||(1)/(sqrt2)costheta-(1)/(sqrt2)sintheta|`
where `theta =arg z`.
`sqrt((x-2)^(2)+(y-1)^(2))=(1)/(sqrt2)|x-y|`,
which represents a parabola
Discussion
173.

If `z^2+z|z|+|z^2|=0,`then the locus `z`isa. a circleb. a straight linec. a pair of straight line d.none of theseA. a circleB. a straight lineC. a pair of straing lineD. none of these

Answer» Correct Answer - C
`z^(2)+z|z|+|z|^(2)=0`
`rArr ((z)/(|z|))^(2)+(z)/(|z|)+1=0`
`rArr (z)/(|z|)=omega,omega^(2)`
`rArr z-omega |z|orz=omega^(2)|z|`
`rArr x+iy=|z|((-1)/(2)+(isqrt3)/(2))orx+iy=|z|((-1)/(2)-(isqrt3)/(2))`
`rArr =-(1)/(2)|z|,y=|z|(sqrt3)/(2)or x=-(|z|)/(2),y=(|z|sqrt3)/(2)`
`rArr y+sqrt3x=0ory-sqrt(3)x=0rArr y^(2)-3x^(2)=0`
Discussion
174.

If `z=x+iy` is a complex number with `x, y in Q and |z| = 1`, then show that `|z^(2n)-1|` is a rational numberfor every `n in N`.

Answer» `|z|=1`
`impliesz=e^(itheta)=x+iy`
`impliesx=costheta,y=sintheta`
Now `costheta" and "sinthetainQ`. Also,
`|z^(2n)-1|^(2)=(z^(2n-1)(bar(z)^(2n)-1)`
`=(zbar(z))^(2n)-z^(2n)-bar(z)^(2n)+1`
`=2-(z^(2n)+bar(z)^(2n))`
`=2-2cos2n theta=4sin^(2)ntheta`
`implies|z^(2n)-1|=2|sinntheta|`
Now, `sinntheta=""^(n)C_(1)cos^(n-1)thetasintheta-""^(n)C_(3)cos^(n-3)thetasin^(3)theta+...`
`=" Rational number "(because sintheta, costheta" are rationals")`
`implies|z^(2n)-1|=" Rational number "`
Discussion
175.

Prove that the roots of the equation `x^4-2x^2+4=0`forms a rectangle.

Answer» `x^(4)-2x^(2)+4=0`
`:." "x^(2)=(2+-sqrt(4-16))/(2)=1+-sqrt(3)i=2(1+-sqrt(3)i)/(2)`
`=2(cos(pi)/(3)+-isin(pi)/(3))`
`:." "x=+-sqrt(2)(cos(pi)/(3)+-isin(pi)/(3))^(1//2)`
`=+-sqrt(2)(cos(pi)/(6)+-isin(pi)/(6))`
`:." "x=sqrt(2)e^(ipi//6),-sqrt(2)e^(ipi//6),-sqrt(2)e^(-ipi//6)`
Clearly these points forms rectangle inscribed in circle of radius `sqrt(2)`.
Discussion
176.

Show that the polynomial `x^(4p)+x^(4q+1)+x^(4r+2)+x^(4s+3)`is divisible by `x^3+x^2+x+1, w h e r ep ,q ,r ,s in ndot`

Answer» Let `f(x) = x ^(4p) + x^(4q+1) + x^(4r +2) + x^(4s + 3)`. Now
`x^(3) + x^(2) + x + 1 = (x^(2) + 1) (x+1)`
` = (x^(2) - i^(2))(x + 1)`
`= (x + i) (x- i)(x +1)`
` f (i) = = i^(4p) + i^(4q+1) + i^(4r + 2) + i^(4s + 3)`
`= 1 + i^(1) + i^(2) + ^(3)`
`= 1 + i -1-i=0`
`f (-i) = (-i)^(4p+1) +(-1)^(4r+2) + (-i)^(4s +3)`
` = 1 + (-i)^(1) +i= 0`
`f (-1) = (-1)^(4p) + (-1)^(4q+1) + (-1) ^(4r+2)+ (-1)^(4s+3) = 0`
Thus, by factor therem f(x) is divisible by `x^(3) + x^(2) + 1`
Discussion
177.

If `ta n dc`are two complex numbers such that `|t|!=|c|,|t|=1a n dz=(a t+b)//(t-c), z=x+i ydot`Locus of `z`is (where a, b are complex numbers)a. line segment b.straight linec. circled. none of theseA. line segmentB. straight lineC. circleD. none of these

Answer» Correct Answer - C
`z=(at+b)/(t-c)rArrt=(b+cz)/(z-a)`
Now, `|t|=1`
`rArr |(b+cz)/(z-a)|=1`
`rArr |(z+(b)/(c))/(z-a)|=(1)/(|c|)" " (ne 1" as "|c|ne|t|)`
`rArr` locus of z is a circle
Discussion
178.

If `z`is complex number, then the locus of `z`satisfying the condition `|2z-1|=|z-1|`isperpendicular bisector of line segment joining 1/2 and 1circleparabolanone of the above curvesA. perpeciular bisector of line segment joining 1/2 and 1B. circleC. parabolaD. none of the above curves

Answer» Correct Answer - B
`2|z-(1)/(2)|=|z-1|`
`therefore (|z-1|)/(|z-(1)/(2)|)=2`
So, locus of z is a circle.
Discussion
179.

In the Argands plane what is the locus of `z(!=1)`such that `a rg{3/2((2z^2-5z+3)/(2z^2-z-2))}=(2pi)/3dot`

Answer» `arg{(3)/(2)((2z^(2)-5z+3)/(3z^(2)-z-2))}=(2pi)/(3)`
or `arg{(3)/(2)((z-1)(2z-3))/((z-1)(3z+2))}=(2pi)/(3)`
or `arg{(3)/(2)((2z-3)/(3z+2))}=(2pi)/(3)`
or `arg((z-3//2)/(z+2//3))=(2pi)/(3)`
Thus, locus of z is minor arc whose end point are `(3)/(2)` and `(-3)/(2)` and included angle is `(2pi)/(3).`
Discussion
180.

If `|z|=1a n dz!=+-1,`then all the values of `z/(1-z^2)`lie ona line not passing through the origin`|z|=sqrt(2)`the x-axis(d) the y-axisA. a line not passing through the originB. `|z|=sqrt(2)`C. the X-axiesD. the Y-axies

Answer» Correct Answer - D
Let `z=cos theta + I sin theta`
`rArr z/(1-z^2)=(cos theta + sin theta )/(1- (cos 2 theta + sin 2 theta))`
`=(cos theta + I sin theta )/(2 sin ^2 theta - 2 I sin theta cos theta)`
`=(cos theta +I sin theta)/(-2 I sin theta (cos theta +i sin theta))=i/(2 sin theta)`
Hence `z/1-z^2 ` lies on the imaginary axis ie, y-axis .
Alternate Solution
Let `E=(z)/(1-z^2)=z/(zbarz-z^2)=1/(barz -z)` which is an imaginary
Discussion
181.

Suppose z and `omega` are two complex number such that `|z + iomega| = 2`. Which of the following is ture about `|z|` and `|omega|`?A. `|z|=|omega|= (1)/(2)`B. `|z|=(1)/(2),|omega|,|omega| = (3)/(4)`C. `|z| = |omega| = (3)/(4)`D. `|z| = |omega| = 1`

Answer» Correct Answer - D
We know that
`|z + iomega| le |z| +|iomega| =|z| +|i| |omega|=1`
Given `|z+iomgea| = 2 rArr|z| =|omega| =1`.
Discussion
182.

Consider a quadratic equaiton `az^(2) + bz + c=0`, where a,b,c are complex number. The condition that the equaiton has one purely real roots isA. `(c bar a -a bar c)^(2) = (b barc + c bar b)(a bar a - bar a b)`B. `(c bar c -a bar c)^(2)= (b barc - c bar a)^(2) (a bar b + bar a b)`C. `(c bar a - a bar c)^(2) = (b barc + c bar b) (a bar b + a bar b)`D. `(c bar a -a bar c)^(2) = (b barc - c bar)(a bar b - bar ab)`

Answer» Correct Answer - D
Let `z_(1)` (purely imginary ) be a root of the given equation Then,
`z_(1) = - barz_(1)`
and `underline(az_(1)^(2) + bz_(1) + c)=0" "(1)`
`rArr az_(1)^(2) + bz_(1) + c = 0`
`rArr bara barz_(1)^(2) + barb barz_(1) + c = 0`
`rArr bar z bar z_(1)^(2) + bar b bar z_(1) + barc = 0`
`rArr bar a bar z_(1)^(2) - bar b barz_(1) + barc = 0" "(as barz_(1) = - z_(1))" "(2)`
Now Eqs. (1) and (2) must have one common root.
`therefore ( cbara-abarc)^(2) = (barbc+ cbarb) (-abarb - barab)`
Let `z_(1)` and `z_(2)` be two purely imaginary roots. Then,
`barz_(1) = -z_(1), barz_(2) = - z_(2)`
Now , `underline(abarz^(2) + bz + c) = 0" "(3)`
or `az^(2) + bz + c=bar0`
or `bara barz_(20 + barb barz + barc =0`
or `bara z^(2) - barbz + barc = 0" "(4)`
Equations (3) and (4) must be identical as their roots are same.
` therefore (a)/(bara) = -(b)/(barb) =(c)/(barc)`
`rArr abarc = barac, + barab = 0` and `b barc +barbc=0`
. Hence, `barac` is purely real and `abarb` and `bbarc` are purely imaginary .
let `z_(1)` (purely real ) be a root of the given equation . Then ,
`z_(1) = barz_(1)` ltbr gt and ``underline(az_(1)^(2) + bz_(1) + c)= bar0" "(5)`
or `az_(1)^(2) + bz_(1) + c=0`
or `baraz_(1)^(2) + bz_(1) + c = bar0`
or `baraz_(1)^(2) + barb z_(1) + c= 0" "(6)`
Now(5) and (6) must have one common root. Hence,
`(cbara - abarc)^(2) = (b barc - cbarb)(abarb-barab)`
Discussion
183.

Consider a quadratic equaiton `az^(2) + bz + c=0`, where a,b,c are complex number. If equaiton has two purely imaginary roots, then which of the following is not ture.A. `abarb` is purely imaginaryB. `b barc` is purely imaginaryC. `cbara` is purely realD. none of these

Answer» Correct Answer - D
Let `z_(1)` (purely imginary ) be a root of the given equation Then,
`z_(1) = - barz_(1)`
and `underline(az_(1)^(2) + bz_(1) + c)=0" "(1)`
`rArr az_(1)^(2) + bz_(1) + c = 0`
`rArr bara barz_(1)^(2) + barb barz_(1) + c = 0`
`rArr bar z bar z_(1)^(2) + bar b bar z_(1) + barc = 0`
`rArr bar a bar z_(1)^(2) - bar b barz_(1) + barc = 0" "(as barz_(1) = - z_(1))" "(2)`
Now Eqs. (1) and (2) must have one common root.
`therefore ( cbara-abarc)^(2) = (barbc+ cbarb) (-abarb - barab)`
Let `z_(1)` and `z_(2)` be two purely imaginary roots. Then,
`barz_(1) = -z_(1), barz_(2) = - z_(2)`
Now , `underline(abarz^(2) + bz + c) = 0" "(3)`
or `az^(2) + bz + c=bar0`
or `bara barz_(20 + barb barz + barc =0`
or `bara z^(2) - barbz + barc = 0" "(4)`
Equations (3) and (4) must be identical as their roots are same.
` therefore (a)/(bara) = -(b)/(barb) =(c)/(barc)`
`rArr abarc = barac, + barab = 0` and `b barc +barbc=0`
. Hence, `barac` is purely real and `abarb` and `bbarc` are purely imaginary .
let `z_(1)` (purely real ) be a root of the given equation . Then ,
`z_(1) = barz_(1)` ltbr gt and ``underline(az_(1)^(2) + bz_(1) + c)= bar0" "(5)`
or `az_(1)^(2) + bz_(1) + c=0`
or `baraz_(1)^(2) + bz_(1) + c = bar0`
or `baraz_(1)^(2) + barb z_(1) + c= 0" "(6)`
Now(5) and (6) must have one common root. Hence,
`(cbara - abarc)^(2) = (b barc - cbarb)(abarb-barab)`
Discussion
184.

`z_(1) and z_(2)` are the roots of the equaiton `z^(2) -az + b=0` where `|z_(1)|=|z_(2)|=1` and a,b are nonzero complex numbers, thenA. `|a| le 1`B. `|a| le 2`C. `arg(a^(2)) = arg(b)`D. `agr a = arg(b^(2))`

Answer» Correct Answer - B::C
`z_(1)` and `z_(2)` are the roots of the equation `z^(2) -az + b=0`. Hence, `z_(1)+z_(2) = a,z_(1)z_(2) = b`
Now `,|z_(1) +z_(2)| le |z_(1)| + |z_(2) |`
` rArr |z_(1) + z_(2)| = |a| le |1+1=2" "(because |z_(1)|=|z_(2)| =1)`
`rArr arg(a) = (1)/(2)[arg(z_(2) + arg(z_(1))]`
Also , arg(b) `= arg(z_(1)z_(2)) = arg(z_(1)) + arg(z_(2))`
`rArr 2 arg (a) = arg (b)`
Discussion
185.

Let `z`be a complex number satisfying the equation `z^3-(3+i)z+m+2i=0,w h e r em in Rdot`Suppose the equation has a real root. Then root non-real root.

Answer» Let`alpha` be the root . Then,
`alpha^(2) -(3+ i)alpha + m + 2i = 0`
`rArr (alpha ^(2) - 3alpha + m) + i(2-alpha)=0`
`rArr alpha^(2) - 3 alpha +m = 0 and 2-alpha =0`
`rArr alpha = 2 and m =2`
Product of the roots is 2(1+i) with one roots as 2. Hence, the nonreal roots is 1+i.
Discussion
186.

If `z_(r):r = 1,2,3,.....50` are the roots of the equaiton `sum_(r=0)^(50) z^(r)= 0`, then find the value of ` sum_(r=1)^(50)1//(z_(r) - 1)`

Answer» Correct Answer - `-25`
`E = (1)/(z_(1)-1) + (1)/(z_(2)-1) +......+ (1)/(z_(50)-1)`
Where `z_(1),z_(2),…..,z_(50)` are the roots of the equation `z^(51) -1=0` other than 1 Let .
`1+z+z^(2)+……+z^(50)) = (z-z_(2)) …..(1-z_(50))`
v`rArr log (1+z+z^(2) +…..+z^(50)) = log [(z-z_(1))(z -z_(2)).....(1-z_(50))]`
Differentiating both sides w.r.t.s and putting z = 1
`(1+2z+3z^(2)+......+50_(z)^(49))/(1+z+z^(2) +......+z^(50))`
`=(1)/(z-z_(1)) +(1)/(z-z_(2))+.......+(1)/(z-z_(50))`
`rArr(50xx 51)/(2xx51)=-[(1)/(z_(1)-1)+(1)/(z_(2)-1)+......+(1)/(z_(50) -1)]`
`therefore sum(1)/(z_(r)-1) = -25`
Discussion
187.

Let `|Z_(r) - r| le r`, for all `r = 1,2,3….,n`. Then `|sum_(r=1)^(n)z_(r)|` is less thanA. nB. 2nC. n(n+1)D. `(n(n+1))/(2)`

Answer» Correct Answer - C
`|sum_(r=1)^(n)z_(r)|lesum_(r=1)^(n)|z_(r)|lesum_(r=1)^(n)|z_(r)-r|+sum_(r=1)^(n)rle2sum_(r=1)^(n)r`
Discussion
188.

Let z be a non - real complex number which satisfies the equation ` z^(23) = 1 `. Then the value of ` sum_(22)^(k = 1 ) (1)/(1 + z ^( 8k) + z ^( 16k )) `

Answer» Correct Answer - 15
`sum_(k=1)^(22) (1)/(1+z^(8k) +z^(16k)) = sum_(k=1)^(22) (z^(8k)-1)/((z^(8k) -1)(z^(16) + z^(8k) +1))`
`= sum_(k=1)^(22)(z^(8k)-1)/(z^(24k)-1)`
`=sum_(k=1)^(22) (z^(8k)-1)/(z^(k)-1)" "(because z^(24k)=z^(23k) z^(k) = 1.z^(k) = z^(k))`
`=sum_(k=1)^(22)(1+z^(k) + z^(2k) +......+z^(7k))` ltbvrgt `=22 +(0-1)xx7`
= 15
Discussion
189.

Suppose n is a natural number such that `|i + 2i^2 + 3i^3 +...... + ni^n|=18sqrt2` where `i` is the square root of `-1`. Then n isA. `9`B. `18`C. `36`D. `72`

Answer» Correct Answer - C
`(c )` `S=i+2i^(2)+3i^(3)+…….+ni^(n)`………..`(i)`
`:. iS=i^(2)+2i^(3)+3i^(4)+…….+ni^(n-1)`…………`(ii)`
Subtracting, we get ,brgt `S(1-i)=i+i^(2)+i^(3)+………..+i^(n)-ni^(n+1)`
`=(i(1-i^(n)))/(1-i)-ni^(n+1)`
Now put the values of `n` and verify.
Discussion
190.

The value of `sum_(n=0)^(100)i^(n!)` equals (where `i=sqrt(-1))`A. `-1`B. `i`C. `2i+95`D. `97+i`

Answer» Correct Answer - C
`(c )` `S=sum_(n=0)^(100)(i)^(n!)`
`=(i)^(0!)+(i)^(1!)+(i)^(2!)+…..`
`=i+i-1+i^(6)+i^(24)+(i)^(5!)+(i)^(6!)+..+(i)^(100!)`
`=95+2i`
Discussion
191.

Let `omega` be the complex number `cos((2pi)/3)+isin((2pi)/3)`. Then the number of distinct complex cos numbers z satisfying `Delta=|(z+1,omega,omega^2),(omega,z+omega^2,1),(omega^2,1,z+omega)|=0` is

Answer» Correct Answer - 1
Let `A=[(1" "omega" "omega^2),(omega" "omega^2" "1),(omega^2" " 1 " "z+omega)]`
`A=[(0" "0" "0),(0" "0" "0),(0" "0" "0)]`and Tr (A)=0,|A|=0
`A^3=0`
`A=[(z+1" "omega" "omega^3),(omega" "z+omega^2" "1),(omega^2" "1" "z+omega)]`=[A+zl]=0
`rArr " "z^3=0`
`rArr z=0 ` the number of z satisfying the given equation is 1.
Discussion
192.

If `omega`is an imaginary fifth root of unity, then find the value of `loe_2|1+omega+omega^2+omega^3-1//omega|dot`

Answer» Here, `omega^(5)=1.` Therefore, `omega^(-1)=omega^(4)`
Also, `1+omega+omega^(2)+omega^(3)+omega^(4)=0`
`:." "log_(2)|1+omega+omega^(2)+omega^(3)-(1)/(omega)|`
`=log_(2)|1+omega+omega^(2)+omega^(3)-omega^(4)|`
`=log_(2)|-2omega^(4)|=log_(2)2=1" "(because|omega|=1)`
Discussion
193.

The value of the expression `1.(2-omega).(2-omega^2)+2.(3-omega)(3-omega^2)+.+(n-1)(n-omega)(n-omega^2),` where omega is an imaginary cube root of unity, is………

Answer» Correct Answer - `((n(n+1))/(2))^(2) - n`
Here `T_r=(r-1)(r-omega)(r-omega)^2]=(r^3-1)`
`therefore " "S_n=underset(r=1)overset(n)Sigma(r^3-1)=[(n(n+1))/2]^2-n`
Discussion
194.

If `x=omega-omega^2-2` then , the value of `x^4+3x^3+2x^2-11x-6` is (where `omega ` is a imaginary cube root of unity)

Answer» Correct Answer - 1
We have `x= omega - omega^(2) - 2 or x + 2 = omega -omega^(2)`
Squaring, `x^(2) + 4x + 4 = omega^(2) + omega^(4) - 2omega^(3) = omega^(2) + omega -2=-3`
`rArr x^(2) + 4x + 7 =0`
Dividing `x^(4) + 3x^(3) + 2x^(2) -11x - 6 by x^(2) + 4x +7`, we get
`x^(4) + 3x^(3) + 2x^(2) - 11x -6 = (x^(2) + 4x + 7) (x^(2) -x-1)+1`
`= (0) (x^(2) -x-1)+ 1=0 +1 =1`
Discussion
195.

if `alpha` and `beta` are imaginary cube root of unity then prove `(alpha)^4 + (beta)^4 + (alpha)^-1 . (beta)^-1 = 0`

Answer» Complex cube roots of unity are `omega,omega^(2)`. Let `alpha = omega, beta = omega^(2)`. Thne
`alpha + beta^(2)+ alpha^(-1)beta^(-1)= omega^(4) +(omega^(2))^(4) +(omega^(-1))(omega^(2))^(-1)`
`omega +omega^(2) + omega^(2) + 1=0`
Discussion
196.

The value of `underset(k=1)overset(6)(sin (2 pi k)/(7) = I cos (2 pi k )/(7)` isA. -1B. 0C. `-i`D. I

Answer» Correct Answer - D
`(underset(k=1)overset(6)Sigmasin"" (2kpi)/7-i cos ""(2kpi)/7)`
`=underset(k=1)overset(6)Sigma-i (cos""(2kpi)/7+sin""(2kpi)/7)`
`=i{underset(k=1)overset(6)Sigma-i`
Discussion
197.

The cube roots of unity when represented on Argand diagram form the vertices of an equilateral triangle.

Answer» Correct Answer - True
Since cube roote of unity are 1, `omega , omega^2` given by
`A(1,0),B(-1/2,sqrt(3)/2),C(-1/2,-sqrt(3)/2)`
`rArr AB= BC=CA= sqrt(3)` ltb rgt Hence , cube roots of untiy from an equilateral triangle .
Discussion
198.

If `a^3+b^3+6a b c=8c^3&omega`is a cube root of unity then:`a , b , c`are in `AdotPdot`(b) `a , b , c ,`are in `HdotPdot``a+bomega-2comega^2=0``a+bomega^2-2comega=0`A. `a,c,b` are in A.PB. a,c,b are in H.PC. `a+bomega - 2comega^(2) = 0`D. `a+ bomega^(2) -2comega = 0`

Answer» Correct Answer - A::C::D
We have `a^(3) + b^(3) - 8c^(3) + 6abc = 0`
`rArr a^(3) + b^(3) + (-2c)^(3) - 3ab(-2c) = 0`
`rArr (a + b- 2c) (a+bomega - 2comega^(2)) (a+ boemga^(2)-2comega) =0`
Discussion
199.

If `omega(!=1)`is a cube root of unity,and `(1""+omega)^7=""A""+""Bomega`. Then (A, B) equals(1) (0, 1) (2) (1, 1) (3) (1, 0) (4) `(-1,""1)`A. 0,1B. 1,1C. 1,2D. `-1,1`

Answer» Correct Answer - B
`(1+omega)^7=(1+omega)(1+omega)^6`
`=(1+omega)(-omega^2)^6=1+omega`
`rArr " "A+Bomega=1+omega`
`rArr" "A=1,B=1`
Discussion
200.

If `omega ` is the imaginary cube roots of unity, then the number of pair of integers (a,b) such that `|aomega + b| = 1` is ______.

Answer» Correct Answer - 6
We have `|aomega + b| = 1`
`rArr |aomega + b|^(2) =1`
`rArr (aomega + b)(abaromega +b) = 1`
`rArr a^(2) + ab (omega + baromega) +b^(2) =1`
`rArr a^(2) -ab + b^(2)=1`
`rArr (a-b)^(2)+ab = 1" "(1)`
When `(a-b)^(2) = 0` and ab = 1, `then (1,1),(-1,-1)`
when `(a-b)^(2) = 1 and ab = 0`, then `(0,1),(1,0),(0,-1),(-1,0)`
Hence, there are 6 ordered pairs.
Discussion