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251.

Complex numbers whose real and imaginary parts `x` and `y` are integers and satisfy the equation `3x^(2)-|xy|-2y^(2)+7=0`A. do not existB. exist and have equal modulusC. form two conjugate pairsD. do not form conjugate pairs

Answer» Correct Answer - B::C
`(b,c)` `(i) xy gt 0`, `3x^(2)-xy-2y^(2)=-7` or `(3x+2y)(x-y)=-7`
`x` and `y` being integers, we can take
`3x+2y=7` and `x-y=-1`
`x=1`, `y=2`
If `x` and `y` are charged to `-x`, `-y`, equation remains same.
`x=-1`, `y=-2` is also a solution pair. ,brgt `3x+2y=-1` and `x+y=7` do not give integral solutions.
`(ii) xy=0` will not give any integral solution.
`(iii) xy lt 0 3x^(2)+xy-2y^(2)=-7`
`(3x-2y)(x+y)=-7`
`3x-2y=-7` and `x+y=1` leads to `x=-1` `y=2`
`3x-2y=-7` and `x+y=-1` leads to `x=1` `y=-2`
Required complex numbers are `1+2i`, `1-2i`, `-1+2i`, `-1-2i` which form two conjugate pairs.
Discussion
252.

If `z=x+iy (x, y in R, x !=-1/2)`, the number of values of z satisfying `|z|^n=z^2|z|^(n-2)+z |z|^(n-2)+1.` `(n in N, n>1)` is

Answer» Correct Answer - B
The given equation is
`|z|^(n)=(z^(2)+z)^(n-2+1)`
`rArr z^(2)+z` is real
`rArr z^(2)+z=barz^(2)+barz`
`rArr (z-z)(z+barz+1)=0`
`rArr z=barz=x " as " z+barz+1ne0(xne-1//2)`
Hence, the given equation reduces to `x^(n)=x^(n)+x|x|^(n-2+1)`
`rArr x|x|^(n-2)=-1`
`rArr x=-1`
So number of solution is 1.
Discussion
253.

Suppose three real numbers `a`, `b`, `c` are in `G.P.` Let `z=(a+ib)/(c-ib)`. ThenA. `z=(ib)/(c )`B. `z=(ia)/(b)`C. `z=(ia)/(c )`D. `z=0`

Answer» Correct Answer - A::B
`(a,b)` Let `r` be the common ratio of `G.P.a,b,c` we have
`:.z=((a)/(b)+i)/((c )/(b)-i)=((1)/(r )+i)/(r-i)=(i)/(r )`
`z=(ib)/(c )` or `(ia)/(b)`
Discussion
254.

If the sum of square of roots of equation `x^2+(p+iq)x+3i=0` is 8, then find |p|+|q| , where p and q are real.

Answer» Correct Answer - p=3, q=1 pr p =-3, q=-1
Let the roots be `alpha` and `beta`
We have `alpha + beta = -(p + iq),alpha beta = 3i`
Given: `alpha^(2) + beta^(2) = 8`
`or (alpha+beta)^(2) - 2alpha beta = 8`
`or (p + iq)^(2) -6i(2pq-6) = 8`
`or p^(2)-q^(2) + (2pq -6) = 8`
`or p^(2) -q^(2) =8 and pq =3`
`rArr p = 3 and q=1 or p=-3 and q=-1`
Discussion
255.

If `(a+b)-i(3a+2b)=5+2i ,`then find `a and b`

Answer» Correct Answer - a=-12,b=17
We have,
`(a+b) -i(3a + 2b) + 5+2i`
`rArr a+b= 5 and - (3a + 2b) = 2`
`rArr a =- 12,b= 17`
Discussion
256.

Find the centre and radius of the circle formed by all thepoints represented by `z = x + iy` satisfying the relation `|(z-alpha)/(z-beta)|= k (k !=1)`, where `alpha and beta` are the constant complex numbers given by `alpha = alpha_1 + ialpha_2, beta = beta_1 + ibeta_2`.

Answer» Correct Answer - Centre = `(alpha - k^(2) beta)/(1 - k^(2))` , Radius = `|(k (alpha - beta))/(1-k^(2))|`
As know we know `|z|^2=z.barz`
Given `(|z-alpha|^2)/(|z-beta|^2)=k^2`
`(z-alpha)(barz-baralpha)=k^2(barz-barbeta)`
`rArr |z|^2-alphabarz-baralphaz+|alpha^2|=k^2(|z|^ 2-betabarz-barbetaz+|beta|^2)`
` rArr |z|^2(1-k^2)-(alpha - k^2beta)barz -(baralpha- betak^2)z`
`rArr |z|^2-((alpha-k^2beta))/((1-k^2))z-((baralpha-barbetak^2))/((1-k^2))z+(|alpha|^2-k^2|beta|^2)/((1-k^2))=0...(i)`
On comparing with equation of circle ,`|z|^2+a barz+barz+0`
Whose center is (-a) and radius `=sqrt(|a|^2-b)`
`therefore` Center for Eq. (i) `=(alpha-k^2beta)/(1-k^2)`
and radius `=sqrt(((alpha-k^2beta)/(1-k^2))((baralpha-k^2beta)/(1-k^2))-(alphabaralpha-k^2betabarbeta)/(1- k^2))`
`=|(k(alpha-beta))/(1- k^2)|`
Discussion
257.

Consider the equation `az + bar(bz) + c =0`, where a,b,c `in`Z If `|a| = |b| ne 0 and az + b barc+ c=0` representsA. an ellipseB. a circleC. a pointD. a straight line

Answer» Correct Answer - D
`az + b barz + c=0" "(1)`
or `barabarz+ baraz+barc = 0" "(2)`
Eliminating `barz` form (1) and (2) we, get
`z = (cbara - b barc)/(|b|^(2)-|a|^(2))`
If `|a|ne|b|`, then z represents one point on the Argand Plane. If
`|a| = |b| and barac ne b barc`, then no such z exists. Adding (1) and (2),
`(bara + b)barz +(a+ barb) z + (c + barc) = 0`
This is of the form `Abarz + Abarz + B = 0` Where `B = c + barc ` is real.
Hence, locus of is a straight line.
Discussion
258.

If `az^2+bz+1=0`, where `a,b in C`, `|a|=1/2` and have a root `alpha` such that `|alpha|=1` then `|abarb-b|=`A. `1//4`B. `1//2`C. `5//4`D. `3//4`

Answer» Correct Answer - D
`(d)` `aalpha^(2)+balpha+1=0` ……….`(i)`
`implies barabaralpha^(2)+barbbaralpha+1=0`
`implies alpha^(2)+barbalpha+bara=0` (as `|alpha|=alphabaralpha=1`) ……….`(ii)`
From `(i)` and `(ii)`
`(alpha^(2))/(barab-barb)=(alpha)/(1-|a|^(2))=(1)/(abarb-b)implies|abarb-b|=1-|a|^(2)=1-(1)/(4)=(3)/(4)`
Discussion
259.

Given that x, `y in R`. Solve: `x/(1+2i)+y/(3+2i)=(5+6i)/(8i-1)`

Answer» `(x)/(1+2i)+(y)/(3+2i)=(5+6i)/(8i-1)`
or `(x (1-2i))/(1-4i^(2))+(y(3-2i))/(9-4i^(2)) = ((5+6i)(8i+1))/((8i)^(2)-1^(2))`
`or (x-2x i)/(5) +(15y -10yi)/(13)+(40 i + 5- 48+6i)/(-64-1)`
`or (13x -26x i+15y - 10yi)/(65) = (-43 + 46i)/(-65)`
or (13x+ 15y) -i(26x+ 10y)= 43-46i`
Equating real and imaginary parts, we get
`13x + 15y = 43 " "(1)`
`13x + 5y = 23" "(2)`
Solving for x and y we, get x = 1 and y =2
Discussion
260.

If `a,b,c,d in R` and all the three roots of `az^3 + bz^2 + cZ + d=0` have negative real parts, thenA. `ab gt 0`B. `bc gt 0`C. `ad gt 0`D. `bc-ad gt 0`

Answer» Correct Answer - A::B::C
`(a,b,c)` Let `z_(1)=x_(1)`, `z_(2)`, `z_(3)=x_(2)+-iy_(2)`
`implies z_(1)+z_(2)+z_(3)=-(b)/(a)`
`impliesx_(1)+2x_(2)=-(b)/(a) lt 0impliesab gt 0`
Also, `z_(1)z_(2)z_(3)=x_(1)[x_(2)^(2)+y_(2)^(2)]=-(d)/(a)lt0impliesad gt 0`
Also `z_(1)z_(2)+z_(2)z_(3)+z_(1)z_(3)=(c )/(a)`
`implies x_(1)(x_(2)+iy_(2))+x_(1)(x_(2)-iy_(2))+x_(2)^(2)+y_(2)^(2)=2x_(1)x_(2)+x_(2)^(2)+y_(2)^(2) gt 0`
`implies (c )/(a) gt 0`
`implies (b)/(a) (c )/(a) gt 0impliesbc gt 0`
Discussion
261.

Find all non zero complex numbers z satisfying `barz=iz^2`

Answer» Correct Answer - `[z =I , pm (sqrt3)/(2) - (i)/(2)]`
Let z= x+iy.
Given `bar z= iz^2`
`rArr bar(x+ iy)=i(x+I y)^2`
`rArr x-iy =i(x^2-y^2+2i xy)`
`rArr x- iy = -2 xy +I ( x^2 -y^2 +2i xy)`
`rArr x-iy= -2 xy +i(x^2 -y^2)`
Note it is a compound equation , therefore we can generate from it more than one primary equation .
On equationg real and imaginary parts equations.
x= -2 xy and `x^2-y^2+y=0`
`rArr x(1 +2y)=0`
`rArr` x=0 or y = -1/2
When y=- 1/2 , `x^2-y^2+y=0`
`rArr x^2 -1/4-1/2=0 rarr x^2 = 3/4`
`rArr x=pm (sqrt3)/(2)`
Therefore z=0 +i 0, 0+ i , `pm sqrt(3)/2-i/2`
`rArr z=i , pm sqrt(3)/2-i/2 " " [ because z ne 0]`
Discussion
262.

Consider the complex numbers `z_(1)` and `z_(2)` Satisfying the relation `|z_(1)+z_(2)|^(2)=|z_(1)| + |z_(2)|^(2)` One of the possible argument of complex number `i(z_(1)//z_(2))`A. `(pi)/(2)`B. `-(pi)/(2)`C. 0D. none of these

Answer» Correct Answer - C
Given that
`|z_(1) + z_(2)|^(2) = |z_(1)|^(2) + |z_(2)|^(2)`
`rArr |z_(1)|^(2) + |z_(2)|^(2) + z_(1)barz_(1) + barz_(1)z_(2) = |z_(1)|^(2)+|z_(2)|^(2)`
`rArr z_(1)barz_(1) + barz_(1)z_(2) = 0" "(1)`
`rArr (z_(1))/(z_(2)) + (barz_(1))/(barz_(2))" "("dividinbg by " z_(2)barz_(2))`
`rArr (z_(1))/(z_(2))+bar((z_(1)/(z_(2)))) = 0" "(2)`
From (1), `z_(2) barz_(2)` is purely imaginary. From (2) `z_(1)//z_(2)` is purely imaginary. Hence,
`arg((z_(1))/(z_(2))) = pm (pi)/(2) or arg(z_(1)) - arg(z_(2)) = pm(pi)/(2)`
Also, `i(z_(1)//z_(2))` is purely real. Hence its possible arguments are 0 and `pi`.
Discussion
263.

Consider the equation `az + bar(bz) + c =0`, where a,b,c `in`Z If `|a| ne |b|`, then z representsA. circleB. straight lineC. one pointD. ellispe

Answer» Correct Answer - C
`az + b barz + c=0" "(1)`
or `barabarz+ baraz+barc = 0" "(2)`
Eliminating `barz` form (1) and (2) we, get
`z = (cbara - b barc)/(|b|^(2)-|a|^(2))`
If `|a|ne|b|`, then z represents one point on the Argand Plane. If
`|a| = |b| and barac ne b barc`, then no such z exists. Adding (1) and (2),
`(bara + b)barz +(a+ barb) z + (c + barc) = 0`
This is of the form `Abarz + Abarz + B = 0` Where `B = c + barc ` is real.
Hence, locus of is a straight line.
Discussion
264.

If `8i z+12 z^2-18 z+27 i=0,t h e n``|z|=3/2`b. `|z|=2/3`c.`|z|=1`d. `|z|=3/4`A. `|z| = (3)/(2)`B. `|z| = (3)/(4)`C. `|z|=1`D. `|z| = (3)/(4)`

Answer» Correct Answer - A
`8iz^(3)+12z^(2)-18z+27i=0`
or `4iz^(2)(2z-3i)-9(2z-3i)=0`
or `(2z-3i)(4iz^(2)-9)=0`
`rArr z =(3i)/(z)andz^(2)=(9)/(4i)`
`rArr |z|=(3)/(2)and|z^(2)|=(9)/(4)`
`rArr |z|=(3)/(2)`
Discussion
265.

If z is a complex number satisfying the equaiton `z^(6) - 6z^(3) + 25 = 0`, then the value of `|z|` isA. `5^(1//3)`B. `25^(1//3)`C. `125^(1//3)`D. `625^(1//3)`

Answer» Correct Answer - A
Let `z^(3)=t`
`therefore " " t^(2)-6t+25=0`
`rArr t=(6pmsqrt(-64))/(2)=3pm4i=z^(3)`
So, `|z^(3)|=5`
`rArr |z|=5^(1//3)`
Discussion
266.

Consider the complex numbers `z_(1)` and `z_(2)` Satisfying the relation `|z_(1)+z_(2)|^(2)=|z_(1)| + |z_(2)|^(2)` Complex number `z_(1)//z_(2)` isA. purely realB. purely imaginaryC. zeroD. none of these

Answer» Correct Answer - B
Given that
`|z_(1) + z_(2)|^(2) = |z_(1)|^(2) + |z_(2)|^(2)`
`rArr |z_(1)|^(2) + |z_(2)|^(2) + z_(1)barz_(1) + barz_(1)z_(2) = |z_(1)|^(2)+|z_(2)|^(2)`
`rArr z_(1)barz_(1) + barz_(1)z_(2) = 0" "(1)`
`rArr (z_(1))/(z_(2)) + (barz_(1))/(barz_(2))" "("dividinbg by " z_(2)barz_(2))`
`rArr (z_(1))/(z_(2))+bar((z_(1)/(z_(2)))) = 0" "(2)`
From (1), `z_(2) barz_(2)` is purely imaginary. From (2) `z_(1)//z_(2)` is purely imaginary. Hence,
`arg((z_(1))/(z_(2))) = pm (pi)/(2) or arg(z_(1)) - arg(z_(2)) = pm(pi)/(2)`
Also, `i(z_(1)//z_(2))` is purely real. Hence its possible arguments are 0 and `pi`.
Discussion
267.

Consider the equaiton of line `abarz + abarz+ abarz + b=0`, where b is a real parameter and a is fixed non-zero complex number. The locus of mid-point of the line intercepted between real and imaginary axis is given byA. `az- bar(az) =0`B. `az + bar(az) =0`C. `az-bar(az) + b =0`D. `az - bar(az) + 2b = 0`

Answer» Correct Answer - B
Given equation of line is `abarz + abarz + b =0AA b in R`.
Let the PQ be the segement intercept between axes.
For intercept on real axis `Z_(R)`.
`z = barz`
`rArr Z_(R)(a+ bara) + b =0`
` rArr Z_(R) = (-b)/(a + bara)`
For intercept on imaginary `Z_(1)`
`z + barz = 0`
`rArr Z_(1)(bara - a) + b=0`
`rArr Z_(1) = (b)/(a+bara)`
For mid-point,
`z= (Z_(R) + Z_(I))/(2)`
`rArr z = (-b)/(2)[(1)/(bara+a)+(1)/(bara +a)]`
`z= (barab)/((a + bara)(a-bara))`
`rArr z = (barab)/(a^(2) -(a)^(2))`
`(z[a^(2)-(a)^(2)])/(bara) = barz((bara)^(2) - (a)^(2))/(a)`
`rArr az + bar(az) =0`
Discussion
268.

Consider the equaiton of line `abarz + abarz+ abarz + b=0`, where b is a real parameter and a is fixed non-zero complex number. The intercept of line on imaginary axis is given byA. `(b)/(bara-a)`B. `(2b)/(bara-a)`C. `(b)/(2(bara - a))`D. `(b)/(a-bara)`

Answer» Correct Answer - D
Given equation of line is `abarz + abarz + b =0AA b in R`.
Let the PQ be the segement intercept between axes.
For intercept on real axis `Z_(R)`.
`z = barz`
`rArr Z_(R)(a+ bara) + b =0`
` rArr Z_(R) = (-b)/(a + bara)`
For intercept on imaginary `Z_(1)`
`z + barz = 0`
`rArr Z_(1)(bara - a) + b=0`
`rArr Z_(1) = (b)/(a+bara)`
For mid-point,
`z= (Z_(R) + Z_(I))/(2)`
`rArr z = (-b)/(2)[(1)/(bara+a)+(1)/(bara +a)]`
`z= (barab)/((a + bara)(a-bara))`
`rArr z = (barab)/(a^(2) -(a)^(2))`
`(z[a^(2)-(a)^(2)])/(bara) = barz((bara)^(2) - (a)^(2))/(a)`
`rArr az + bar(az) =0`
Discussion
269.

Consider the equaiton of line `abarz + abarz+ abarz + b=0`, where b is a real parameter and a is fixed non-zero complex number. The intercept of line on real axis is given byA. `(-2b)/(a+bara)`B. `(-b)/(2(a+ bara))`C. `(-b)/(a+bara)`D. `(b)/(a+bara)`

Answer» Correct Answer - C
Given equation of line is `abarz + abarz + b =0AA b in R`.
Let the PQ be the segement intercept between axes.
For intercept on real axis `Z_(R)`.
`z = barz`
`rArr Z_(R)(a+ bara) + b =0`
` rArr Z_(R) = (-b)/(a + bara)`
For intercept on imaginary `Z_(1)`
`z + barz = 0`
`rArr Z_(1)(bara - a) + b=0`
`rArr Z_(1) = (b)/(a+bara)`
For mid-point,
`z= (Z_(R) + Z_(I))/(2)`
`rArr z = (-b)/(2)[(1)/(bara+a)+(1)/(bara +a)]`
`z= (barab)/((a + bara)(a-bara))`
`rArr z = (barab)/(a^(2) -(a)^(2))`
`(z[a^(2)-(a)^(2)])/(bara) = barz((bara)^(2) - (a)^(2))/(a)`
`rArr az + bar(az) =0`
Discussion
270.

Consider the complex numbers `z_(1)` and `z_(2)` Satisfying the relation `|z_(1)+z_(2)|^(2)=|z_(1)|^(2) + |z_(2)|^(2)` Possible difference between the argument of `z_(1)` and `z_(2)` is

Answer» Correct Answer - C
Given that
`|z_(1) + z_(2)|^(2) = |z_(1)|^(2) + |z_(2)|^(2)`
`rArr |z_(1)|^(2) + |z_(2)|^(2) + z_(1)barz_(1) + barz_(1)z_(2) = |z_(1)|^(2)+|z_(2)|^(2)`
`rArr z_(1)barz_(1) + barz_(1)z_(2) = 0" "(1)`
`rArr (z_(1))/(z_(2)) + (barz_(1))/(barz_(2))" "("dividinbg by " z_(2)barz_(2))`
`rArr (z_(1))/(z_(2))+bar((z_(1)/(z_(2)))) = 0" "(2)`
From (1), `z_(2) barz_(2)` is purely imaginary. From (2) `z_(1)//z_(2)` is purely imaginary. Hence,
`arg((z_(1))/(z_(2))) = pm (pi)/(2) or arg(z_(1)) - arg(z_(2)) = pm(pi)/(2)`
Also, `i(z_(1)//z_(2))` is purely real. Hence its possible arguments are 0 and `pi`.
Discussion
271.

Find the real values of `x a n d y , if:((1+i)x-2i)/(3+i)+((2-3i)y+i)/(3-i)=i`

Answer» Correct Answer - (x = 3 and y = -1)
`((1 +i)x-2i)/(3+i)+((2-3i)y+i)/(3-i)=i`
`rArr 4x+2i x-6 i-2 `
Discussion
272.

Let z be a complex number satisfying `z^(2) + 2zlambda + 1=0` , where `lambda ` is a parameter which can take any real value. The roots of this equation lie on a certain circle ifA. `-1 lt lambda lt 1`B. `lambda gt 1`C. `lambda lt 1`D. none of these

Answer» Correct Answer - A
`z = - lambda pm sqrt(lambda^(2) -1)`
Case I:
When `-1 lt lambdalt 1`, we have
`lambda^(2) lt 1 rArr lambda^(2) - 1 lt 0`
`z = - lambda pm isqrt(1-lambda^(2))`
`rArr y^(2) = 1 - x^(2) or x^(2) y + y^(2) = 1`
Case II:
`lambda gt 1 rArr lambda^(2) - 1 gt0`
` z=- lambda pm sqrt(lambda^(2) -1)`
`or x = - lambda pm sqrt(lambda^(2) -1),y = 0`
Roots are`(-lambda + sqrt(lambda^(2) -1,0),(-lambda - sqrt(lambda^(2))-1,0)`.One root lies inside the units circle and the other root lies outside the unit circle.
Case III: When `lambda` is very large, then
`z = - lambda- sqrt(lambda^(2) -1) ~~ - 2lambda`
`z=- lambda+ sqrt(lamda^(2) -1) =((-lambda + sqrt(lambda^(2) -1))(-lambda -sqrt(lambda^(2)-1)))/((-lambda - sqrt(lambda^(2) -1)))`
`= (1)/(-lambda-sqrt(lambda^(2)-1)) =- (1)/(2lambda)`
Discussion
273.

Express `1/(1+cos theta-i sin theta)` in the form of `a +ib`.

Answer» Correct Answer - `A + iB = (1)/(2(1 + 3 cos^(2) (theta)/(2))) -I (cot (theta//2))/(1 + 3 cos^(2) (theta//2))`
Now `(1)/((1-cos theta)+2i sin theta)=(1)/(2 sin^2 ""theta/2+4i sin""theta/2 cos""theta/2)`
`=1/(2sin""theta/2(sin^2""theta/2+4 cos ^2""theta/2))xx(sin""theta/2-2 i cos ""theta/2)/((sin""theta/2-2 icos ""theta/2))`
`=(sin""theta/2-2i cos""theta/2)/(2 sin""theta/2(1 +3 cos^2""theta/2))`
`rArr A + iB =(1)/(2(1+3 cos^2""theta/2))-i(cot""theta/2)/(1+3 cos^2 ""theta/2)`
Discussion
274.

If `i z^4+1=0,`then prove that `z`can take the value `cospi//8+is inpi//8.`

Answer» `iz^(4) = -1`
`z^(4) = (-1)/(i)`
`or z^(4) =i`
`or z = (i) ^(1//4)`
`or z = (0+i)^(1//4)`
` or z=(0+i)^(1//4)`
`or z = (cos.(pi)/(2) + isin.(pi)/(2))^(1//4) = cos.(pi)/(8)+isin.(pi)/(8)`
Discussion
275.

Express the following in `a + ib` form: (a) `((cos alpha + i sin alpha)^(4))/((sin beta + i cos beta)^(5))` (b) `((1+ cos phi + i sin phi)/(1 + cos phi - isin phi))^(n)` (c) `((cos alpha + i sin alpha)(cos beta + i sin beta))/((cos gamma + i sin gamma)(cos delta + i sin delta))`

Answer» Correct Answer - (a) ` sin(4 alpha + 5beta) - i cos ( 4alpha + 5 beta)`
(b) ` cos nphi+ isin nphi `
(c) `cos (alpha + beta -gamma -delta) + i sin (alpha + beta- gamma - delta)`
`((cos alpha+sin alpha)^(4))/((sin beta +icos beta)^(5)) = (cos4alpha + isin 4alpha)/(i^(5)(cos beta -isin beta)^(5))`
`=-i(cos 4alpha + isin 4alpha)(cos beta -isin beta)^(5)`
` =- i[cos 4alpha + isin 4alpha][cos 5 beta + isin 5 beta]`
`= -i[cos(4alpha + 5beta)+isin(4alpha + 5beta)]`
`= sin (4alpha + 5beta)-icos (4alpha + 5beta)`
(b) `((1+cos phi + isin phi)/(1+cos phi -isin phi))^(n)`
`=[(2cos^(2) (phi//2) + isin (phi//2) cos(pi//2))/(2cos^(2)(phi//2) -2isin(phi//2) cos(phi//2))]^(n)`
`= [(cos(phi//2) +isin (phi//2) )/(cos(phi//2) -isin (phi//2))]^(n)`
`=[((cos phi+ sin phi)^(1//2))/((cos phi +isin phi)^(-1//2))]^(n)`
`=[cos phi + isin phi]^(n)`
`= cos n phi + isinnphi`
Discussion
276.

Find the value of following expression:`[(1-cospi/(10)+isinpi/(10))/(1-cospi/(10)-isinpi/(10))]^(10)`

Answer» Correct Answer - `-1`
Let `cos.(pi)/(10) - isin.(pi)/(10) = z`
`rArr cos.(pi)/(10) + isin.(pi)/(2) = (1)/(z)`
`rArr [(1-cos.(pi)/(10)+isin.(pi)/(10))/(1-cos.(pi)/(10) -isin.(pi)/(10))]^(10) =((1-z)/(1-(1)/(z)))^(10)`
`={(-(z-1)z)/((z-1))}^(10)`
`=(-z)^(10)`
=`z^(10)`
`=(cos.(pi)/(10) -isin.(pi)/(10))^(10)`
`=cos pi -isin pi`
`=-1`
Discussion
277.

If `2z_1//3z_2`is a purely imaginary number, then find the value of `"|"(z_1-z_2")"//(z_1+z_2)|dot`

Answer» As given, let
`(2z_(1))/(3z_(2))=iy" or "(z_(1))/(z_(2))=(3)/(2)iy`
so that `|(z_(1)-z_(2))/(z_(1)+z_(2))|=|((z_(1))/(z_(2))-1)/((z_(1))/(z_(2))+1)|=|((3)/(2)iy-1)/((3)/(2)iy+1)|=|(1-(3)/(2)iy)/(1+(3)/(2)iy)|=1`
`" "[because|z|=|bar(z)|]`
Discussion
278.

Find the value of `1+i^2+i^4+i^6++i^(2n)`

Answer» `S = 1 + i^(2) + i^(4) + …….+ i^(2n) = 1 - 1 + 1-1 +…….+ (-1)^(n)`
Obsviously it depends on n . Hence,it cannot be determined unless n is known.
Discussion
279.

If `z+1//z=2costheta,`prove that `|(z^(2n)-1)//(z^(2n)+1)|=|tanntheta|`

Answer» `z+(1)/(2)=2costheta`
or `z^(2)-2costhetaz+1=0`
or `z=(2costheta+-sqrt(4cos^(2)theta-4))/(2)`
`=costheta+-isintheta`
Taking positive sign, we get
`z=costheta+isintheta`
`:." "(1)/(z)=(costheta-isintheta)`
`:." "(z^(2n)-1)/(z^(2n)+1)=(z^(n)-(1)/(z^(n)))/(z^(n)+(1)/(z^(n)))`
`=((costheta+isintheta)^(n)-(costheta-isintheta)^(n))/((costheta+isintheta)^(n)+(costheta-isintheta)^(n))`
`=(2isinntheta)/(2cosntheta)`
`=itan ntheta`
Taking negative sign, we get
`(z^(2n)-1)/(z^(2n)+1)=(-2isinntheta)/(2costheta)=-itanntheta`
`implies|(z^(2n)-1)/(z^(2n)+1)|=|+-itantheta|=tanntheta`
Discussion
280.

Let `C_1 and C_2` are concentric circles of radius ` 1and 8/3` respectively having centre at `(3,0)` on the argand plane. If the complex number `z` satisfies the inequality `log_(1/3)((|z-3|^2+2)/(11|z-3|-2))>1,` thenA. z lies outside `C_(1)` but inside `C_(2)`B. z line inside of both ` C_(1)` and `C_(2)`C. z line outside both `C_(1)` and `C_(2)`D. none of these

Answer» Correct Answer - A
`log_(1//3)((|z-3|^(2)+2)/(11|z-3|-2))gt1,11|z-3|-2|gt0`
`rArr (|z-3|^(2)+2)/(11|z-3|-2)lt(1)/(3)`
`rArr (3t-8)(t-1)lt0 " "("where "|z-3|=t)`
`rArr 1lt|z-3|lt8//3`
Discussion