InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Let A,B,C be three sets of complex number as defined below `A={z:lm (z) ge 1}` `B={z:|z-2-i|=3}` `C={z:Re((1-i)z)=sqrt(2)}` Let z be any point in `A cap B cap C` and let w be any point satisfying `|w-2-i|lt 3` Then `|z|-|w|+3` lies betweenA. `-6 and 3`B. `-3 and 6`C. `-6 and 6`D. `-3 and 9 ` |
|
Answer» Correct Answer - D Since `|w-(2+i)| lt 3 rArr |w|-|2+i|lt 3 ` `rArr -3+sqrt(5)lt |w| lt 3 + sqrt(5)` ltbr `rArr -3-sqrt(5)lt -|w|lt 3 -sqrt(5)` Also `|z-2(2+i)|=3` `rArr -3 + sqrt(5) le |z| le 3 +sqrt(5)` `therefore -3 lt |z| -|w|+3 lt 9 ` |
|
| 102. |
Let `S=S_1 cap S_2 cap S_3` where `S_1={z in C:|z| lt 4"}",S_2={z in C: lm[(z-1+sqrt(3)i)/(1-sqrt(3)i)]gt0}` and `S_3:{z in : Re z lt 0 }` Let z be any point in `A cap B cap C` The `|z+1-i|^2+|z-5-i|^2` lies betweenA. 25 and 29B. 30 and 34C. 35 and 39D. 40 and 44 |
|
Answer» Correct Answer - C `|z+1-i|^2+|z-5-i|^2` = `(x+1)^2 +(y-1)^2+(x-5)^2+(y-1)^2` `=2(x^2+y^2-4x-2y)+28` `=2(4)+28=36 " "[ because x^2+y^2-4x-2y =4]` |
|
| 103. |
Find nonzero integral solutions of `|1-i|^x=2^xdot` |
|
Answer» We have, `|1-i|^(x) = 2^(x)` `(sqrt(1^(2) +(-1)^(2)))^(x) = 2^(x)` `rArr (sqrt(2))^(x) = 2^(x)` `or 2^(x//2) = 1` `or 2^(x//2) = 2^(x)` `or 2^(x//2) = 2^(0) or (x)/(2) = 0 or x = 0` Hence, the given equation has no solution. |
|
| 104. |
Let z be a complex number satisfying `|z| = 3 |z-1|`. Then prove that `|z-(9)/(8)| = (3)/(8)` |
|
Answer» Let `z = x + iy` Now, `|z| = 3|z-1|` `rArr|z|^(2) = 9|z-1|^(2)` `rArr |x+iy|^(2) = 9|(x -1)+iy|^(2)` `rArr x^(2) +y^(2) = 9 [(x-1)+iy|^(2)` `rArr 8x^(2) + 8y^(2) - 18x x + 9 = 0` `rArr (x-(9)/(8))^(2) + y^(2) =(9)/(64)` `rArr |(x-(9)/(8)) + iy| =(3)/(8)` `rArr |z-(9)/(8)| = (3)/(8)` |
|
| 105. |
Match the statements of Column I with those of Column II Here ,z takes values in the complex plane and Im (z) and Re(z) denote respectively , the imaginary part and the real part of z |
|
Answer» Correct Answer - `A rarr q,r ; B rarr p ; C to p , s , t ; D to q ,r , st ` A.Let Z=x+iy `rArr we get ysqrt(x^+y^2)=0` `rArr y=0` `rArr I_m(z)=0` B. We have `rArr 10_e =8` `rArr e=4/5` `rArr b^2=25(1-16/25)=9` `therefore x^2/25+y^2/9=1` C. Let `w=2(cos theta +I sin theta)` `therefore z=2 (cos theta+ i sin theta)-(1)/(2 cos theta _ i sin theta)` `=2 (cos theta + i sin theta)-1/2(cos theta - sin theta)` `=3/2 cos theta + 5/2 i sin theta` Let z=x+iy `rArr x=3/2 cos theta and y= 5/2 sin theta` `rArr ((2x)/(3))^2+((2y)/(5))^2=1` `rArr (x^2)/(9//4)+(y^4)/(25//4)=1` `therefore e=sqrt(1-9//4/25//4=4/5` D Let `W= cos theta + i sin theta` Then `z=x+iy = cos theta + i sin theta +1/( cos theta + i sin theta)` `= 2 cos theta` `rArr x=2 cos theta y=0` |
|
| 106. |
If `z=x+iy` and `w=(1-iz)/(z-i)`, then `|w|=1` implies that in the complex plane(A)`z` lies on imaginary axis (B) `z` lies on real axis (C)`z` lies on unit circle (D) None of theseA. z lies on the imaginary axisB. z lies on the real axisC. z lies on the unit circleD. None of these |
|
Answer» Correct Answer - B Since `|w| =1 rArr |(1-iz)/(z-i)|=1` `rArr |z-i|=|1-iz|` `rArr |z-i|=|z+i| [ therefore |1-iz|=|-i||z+i|=|z+i|]` `therefore` It is a perpendicular bisector of (0,1) i.e. X-axis .Thus z lies on the real axis |
|
| 107. |
If `alpha,beta,gamma` are the cube roots of p, then for any x,y,z `(x alpha + y beta + z gamma)/(x beta + y gamma + z alpha` = |
|
Answer» Correct Answer - `omega^(2)` `(xalpha+ybeta+zgamma)/(zbeta+ygamma+zalpha)=(x(p)^(1//3)+y(p)^(1//3)+omega+z(p)^(1//3)omega^2omega^2x(+yomega+zomega^2))/(x(p)^(1//3)omega^2+y(p)^(1//3)omega^(3)+z(p)^(1//3)omegaomega^2(xomega+yomega^2+z))` `(omega^2(x+yomega+zomega^2))/(x+yomega +zomega^2)=omega` |
|
| 108. |
If the expression`([s in(x/2)+cos(x/2)-i t a n(x)])/([1+2is in(x/2)])`is real, then the set of all possible values of `x`is......... |
|
Answer» Correct Answer - x = `2 n pi + 2 alpha , alpha = tan^(-1) k ` , where k `in (1 ,2 )` or x = 2 `n pi` `((sin x/2+cos x/2)- I tan x )/(1+2 i sin x/2)in R` `=((sin"" x/2+ cos""x/2-i tanx)(1-2 i sin ""x/2))/(1+4 sin^2""x/2)` Since , it is real so imaginary part will be zero `therefore -2 sin ""x/2(sin ""x/2+ cos ""x/2)-tan x=0` `rArr -2 sin ""x/2(sin ""x/2+ cos ""x/2)cos x+2 sin ""x/2cos ""x/2=0` `rArr sin ""x/2[(sin""x/2cos ""x/2)(cos^2""x/2- sin ^2""x/2)+cos""x/2]=0` `therefore sin ""x/2=0` `rArr x= 2 npi` or `(sin ""x/2+cos ""x/2)(cos^2""x/2- sin ^2""x/2)+cos""x/2=0` on dividing by `cos^3""x/2` we get `(tan ""x/2+1)(1-tan^2""x/2)+(1+tan^2""x/2)=0` `rArr tan^3""x/2-tan""x/2-2=0` Let `tan""x/2=t` and `f(t)=t^3-t-2` then `f(1)=-2 lt 0` and `f(2)=4 gt 0 ` Thus f(t) changes sign from negative to positve in the interval (1,2) `therefore` Let t= k be the root for which f(k)=0 and `k in (1,2)` `therefore` t=k or `tan ""x/2=k = tan alpha` `rArr x//2 = npi+alpha` `rArr {:{(x=2npi+2alpha ","alpha=tan^(-1)k",where k" in "(1,2)"),(" "or x=2npi):}` |
|
| 109. |
Find the square roots of the following:(i) `7-24 i`(ii) `5+12 i` |
|
Answer» (i) Let `sqrt(7-42i) = x + iy ,` Then `sqrt(7-24i) = x+iy` ` or 7-24i = (x+iy)^(2)` `or 7 -24i =(x^(2) -y^(2)) + 2iy` `or x^(2) - y^(2) = 7" "(1)` ` and 2xy = - 24" "(2)` Now, `(x^(2)+y^(2) )^(2) = (x^(2)-y^(2))^(2)+4x^(2)y^(2)` ` or (x^(2)+y^(2))^(2) = 49+576= 625` ` or x^(2) +y^(2)" "[because x^(2)+y^(2) gt 0]" "(3)` on solving (1) and (3), we get `x^(2) = 16 and y^(2) = 9 rArr x = pm 4 and y = pm 3` From (2), 2xy is negative. So, x and y are of opposite signs. Hence, x =4 and y = - 3 or x= - 4 and y = 3 Hence, `sqrt(7-24i)= pm (4-3i)` (ii) Let `sqrt(5+12i) = x + iy`, Then `sqrt(5+12i) = x+iy` `or 5+12i=(x+iy)^(2)` `or 5+12i=(x^(2) -y^(2)) + 2ixy` `or x^(2) - y^(2) = 5" (1)` `and 2xy = 12" "(2)` Now,`(x^(2) +y^(2))^(2)=(x^(2)-y^(2))^(2)+4x^(2)y^(2)` `or (x^(2) +y^(2))^(2) = 5^(2) + 12^(2) = 169` `or x^(2)+y^(2) = 13" "(because x^(2)+y^(2) gt 0)" (3)` On sloving (1) and (2) , we get `x^(2) = 9 and y^(2) = 4 rArr x = pm3 and y = pm 2` From (2), 2xy is positive . So, x and y are of the same sign. Hence, `x = 3 and y = 2 or x = - 3 and y = -2` Hence, `sqrt(5+12i) = pm (3+2i)` (iii) Let `sqrt(-15 -8i) = x+iy`. Then `sqrt(-15-8i) = x + iy` ` or -5-8i = (x+iy)^(2)` `or -15-8i= (x+iy)^(2)` `or -15 = x^(2) -y^(2)" "(1)` `and 2xy = - 8 " "(2)` Now `,(x^(2)+y^(2))^(2) = (x^(2) -y^(2))^(2) + 4x^(2) y^(2)` `or (x^(2)+y^(2))^(2) = (-15)^(2) + 64 = 289` `or x^(2)+y^(2) = 17` On sloving (1) and (3), we get `x^(2) = 1 and y^(2) = 16 rArr x = pm 1 and y = pm 4` From (2), 2xy is negative . So, x and y are of oppsite signs. Hence, x = 1 and y = - 4 or x = -1 and y = 4 Hence, `sqrt(-15-8i) =pm (1-4i)`. |
|
| 110. |
Find all possible values of `sqrt(i)+sqrt(-i)dot` |
|
Answer» `sqrt(i) + sqrt(-i) = sqrt(0+1.i)+sqrt(0-1.i)` Now`sqrt(a+i.b) = {sqrt((1)/(2){sqrt(a^(2)+b^(2))+a})+ isqrt((1)/(2){sqrt(a^(2)+b^(2))-a})}` `and sqrt(a-i.b) =om{sqrt((1)/(2){sqrt(a^(2)+b^(a))+a}-i)sqrt((1)/(2){sqrt(a^(2)+b^(2))-a})}` `rArr sqrt(0+1.i) = pm{sqrt((1)/(2){sqrt(0 +1^(2)) +0})+isqrt((1)/(2){sqrt(0+1^(2)) -0})}` `= pm (1)/(sqrt(2)) (1-i)` ` and sqrt(0-1.i)= pm {sqrt((1)/(2) {sqrt(0+1^(2))+0})-isqrt((1)/(2){sqrt(0+1^(2))-0})}` ` = pm (1)/(sqrt(2)) (1-i)` Now ` sqrt(i) + sqrt(-i) = pm (1)/(sqrt(2)) (1+i) pm (1)/(sqrt(2))(1-i)` `or sqrt(i) + sqrt(-i) =pmsqrt(2) + 0.ior 0pm sqrt(2i)` |
|
| 111. |
For any two complex number `z_1,z_2` and any real numbers a and b, `|az_1-bz_2|^2+|bz_1+az_2|^2`=…. |
|
Answer» Correct Answer - `(a^(2) + b^(2)) (|z_(1)|^(2) + |z_(2))|^(2))` `|az_1-bz_2|+|bz_1+az_2|^2` `=[|a|^2|z_1|^2+b^2|z_2|^2-2ab Re (z_1barz_2)]+[a^2|z_1|^2+a^2|z_2|^2+2abRe(z_1barz_2)]` `=(a^2+b^2)(|z_1|^2+|z_2|^2)` |
|
| 112. |
If one root of the equation `z^2-a z+a-1=0i s(1+i),w h e r ea`is a complex number then find the root. |
|
Answer» Correct Answer - z=1 `z^(2) -az + a 1 =0` Putting z = 1 + i a=2+i `rArr z^(2) - (2+i)z + 1 + i=0` is the equaiton `rArr z = 1` is the other roots |
|
| 113. |
If `z=(sqrt3/2+i/2)^5+(sqrt3/2-i/2)^5`, thenA. `Re(z) =0`B. `Im ( z) =0`C. `Re(z) gt 0 ,Im(z) gt 0 `D. `Re(z) gt 0,Im (z) lt 0` |
|
Answer» Correct Answer - B Given `z=(sqrt(3)/2+i/2)^5+(sqrt(3)/2-i/2)^5` `[because omega=(-1+isqrt(3))/(2) and omega^2=(-1-isqrt(3))/(2)]` Now , `sqrt(3+i)/2=-i((-1+isqrt(3))/2)=-iomega` and `(sqrt(3)-1)/2=((-1-isqrt(3))/2)=i omega^2` `therefore z=(-iomega)^5+(iomega^2)^5=iw^2+iw` `=i(omega-omega^2)=i(isqrt(3))=-sqrt(3)` `rArr Re(z) lt 0 andlm (z)=0` Alternate Solution We know that `z+bar z =2 Re (z)` If ` z=(sqrt(3)/2+i/2)^5 +(sqrt(3)/2-i/2)^5` then z is purely real ,i,e Im (z)=0 |
|
| 114. |
Find the least positive integer `n`such that `((2i)/(1+i))^n`is a positive integer. |
|
Answer» Correct Answer - n =8 `((2i)/(1+i))+((2i(1-i))/((1+i)(1-i)))^(n)= ((2(i-i^(2)))/(2)))^(n)` `= (i+1)^(n)` `=(2i)^(n//2)` Hence , n= 8 is the least positive interger for which the given complex number is a positive integer. |
|
| 115. |
Let ` z_1 and z_2` be complex numbers of such that `z_1!=z_2 and |z_1|=|z_2|. If z_1` has positive real part and `z_2` has negative imginary part, then which of the following statemernts are correct for te vaue of `(z_1+z_2)/(z_1-z_2)` (A) 0 (B) real and positive (C) real and negative (D) purely imaginaryA. zeroB. real and positiveC. real and negativeD. purely imaginary |
|
Answer» Correct Answer - A::D Given `|z_1|=|z_2|` Now `(z_1+z_2)/(z_1-z_2)xx(barz_1-barz_2)/(z_1-z_2)=(z_1barz_1-z_1barz_2+z_2barz_1-z_2barz_2)/(|z_1-z_2|)` `=(|z|^2+(z_2barz_1-z_1barz_2)-|z_2|^2)/(|z_1-z_2|^2)` `=(z_2barz_1-z_1barz_2)/(|z_1-z_2|^2)" "[because |z_1|^2=|z_2|^2]` As we know `z-bar z `=2 i Im (z) `therefore ( z_2barz_1-z_1barz_2=2i Im (z_2 bar z_1)` `therefore (z_1+z_2)/(z_1-z_2)=(2i Im(z_2 bar z_1))/(|z_1-z_2|^2)` Which is purely imaginary or zero. |
|
| 116. |
Express the following complex numbers in `a+i b`form:`((3-2i)(2+3i))/((1+2i)(2-i))`(ii) `(2-sqrt(-25))/(1-sqrt(-16))` |
|
Answer» Correct Answer - (a) `(63)/(25) -(16)/(25)i` (b) `(12)/(17) + (3)/(17)i` (a) `((3-2i)(2+3i))/((1+2i)(2-i)) = ((6+6)+i(-4+9))/((2+2)+i(4-1))` `- (12+5i)/(4+3i)` ` (12 + 5i)/(4+3i)(4-3i)/(4-3i)` `= ((48 = 15)+i(-36+20))/(16-9i^(2))` `= (63)/(25)-(16)/(25i)` (b) `(2-sqrt(25))/(1-sqrt(-16)) = (2-5i)/(1-4i)` `= (2-5i)/(1-4i)xx(1+4i)/(1+4i)` `= ((2+20)+i(8-5))/(1-16i^(2))` `= (22+ 3i)/(17) = (22)/(17) + (3)/(17) i` (c) `((2i)/( 1+i))^(2) = ((2i(1-i))/((1+i)(1-i))^(n)` ` = ((2(i-i)^(2))/(2))^(n)` `= (i+1)^(n)` ` = (2i)^(n//2)` Hence n = 8 is the least positive integer for which the given complex number is a positive integer. |
|
| 117. |
If `z=r e^(itheta)`, then prove that `|e^(i z)|=e^(-r s inthetadot)` |
|
Answer» `z = re^(i0) = r (cos theta + i sin theta)` `rArr iz = ir (cos theta + i sin theta)` `=- rsin theta + ir cos theta` `rArr e^(1z) = e^((-rsin theta+ ir cos theta))` `= e^((-rsin theta)) e^((ri cos theta))` `rArr |e^(iz)|=|e^(-rsin theta)||e^(r icos theta)|` `= e^((- r sin theta))|e^(ialpha)|," ""where"alpha = r cos theta` `= e^(-rsin theta)` |
|
| 118. |
The polynomial `x^6+4x^5+3x 64+2x^3+x+1`is divisible by_______ where `w`is the cube root of units`x+omega`b. `x+omega^2`c. `(x+omega)(x+omega^2)`d. `(x-omega)(x-omega^2)`where `omega`is one of the imaginary cuberoots of unity.A. `x + omega `B. ` x + omega^(2)`C. `( x+ omega) (x + omega^(2))`D. `(x + omega)(x - omgea^(2))` |
|
Answer» Correct Answer - D Let `f(x) = x^(6) + 4x^(5) + 3x^(4) + 2x^(3) + x+ 1`. Hence, `f(omega) = omega^(6) + 4omega^(5) + 3omega^(4) + 2omega^(3) + omega + 1` `= 1+ 4 omega ^(2)+ 3omega^(4) + 3omega + 2 + omega +1` `= 4(omega^(2) + omega + 1)` `= 0` Hence, `f(x)` is divisible by `x-omega` Then `f(x)` is also divisible by `x - omega^(2)` (as complex roots occur in conjugate pairs). `f(-omega) = (-omega)^(6) + 4(-omeaga)^(5) + 3(-omega)^(4) + 2(-omega)^(3) + (-omega) +1` `= omega^(6) - 4omega^(5) + 3omega^(4) - 2omega^(3) - omega + 1` ` = 1 - 4omega^(2) + 3omega- 2 - omega + 1` ` ne 0`. |
|
| 119. |
If `sqrt(5-12 i)+sqrt(5-12 i)=z`, then principal value of `a rgz`can be`pi/4`b. `pi/4`c. `(3pi)/4`d. `-(3pi)/4`A. `-(pi)/(4)`B. `(pi)/(4)`C. `(3pi)/(4)`D. `-(3pi)/(4)` |
|
Answer» Correct Answer - A::B::C::D `sqrt(5-12i) = sqrt((3-2i)^(2)) = pm (3-2i)` `sqrt(-5-12i)= sqrt((2-3i)^(2)) = pm (2-3i)` `rArr z = sqrt(5-12i) + sqrt(5-12i) = sqrt((3-2i)^(2)) = pm (3-2i)` `sqrt(-5-12i)= sqrt((2-3i)^(2)) = pm (2-3i)` `rArr z = sqrt(5-12i) + sqrt(-5-12i)` `=- 1-i, -5 + 5i,5-5i, 1+i` Therefore, principal values of arg are `- 3pi//4, 3pi//4,-pi//4, pi//4`. |
|
| 120. |
If `x` and `y`are complex numbers, then the system of equations `(1+i)x+(1-i)y=1,2i x+2y=1+i`hasA. unique solutionB. no solutionC. infinte number of solutionsD. none of theses |
|
Answer» Correct Answer - C Observing carefully the system of equations , we find `(1+i)/(2i) = (1-i)/(2) =(1)/(1+i)` Hence,there are infinite number of solutions. |
|
| 121. |
प्रश्न 11 से 13 तक कि सम्मिश्र संख्याओं में प्रत्येक का गुणात्मक प्रतिलोम ज्ञात कीजिए । `4-3i` |
|
Answer» Correct Answer - `(4)/(25) +(3)/(25) i` We have `z = 4-3i` Multiplcative inverse of z is `(1)/(z)=(1)/(4-3i)=(4+3i)/((4-3i)xx(4+3i)) = (4+3i)/(16-9i^(2)) = (4+3i)/(25)=(4)/(25) +(3)/(25)i` |
|
| 122. |
The value of `i^(1+3+5++(2n+1))`is, If n is odd. |
|
Answer» Correct Answer - z=`{:{(,"1, if n is odd"),(,"i, if n is even"):}` Let `z = i^([1+3+5+......+(2n+1)])` Now , `1+3+5+…..+(2n+1)` are `n+1` terms of an A.P whose sum is `(n+1)/(2)[1+2n +1]=(n+1)^(2)` Hence, `z=i^((n+1))`. Now put n = 1,2,3,4,5.... `n =1,z=i^(4) = 1,n=2, z = i^(9) = i` `n=3,z=i^(16)=1,n=4` `z=i^(25)=i,n = 5,z =i^(12) = 1,......` Thus, `z{{:(,"1,if n is odd"),(,"i, if n is even"):}` |
|
| 123. |
Suppose `A`is a complex number and `n in N ,`such that `A^n=(A+1)^n=1,`then the least value of `n`is`3`b. `6`c. `9`d. `12`A. 3B. 6C. 9D. 12 |
|
Answer» Correct Answer - B Let `A=x+iy.`Given, `|A|=1rArrx^(2)+y^(2)=1` and `|A+1|=1rArr(x+1)^(2)+y^(2)=1` `rArr x=-(1)/(2)andy=pm(sqrt3)/(2)` `rArr A=omegaoromega^(2)` `rArr (omega)^(n)=(1+omega)^(n)=(-omega^(2))^(2)` Therefore, n must be even and divisible by 3. |
|
| 124. |
If `a m p(z_1z_2)=0a n d|z_1|=|z_2|=1,t h e n``z_1+z_2=0`b. `z_1z_2=1`c. `z_1=z _2`d. none of theseA. `z_(1)+z_(2) = 0`B. `z_(1)z_(2) = 1`C. `z_(1)=barz_(2)`D. none of these |
|
Answer» Correct Answer - B::C `amp(z_(1)z_(2)) = 0` ` rArr amp z_(1) + ampz_(2) = 0` ` therefore amp z_(1) + amp z_(2) = amp barz_(2)` Since `|z_(1)| = |z_(2)|`, we get `|z_(1)| = |barz_(1)|`. So, `z_(1) = barz_(2)`. Also, `barz_(1)z_(2) = barz_(2)z_(2) = |z_(2)|^(2) = 1` because `|z_(2)|=1` |
|
| 125. |
lf `z = ilog(2-sqrt(-3))`, then `cos z =`A. `-1`B. `-1//2`C. 1D. 2 |
|
Answer» Correct Answer - D `z=ilog(2=sqrt3)` `rArr e^(iz)=e^(i"^(2)log(2-sqrt3))=e^(-log(2-sqrt3))` `=e^(log(2-sqrt3)^(-1))=e^(log(2+sqrt3))=(2+sqrt3)` `rArr cosz=(e^(iz)+e^(-iz))/(2)=((2+sqrt3)+(2-sqrt3))/(2)=2` |
|
| 126. |
If `z=(i)^{(i)^(i)} w h e r e i=sqrt(-1),t h e n|z|`is equal toA. 1B. `e^(-pi//2)`C. `e^(-pi)`D. none of these |
|
Answer» Correct Answer - A `i=cos.(pi)/(2)+isin.(pi)/(2)=e^((ipi)/(2))` `rArr i^(i)=(e^((ipi)/(2)))^(i)=e^((pi)/(2))` `rArrz=(i)^("(i)"^(i))=i^(e^(-(pi)/(2)))` `rArr |z|=1` |
|
| 127. |
If `|z_1/z_2|=1` and `arg (z_1z_2)=0` , thenA. `z_(1) = z_(2)`B. `|z_(2)|^(2) = z_(1)z_(2)`C. `z_(1)z_(2)= 1`D. more than 8 |
|
Answer» Correct Answer - B Let `z_(1)=|z_(1)|(costheta_(1)+I sintheta_(1))`. Now, `|(z_(1))/(z_(2))|=1rArr|z_(1)|=|z_(2)|` Also, `arg(z_(1)z_(2))=0orarg(z_(1)+arg(z_(2))=0` `rArr arg(z_(2))=-theta_(1)` `rArr z_(2)=|z_(2)|(cos(-theta_(1))+i sin(-theta_(1)))` `=|z_(1)|(costheta_(1)-isintheta_(1))=barz_(1)` `rArr |z_(2)|=bar((barz_(1)))=z_(1)` `rArr|z_(2)|^(2)=z_(1)z_(2)` |
|
| 128. |
Given `alpha,beta,`respectively, the fifth and the fourth non-real roots of units, thenfind the value of`(1+alpha)(1+beta)(1+alpha^2)(1+beta^2)(1+alpha^4)(1+beta^4)` |
|
Answer» As ` alpha` is the fifth nonreal root of unity, we have `alpha^(4) + alpha^(3) + alpha^(2) + alpha + 1=0` `beta ` is the fouth nonreal root of unity . Therefore, `beta^(3) + beta^(2) + beta + 1=0` Now, `( 1 + alpha )(1 + alpha^(2))(1 + alpha^(4))(1+ beta)(1+ beta^(2))(1 + beta^(3))` `= (1+ alpha + alpha^(2)+ alpha^(3) ) (1+alpha^(4)) (1+ beta + beta^(2) + beta^(3))(1+ beta^(3)) (because 1+ beta + beta^(2) + beta^(3) = 0)` `= 0` |
|
| 129. |
If `a ,b ,c`and `u ,v ,w`are the complex numbers representing the vertices of two triangles suchthat `(c=(1-r)a+r b`and `w=(1-r)u+r v ,`where `r`is a complex number, then the two triangleshave the same area(b) are similarare congruent(d) None of theseA. have the same areaB. are similarC. are congruentD. Non of these |
|
Answer» Correct Answer - B Since a,b,c and u,v ware the vertice of two triangles Also `c=(1-r) a+ rb ` and `w= (1-r ) u+rb ` Applying `R_3rarr R_3-{(1-r)R_1+rR_2)}`s `=|{:(" "a" "u" "1),(" "b" "v" "1),(c-(1-r)a-rb w-(1-r)u-rv 1 - (1-r)-r):}|` `=|(a,u,1),(b,v,1),(0,0,0)|=0 " "[from Eq. (i) ]` |
|
| 130. |
Find the value of `x^4+9x^3+35 x^2-x+4`for `x=-5+2sqrt(-4).` |
|
Answer» Correct Answer - `-160` we have, `x = -5 + 4i` `or (x+5)^(2) =- 16` `or x^(2) + 10 x + 41 = 0` Now, `x^(4) + 9x^(3) +35x^(2) - x +4` `= x^(2)(x^(2)+10x+41)-x(x^(2)+10x + 41) + 4(x^(2) + 10 + 41) - 160` `=0x^(2) = 0x + 4 xx 0 - 160=-160" "["Using(1)"]` |
|
| 131. |
Find the principal argument of the complex number `((1+i)^5(1+sqrt(3i))^2)/(-1i(-sqrt(3)+i))` |
|
Answer» `z = ((1+i)^(5)(1+sqrt(3i))^(2))/(-2i(-sqrt(3)+i))` `(sqrt(2)^(5)((1)/(sqrt(2))+(i)/(sqrt(2))).2^(2)((1)/(2)+(sqrt(3))/(2)i))/((2i)2(sqrt(3)/(3)-(i)/(2)))` `5 arg((1)/(sqrt(2))+(1)/(sqrt(2)))+2 arg((1)/(2)+(sqrt(3))/(2)i)` `therefore arg(Z) - arg(i) - arg((sqrt(3))/(2)-(i)/(2))" "(because arg(sqrt(2)^(5)),arg(2^(2)), arg (4)=0)` `=(5pi)/(4) + (2pi)/(3)-(pi)/(2)+(pi)/(6) = (19pi)/(12)= 2pi - (5pi)/(12)` Therefore, Z lies in the fourth quadrant. Hence, pricncipal argument is `- (5pi)/(12)` |
|
| 132. |
If for complex numbers `z_1 and z_2, arg(z_1) -arg(z_2)=0` then `|z_1-z_2|` is equal toA. `|z_(1)|+|z_(2)|`B. `|z_(1)| - |z_(2)|`C. `||z_(1)|-|z_(2)||`D. 0 |
|
Answer» Correct Answer - C We have `|z_(1)-z_(2)|^(2)=|z_(1)|^(2)+|z_(2)|^(2)-2|z_(1)||z_(2)|cos(theta_(1)-theta_(2))` where `theta_(1)=arg(z_(1))and theta_(2)=arg(z_(2))`. Given, `arg(z_(1))-arg(z_(2))=0` `rArr|z_(1)-z_(2)|^(2)=|z_(1)|^(2)+|z_(2)|^(2)-2|z_(1)||z_(2)|` `=(|z_(1)|-|z_(2)|)^(2)` `rArr|z_(1)-z_(2)|=||z_(1)|-|z_(2)||` |
|
| 133. |
For any integer k, let `alpha _h= cos. (k pi)/(7)+ I sin. (k pi)/(7) " where if "=sqrt(-1).` The value of the expression `(Sigma_(k=1)^(12) |alpha_(k+1)-alpha_k|)/(Sigma_(k=1)^(3) |alpha_(4k+1)-alpha_(4k-2)|)`is |
|
Answer» Correct Answer - 4 Given `alpha_(k) = cos ((kpi)/(7)) + i sin ((k pi)/(7))` `= cos ((2k pi)/(14)) + i sin ((2 kpi)/(14))` `therefore alpha_(k)` are vertices of regular polygon having 14 sides . Let the side length of regular polygon be `alpha`. `therefore |alpha_(k + 1) - alpha_(k)| ` = length of a side of the regular polygon `= alpha " " ... (i)` and `|alpha_(4k-1) - alpha_(4k-2)|` = length of a side of the regular polygon ` = alpha " " .... (ii)` `therefore ( sum_(h=1)^(12) |alpha_(k+1) - alpha_(k)|)/(sum_(h=1)^(3) |alpha_(4k-1) - alpha_(4k-2)|) = (12(a))/(3(a)) = 4` |
|
| 134. |
If `z=costheta+isintheta`is a root of the equation `a_0z^n+a_2z^(n-2)++a_(n-1)z^+a_n=0,`then prove that`a_0+a_1costheta+a_2^cos2theta++a_ncosntheta=0``a_1"sin"theta+a_2^sin2theta++a_nsinntheta=0` |
|
Answer» Dividing the given equation by `z^(n),` we get `a_(0)+a_(1)z^(-1)+a_(2)z^(-2)+…+a_(n-1)z^(1-n)+a_(n)z^(-n)=0` Now, `z=costheta+isintheta=e^(itheta)` satisfies the above equation. Hence, `a_(0)+a_(1)e^(-itheta)+a_(2)e^(-2itheta)+…+a_(n-1)e^(-i(n-1)theta)+a_(n)e^(-intheta)=0` `implies(a_(0)+a_(1)costheta+a_(2)cos2theta+...+a_(n)cosntheta)` `-i(a_(1)sintheta+a_(2)sin2theta+...+a_(n)sinntheta)=0` `impliesa_(0)+a_(1)costheta+a_(2)cos2theta+...+a_(n)cosntheta=0` and `a_(1)sintheta+theta_(2)sin2theta+...+a_(n)sinntheta=0` |
|
| 135. |
Match the conditions /expression n Column I with statement in Column II `(z ne 0` is a complex number ) |
|
Answer» Correct Answer - `A to q ; B to p` Let z=a+ ib Given `Re(z)=0 rArr a=0` Then `z=ib rArr z^2 =- b^2 or lm(z^(2))=0` Therefore sds`A rarr q` Also, given ,arg (z)`= pi/2` Let `z=r(cos"" pi/4 - I sin""pi/4)` Then `z^2r^2( cos^2 "" pi/4 - sin^2 L"" pi/4)+2 ir^3 ""pi/4 sin "" pi/4` `=ir^2 sin pi//2 = ir^2` Therefore , `Re (z^2)=0 rArr B rArr p.` `rArr a= b= 2 - sqrt(3) " " [ because a, b larr (0,1)] ` |
|
| 136. |
If `z=z_0+A( z -( z )_0), w h e r eA`is a constant, then prove that locus of `z`is a straight line. |
|
Answer» `bar(z)=bar(z)_(0)+A(z-z_(0))` `impliesAz-bar(z)-Az_(0)+bar(z)_(0)=0" "(1)` `impliesbar(A)bar(z)-z-bar(A)bar(z)_(0)+z_(0)=0" "(2)` Adding (1) and (2), we get `(A-1)z+(bar(A)-1)bar(z)-(Az_(0)+bar(A)bar(z)_(0))+z_(0)+bar(z)_(0)=0` This is of the form `bar(a)z+abar(z)+b=0," where "a=bar(A)-1" and "b=-1(Az_(0)+bar(A)bar(z)_(0))+z_(0)+bar(z)_(0)inR.` Hence, locus of z is a straight line. |
|
| 137. |
If `z_1, z_2, z_3`are three complex numbers such that `5z_1-13 z_2+8z_3=0,`then prove that`|z_1( z )_1 1z_2( z )_2 1z_3( z )_3 1|=0` |
|
Answer» `5z_(1)-13z_(2)+8z_(3)=0` or `(5z_(1)+8z_(2))/(5+8)=z_(3)` This mean that `z_(3)` divides segment joining `z_(1)" and "z_(2)` in the ratio 8 : 5. Hence, `z_(1),z_(2),z_(3)` are collinear. `implies|{:(z_(1),bar(z)_(1),1),(z_(2),bar(z)_(2),1),(z_(3),bar(z)_(3),1):}|=0" "("condition of collinear points")` |
|
| 138. |
If `z = (3)/( 2 + cos theta + I sin theta)`, then prove that z lies on the circle. |
|
Answer» Given `z = (3)/(2 + cos theta + i sintheta)` ` rArr cos theta + i sin theta = (3)/(z) - 2 = (3-2z)/(z)` `rArr 1= (|3-2z|)/(|z|)" "["taking modulus"]` `rArr (|z-(3)/(2)|)/(|z|) = (1)/(2)` Therefore, locus of z is circle. |
|
| 139. |
Prove that `|Z-Z_1|^2+|Z-Z_2|^2=a`will represent a real circle [with center `(|Z_1+Z_2|^//2+)`] on the Argand plane if `2ageq|Z_1-Z_1|^2` |
|
Answer» `|z-z_(1)|^(2)+|z-z_(2)|^(2)=a` `implies(z-z_(1))(bar(z)-bar(z)_(1))+(z-z_(2))(bar(z)-bar(z)_(2))=a` `implies2zbar(z)-z(bar(z)_(1)+bar(z)_(2))-bar(z)(z_(1)+z_(2))+z_(1)bar(z)_(1)+z_(2)bar(z)_(2)=a` `implieszbar(z)=z((bar(z_(1))+bar(z_(2)))/(2))-bar(z)((z_(1)+z_(2))/(2))+(z_(1)bar(z)_(1)+z_(2)bar(z)_(2)-a)/(2)=0" "(1)` Above equation is of the form `zbar(z)+alphabar(z)+bar(alpha)z+beta=0`, which represents the real circle if `alphabar(alpha)-betage0` `implies((z_(1)+z_(2))(bar(z_(1))+bar(z_(2))))/(4)ge(z_(1)bar(z_(1))+z_(2)bar(z_(2))-a)/(2)` `implies2agez_(1)bar(z_(1))+z_(2)bar(z_(2))-z_(1)bar(z_(2))-z_(2)bar(z_(1))` `implies2age|z_(1)-z_(2)|^(2)` |
|
| 140. |
For three non-colliner complex numbers `Z,Z_(1)` and `Z_(2)` prove that `|Z-(Z_(1)+z_(2))/(2)|^(2) + |(Z_(1) -Z_(2))/(2)|= (1)/(2) | Z - Z_(1)|^(2) +(1)/(2)|Z- Z_(2)|^(2)` |
|
Answer» Consider the formed by A(Z), `B(Z_(1))` and `C(Z_(2))`. Let midpoint of BC be D having complex number `(Z_(1)+Z_(2))/(2)` By Apollononius therorem, we have `AB^(2) + AC^(2) + 2(AD^(2) + BD^(2))` `therefore |Z-(Z_(1) +Z_(2))/(2)|^(2) + |(Z_(1) +Z_(2))/(2)|` `AD^(2) + BD^(2)` `=(1)/(2)AB^(2) + (1)/(2)Ac^(2)` `= (1)/(2)|Z- Z_(1)|^(2)+(1)/(2)|Z-Z_(2)|^(2)` |
|
| 141. |
If `omega = z//[z-(1//3)i] and |omega| = 1`, then find the locus of z. |
|
Answer» Correct Answer - Perpendicular bisector of the line joininig 0+ 0i and `0 + (1//3)i` `omega = (z)/(z-(1)/(2)i)` or ` |omega| = |(z)/(z-(1)/(3)i)|` or ` |z| = |omega||z-(1)/(3)i|` or ` |z| = |z-(1)/(3)i|" "(because |omega|=1)` Hence, locous of z is perpendicular bisector of the line joining `0 + 0i` and `0+(1//3)i`. Hence z lies on a straight line . |
|
| 142. |
If `|z|=1` and let `omega=((1-z)^2)/(1-z^2)`, then prove that the locus of `omega`is equivalent to `|z-2|=|z+2` |
|
Answer» Given `omega=((1-z)^(2))/(1-z^(2))=(1-z)/(1+z)` `implies" "omega=(zbar(z)-z)/(zbar(z)+z)=(bar(z)-1)/(bar(z)+1)=-((bar(1-z))/(1+z))=-bar(omega)` `:." "omega+bar(omega)=0` `implies omega" is purely imaginary. Hence, "omega" lies on the y-axis."` Also `|z-2|=|z+2|impliesz` lies on perpendicular bisector of line segment joning 2 and -2, which is the imaginary axis. |
|
| 143. |
Identify the locus of `z`if ` z = a +(r^2)/(z-a),> 0.` |
|
Answer» `bar(z)=bar(a)+(r^(2))/(z-a)` or `bar(z)-bar(a)=(r^(2))/(z-a)` or `(z-a)(bar(z)-bar(a))=r^(2)` or `|z-a|^(2)=r^(2)` or `|z-a|=r` Hence, locus of z is circle having center a and radius r. |
|
| 144. |
If `|z| |
|
Answer» Correct Answer - 9 `|iz + 3 - 4i| le |iz| + | 3 - 4i| ` ` " " = |z| + 5 le 4 + 5 = 9` Hence, `|z|_(max) = 9`. |
|
| 145. |
`Z in C`satisfies the condition `|Z|>-3.`Then find the least value of `|Z+1/Z|` |
|
Answer» Correct Answer - `8//3` `|Z + (1)/(Z)| = |Z - (-(1)/(Z))| ge |Z| ~|- (1)/(Z)| ge 3~ (1)/(3) = (8)/(3)` |
|
| 146. |
Let `z in C` be such that `|z| lt 1 `.If `omega =(5+3z)/(5(1-z)` thenA. `4 Im ( omega) gt 5`B. `5 Re (omega) gt 1`C. `5 Im(omega) lt 1 `D. |
|
Answer» Correct Answer - B Given complex number `omega=(5+3z)/(5(1-z))` `rArr 5 omega -5 omega z=5+3z` `rarr (3+5 omega)z=5 omega-5` `rArr |3+5 omega||z|=|5 omega- 5|` [applying modulus both sides and `|z_1z_2|=|z_2||z_2|]` `therefore | z| lt 1 ` `therefore |3+5 omega| gt |5 omega - 5|" "["from Eq.(i)"]` `rArr " "|omega+3/5| gt | omega-1|` Let `omega` =x+iy, then `(x+3/5)^2+y^2 gt (x-1)^2+y^2` `rArr x^2+9/25+6/5 x lt x^2 +1 -2x` `rArr 16x/5 gt 16/25 rArr x gt 1/5 rArr lt 1` `rArr 5 Re(omega ) gt 1` |
|
| 147. |
If `I m((z-1)/(e^(thetai))+(e^(thetai))/(z-1))=0`, then find the locus of `zdot` |
|
Answer» Correct Answer - Circle having centre at `1 + i0` and radius 1 Let `u = (z -1)/(e^(thetai)) rArr (e^(thetai))/(z-1) = (1)/(u)` Now `(u +(1)/(u)) - (baru +(1)/(baru))= 0` ` rArr (u- baru)(1(1)/(ubaru))=0` If u is not purely real, then `ubaru = 1` `rArr |(z-1)/(e^(thetai))| = 1 rArr |z-1|=1` Hence, z lies on circle haivng centre at 1+ i0 and radius 1. |
|
| 148. |
If `|z|=1,`then the point representing the complex number `-1+3z`will lie ona. a circleb. a parabolac.``a straight lined. a hyperbolaA. a circleB. a straight lineC. a parabolaD. a hyperbola |
|
Answer» Correct Answer - A `|z|=1," let "alpha=-1+3z` `rArr alpha +1=3z` `rArr |alpha+1|=3|z|=3` Hence, `alpha` lies on a circle centered at `-1` and radius equal to 3. |
|
| 149. |
If `z`is any complex number such that `|3z-2|+|3z+2|=4`, then identify the locus of `zdot` |
|
Answer» `|3z-2|+|3z+2|=4` `implies|z-(2)/(3)|+|z+(2)/(3)|=(4)/(3)" "(1)` If P(z) be any point `A-=(-2//3,0),` then (1) represents `PA+PB=4` Clearly, `AB=4//3impliesPA+PB=AB` `impliesP" lies on the line segment AB."` |
|
| 150. |
All the points in the set `S={(alpha+i)/(alpha-i):alpha in R } (i= sqrt(-1))lie` on aA. cicle whose radius is `sqrt(2)`B. straight line whose slope is -1C. circle whose radius is 1.D. straight line whose slope is 1. |
|
Answer» Correct Answer - C Let x+I y = `(alpha +i)/(alpha- i)` `rArr x+iy=((alpha+i)^2)/(alpha ^2+1) and y=(2 alpha )/(alpha^2n+1)` Now `x^2+y^2=((alpha^2-1)/(alpha ^2+1))^2+((2alpha )/(alpha)^2+I)^2` `=(alpha^4+1-2alpha^2+4 alpha^2)/(alpha^2 +1)^2=(alpha ^2+1)^2/(alpha^2+1)^2=1` `rArr x^2+y^2=1` Which is an equations of circle with centre (0,0) and radius 1 unit . So,`S={(alpha+i)/(alpha-1),alpha ne R}` lies on a circle with radius 1. |
|