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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.
| 201. |
Let ` omega= e^((ipi)/3) and a, b, c, x, y, z` be non-zero complex numbers such that `a+b+c = x, a + bomega + comega^2 = y, a + bomega^2 + comega = z`.Then, the value of `(|x|^2+|y|^2|+|y|^2)/(|a|^2+|b|^2+|c|^2)` |
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Answer» Correct Answer - 3 Priniting error `=e^i(2pi)/(3)` Then, `(|x^2|+|y|^2+|z|^2)/(|a|^2+|b|^2+|c|^2)=3` |
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| 202. |
If `omega ne 1` is a cube root of unity and `a+b=21`, `a^(3)+b^(3)=105`, then the value of `(aomega^(2)+bomega)(aomega+bomega^(2))` is be equal toA. `3`B. `5`C. `7`D. `35` |
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Answer» Correct Answer - B `(b)` `(aomega^(2)+bomega)(aomega+bomega^(2))=a^(2)-ab+b^(2)` `= ((a+b)(a^(2)-ab+b^(2)))/(a+b)` `=(a^(3)+b^(3))/(a+b)=(105)/(21)=5` |
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| 203. |
`omega` is an imaginary root of unity. Prove that (i) `( a + bomega + comega^(2))^(3) + (a+bomega^(2) + comega)^(3) = (2a-b-c)(2b -a -c)(2c -a-b)` (ii) If `a+b+c = 0` then prove that `(a + bomega + comega^(2))^(3)+(a+bomega^(2) + comega)^(3) = 27abc`. |
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Answer» (i) Let `a + bomega + comega^(2) = x` and `a + bomega^(2) + comega = y` , Then `(a+bomega + comega^(2))^(3) + (a+ bomega^(2) + comega)^(3) = x^(3) + y^(3)` `=(x+y)(x+ omegay)(x+ omega^(2)y)` Now, `x + y = (a+ bomeag + comega^(2) )+ (a+ bomega^(2) + comega)` `=2a + b(omega+omega^(2))` `=2-b-c` `x + omegay= (a+ bomega+ comega^(2)) + omega(a+bomega^(2) + comega)` `= (1+omega) a+ 1(1+ omega)b + 2omega^(2)c` `omega^(2) (2c-a-b)` Similarly, `x + omega^(2)y = omega(2b-a-c)` `rArr (x+y)(x+ omegay)(x+ omega^(2)y)` `=omega^(3)(2a-b-c)(2c-a-b)(2b -a-c)` `=(2a-b-c)(2c-a-b)(2b-a-c)` (ii) `a+ b+c=0` . `rArr b+c = -a,c+a= -b` and `a+ b= -c` Putting these values of the R.H.S of result (i), we get `(a+ bomega + comega^(2))^(3) + (a+bomega^(2)+ comega)^(3) = 27abc` |
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| 204. |
If `z=x+i ya n dw=(1-i z)/(z-i)`, show that `|w|=1 z`is purely real. |
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Answer» We have `|w| = 1` `rArr |(1-iz)/(z-i)| = 1 or (|1-iz|)/(|z-i|)=1` `or | 1 - iz | = |z-i|` or `|1-i(x+iy)|=|x+iy -i|,` where z = x + iy `or |1+y-ix|=|x +i(y-1)|` `or sqrt((1+y)^(2) +(-x)^(2))=sqrt(x^(2) + (y-1)^(2))` `or (1+y)^(2) + x^(2) =x^(2) +(y-1)^(2)` ` or y = 0` `rArr z = x +i0 = x`, which is purely real |
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| 205. |
If the imaginary part of `(2z+1)//(i z+1)`is -2, then find the locus of the point representing in the complex plane. |
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Answer» Let x + iy `rArr (2z +1)/(iz +1)= (2(x +iy)+1)/(i(x+iy)+1)` `=((2x +1)+i2y)/(1-y +ix)` `= ((2x+1)+i2y)/((1-y)+ix)((1-y)-ix)/((1-y)-ix)` `=((2x+1)(1-y)+2xy +i[-x(2x+1)+2y (1-y)])/((1-y)^(2)+x^(2))` Since imaginary part of `(2z +1)//(iz +1)` is `-2`, we have `(-x(2x +1)+2y(1-y))/((1-y)^(2) +x^(2))=-2` or ` -2x^(2) -x+2y _2y^(2) = - 2[1+y^(2) -2y+x^(2)]` or `x +2y -2=0`, which is a straight line |
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| 206. |
Let `|z|=2a n dw-(z+1)/(z-1),w h e r ez ,w , in C`(where `C`is the set of complex numbers). Then product of least and greatestvalue of modulus of `w`is__________. |
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Answer» Correct Answer - 1 Let `z = a+ib` Given `|z| = 2` `rArr a^(2) + b^(2) = 4 rArr a,b in [-2,2]` Now `w ((a+1)+ib)/((a-1)+ib),` `rArr |w| = sqrt(((a+i)^(2) + b^(2))/((a-1)^(2) + b^(2)))=sqrt((a^(2) + b^(2)+2a+1)/(a^(2) +b^(2) -2a+1)) = sqrt((5+2a)/(5-2a))` `|w|_("min") = sqrt((5+4)/(1)) = 3` (when a =2) `|w|_("min") = sqrt((5-4)/(9)) = (1)/(3)` (when a = -2) Hence, required product is 1. |
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| 207. |
If `|z_1-z_0|=z_2-z_1=pi//2`, then find `z_0dot` |
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Answer» Correct Answer - `(1)/(2){(i +1)z_(1) +(1-i)z_(2)}` Here, `|(z_(2)-z_(0))/(z_(0)-z_(1))| = 1 and amp ((z_(2)-z_(0))/(z_(0)-z_(1))) = (pi)/(2)` `therefore (z_(2)-z_(0))/(z_(0)-z_(1)) = 1{cos.(pi)/(2) + i sin .(pi)/(2)} =i` `rArr z_(2) -z_(0) = iz_(0) = iz_(0)` `rArr z_(2)+iz_(1) = (i+1)z_(0)` `or z_(0) = (z_(2) + iz_(1))/(1+i)` `=((z_(2) + iz_(1))(1-i))/(1^(2)+ 1^(2))` `=(1)/(2){(i+ 1)z_(1)+(1-i)z_(2)}` |
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| 208. |
Let `z`be a complex number satisfying the equation `(z^3+3)^2=-16`, then find the value of `|z|dot` |
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Answer» Correct Answer - `5^(1//3)` `(z^(3)+ 3)^(2) = 16i^(2)` `z^(3) + 3 = 4i or - 4i` `z^(3) = -3 + 4i or - 3 - 4i` `z^(3) = -3 + 4i or - 3-4i` `|z^(3)| = |-3+4i| = 5` `|z|^(3) = 5` `rArr |z| = 5^(1//3)` |
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| 209. |
If `theta`is real and `z_1, z_2`are connected by `z1 2+z2 2+2z_1z_2costheta=0,`then prove that the triangle formed by vertices `O ,z_1a n dz_2`is isosceles. |
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Answer» `z_(1)^(2) + z_(2)^(2) + 2z_(1)z_(2) cos theta = 0` `or ((z_(1))/(z_(2)))^(2) + 2(z_(1)/(z_(2))) cos theta + 1 =0` `or ((z_(1))/(z_(2)) + cos theta )^(2) = - (1-cos^(2) theta) = - sin^(2) theta` ` or (z_(1))/(z_(2)) = - cos theta pm i sin theta` `or |(z_(1))/(z_(2))| = sqrt((-costheta)^(2) + sin^(2) theta)=1` `or |z_(1)|=|z_(2)|` `or |z_(1) -0|=|z_(2)-0|` Thus, triangle with vertices O,`z_(1),z_(2)` is iscosceles. |
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| 210. |
Given `z`is a complex number with modulus 1. Then the equation `[(1+i a)//(1-i a)]^4=z`hasall roots real and distincttwo real and two imaginarythree roots two imaginaryone root real and three imaginaryA. all roots real and distinctB. two real and tw imaginaryC. three roots real and one imaginaryD. one root real and three imaginary |
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Answer» Correct Answer - A `((1+ia)/(1-ia))^(4) =z` or ` |(1+ia)/(1-ia)|^(4) = |z|` or `|(a-i)/(a+i)|^(4) = 1` ` or |a -i|= |a + i|` Therefore, a lies on the perpendicular bisector of iand -i. which is the real axis. Hence,all the roots are real.Obvioulsy roots are distinct. |
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| 211. |
If `pa n dq`are distinct prime numbers, then the number of distinct imaginarynumbers which are pth as well as qth roots of unity are.`min(p ,q)`b. `min(p ,q)`c. `1""`d. `z e ro`A. min(p,q)B. max(p,q)C. 1D. zero |
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Answer» Correct Answer - D We have, `e^((2piri)/(rho)) = e^((2pimi)/(q))` `r = 0,1,.....,q-1` `m = 0,1,......,q-1` This in possible iff r=m = 0 but for r=m=0 we get 1 which is not an imaginary number. |
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| 212. |
Let `x_(1),x_(2)` are the roots of the quadratic equation `x^(2) + ax + b=0`, where a,b, are complex numbers and `y_(1), y_(2)` are the roots of the quadratic equation `y^(2) + |a|yy+ |b| = 0`. If `|x_(1)| = |x_(2)|=1`, then prove that `|y_(1)| = |y_(2)| =1` |
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Answer» `x^(2) + ax +b = 0` has roots `x_(1)` and `x_(2)` . Then `x_(1) + x_(2) = -a` `x_(1) + x_(2) = -a` `and x_(1)x_(2) = b` From (2), `|x_(1)| |x_(2)| = |b|` `rArr |b|=1` Also, `|-a| = |x_(1) +x_(2)|` `rArr |a| le |x_(1)|+|x_(2)|` `or |a| le 2` Now `y^(2) + |a|y + |b| = 0` has roots `y_(1)` and `y_(2)` . Then `y_(1),y_(2) =(-|a| pm sqrt(|a|^(2) -4|b|))/(2)` `= (-|a| pm (sqrt(4-|a|^(2)))i)/(2)` `rArr |y_(1)|,|y_(1)| = (sqrt(|a|^(2)+4-|a|^(2)))/(2) = 1` Hence, `|y_(1)|=|y_(2)|= 1` |
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| 213. |
If `y_(1)=max||z-omega|-|z-omega^(2)||`, where `|z|=2` and `y_(2)=max||z-omega|-|z-omega^(2)||`, where `|z|=(1)/(2)` and `omega` and `omega^(2)` are complex cube roots of unity, thenA. `y_(1)=sqrt(3)`, `y_(2)=sqrt(3)`B. `y_(1) lt sqrt(3)`, `y_(2)=sqrt(3)`C. `y_(1)=sqrt(3)`, `y_(2) lt sqrt(3)`D. `y_(1) gt 3`, `y_(2) lt sqrt(3)` |
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Answer» Correct Answer - C `(c )` We have `||z_(1)|-|z_(2)|| le |z_(1)-z_(2)|` and equality holds only when `argz_(1)=argz_(2)` `implies||z-w|-|z-w^(2)|| le |w^(2)-w| le sqrt(3)` and equality canhold only when `|z|=2` and not when `|z|=(1)/(2)` |
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| 214. |
Solve `:+z^2+|z|=0`. |
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Answer» `z^(2) + |z|=0` `rArr x^(2) -y^(2) + i(2xy) + sqrt(x^(2) + y^(2)) = 0` `rArr x^(2) - y^(2) + sqrt(x^(2) + y^(2)) = 0" "(1)` and ` 2xy = 0" "(2)` From (2), let x= 0 . From (1), `-y^(2) + sqrt(y^(2)) = 0` `rArr -|y|^(2) + |y| = 0` `rArr |y| = 0 or 1` `rArr y = 0 or y = pm 1` From (2), if y = 0 , then from (1), `x^(2)+ sqrt(x^(2)) = 0` `rArr |x|^(2) +|x|=0` `rArr x = 0` Hence, complex numbers are `0 + i0, 0+i,0-i` |
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| 215. |
If `z_1a n dz_2`are complex numbers and `u=sqrt(z_1z_2)`, then prove that`|z_1|+|z_2|=|(z_1+z_2)/2+u|+|(z_1+z_2)/2-u|` |
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Answer» `|(z_(1)+z_(2))/(2)+u|+|(z_(1) +z_(2))/(2) -u|` `=|(z_(1)+z_(2))/(2) +sqrt(z_(1)z_(2))|+|(z_(1)+z_(2))/(2) =sqrt(z_(1)z_(2))|` `=|((sqrt(z_(1))+sqrt(z_(2))^(2)))/(2)|+|((sqrt(z_(1))-sqrtz_(2)))/(z)|` `|((p+q)^(2))/(2)|+|((p-q)^(2))/(2)|" "("Where" p = sqrt(z_(1)) and q = sqrt(z_(2)))` `= (1)/(2) [|p+q|^(2) +|p-q|^(2)]` `=(1)/(2) [2|p|^(2)+2|q|^(2)]` `|p|^(2) +|q|^(2)` `=|p^(2)| +|q^(2)|` `=|z_(1)| +|z_(2)|` |
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| 216. |
Let I, `omega` and `omega^(2)` be the cube roots of unity. The least possible degree of a polynomial, with real coefficients having `2omega^(2), 3 + 4 omega, 3 + 4 omega^(2) ` and `5- omega - omega^(2)` as roots is -A. `4`B. `5`C. `6`D. `7` |
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Answer» Correct Answer - B `(b)` Roots are `alpha=2omega^(2)`, `beta=3+4omega`, `gamma=3+4omega^(2)`, `delta=5-omega-omega^(2)` If `alpha=2omega^(2)` is a root, then `2omega` has to be a root too. |
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| 217. |
Real part of `(e^e)^(iotatheta)` is |
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Answer» `e^(e^(itheta)) = e^(cos theta + sin theta )= e^(cos theta)[e^(isin theta)]` `=e^(cos theta)[cos(sin theta) + i sin (sin theta)]` Therefore, the real part is `e^(cos theta) [cos (sin theta)]`. |
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| 218. |
if`omegaa n domega^2`are the nonreal cube roots of unity and `[1//(a+omega)]+[1//(b+omega)]+[1//(c+omega)]=2omega^2`and `[1//(a+omega)^2]+[1//(b+omega)^2]+[1//(c+omega)^2]=2omega^`, then find the value of `[1//(a+1)]+[1//(b+1)]+[1//(c+1)]dot` |
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Answer» The given relation can be rewritten as `(1)/(a+omega) + (1)/(b+omega) + (1)/(c+omega) = (2)/(omega)` and `(1)/(a+omega^(2)) + (1)/(b+omega^(2)) + (1)/(c+ omega^(2)) = (2)/(omega^(2))` `rArr omega and omega^(2)` are roots of `(1)/(a+x) +(1)/(b+omega^(2)) +(1)/(c+x) = (2)/(x)` Tow roots of the (1) are `omega` and `omega^(2)`.Let the third root be `alpha`.Then, `alpha +omega+ omega^(2) =0 or aklpha = - omega -omega^(2) = 1` Therefore `alpha = 1` will satisfy Eq.(1) Hence, `(1)/(a+1)+(1)/(b+1)+(1)/(c+1) = 2` |
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| 219. |
Let z and w be two complex numbers such that `|Z| |
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Answer» Correct Answer - C Given `|z+I w|=|z-ibarw|=2` `rArr |z-(-iw)|=|z-(bar iw)|=2` `rArr |z-(-iw)|=|z-(-ibarW)|` `therefore` z lies on the perpendicular bisector of the line joininig -iw and - `Ibarw ."since " - bar w` and y=0 Now `|z| le 1 rArr x^2 +0^2 le 1 rArr -1 le x le 1` `therefore` z may take values given in opton ( C) Alteranate Solution |z+ iw | le |z| + |iw | = |z| + |w| `le 1+1 =2` `therefore |z+i w | le 2 ` `rArr` |z+iw|= 2 holds when argz- arg i w =0 `rArr arg z-arg i w =0 ` ` rArr arg z/(iw)=0` ` z/(iw) ` is purely real. `rArr z/w ` is purely imaginary Similarly when `|z-i bar w|=2 "then " z/w` is purely imaginary Now , given relation`|z+iw|=|z-barw|=2` Put w=i ,we get `|z+ i^2|=|z+i^2|=2` `rArr |z+1|=2 rArr z=1 [therefore |z| le 1]` `therefore ` z=1 or -1 is the correct option. |
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| 220. |
If `z and w` are two complex number such that `|zw|=1 and arg (z) – arg (w) = pi/2,` then show that `overline zw = -i.`A. `bar z w =i`B. `zbarw=(1-i)/sqrt(2)`C. `bar z w=i`D. `z bar w = (-1 +i)/sqrt(2)` |
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Answer» Correct Answer - A It is given that there, are two complex numbers z and w sunc that `|z w|=1 and arg(w) = pi//2` `therefore |z||w|=1 [ because |z_1 z_2|= | z_1||z_2|]` and `arg(z) = pi/2 + arg (w)` Let `|z| =r , then | w|= 1/r` and let arg (w) `= theta " then arg(z) pi/2 + theta ` So we can assume `z= re^(i pi//2+ theta)` `[ because ` if z= x+ iy is a complex number then it can be written as z `=re^(i theta) where r= |z| and theta = arg (z) ]` and `w =1/re^(it theta)` Now ` barz.w = re^(-i(pi//2theta)). 1/re^(i theta)` `=e^(i(-pi//2- theta+ theta))=e^(-i(pi//2))-i " " [ because e^(-i theta) = cos theta - i sin theta]` and `z bar w = re^(i(pi//2 + theta)).1/r e^(- i theta)` `= e^(i(pi//2+ theta - theta))=e^(i(pi//2))=i` |
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| 221. |
If `z_1` and `z_2` are two non- zero comlex numbers such that `|z_1+z_2|=|z_1|+|z_2|` then arg `(z_1)- arg (z_2)` is equal toA. `-pi`B. `-pi/2`C. 0D. `pi/2` |
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Answer» Correct Answer - C Given `|z_1+z_2|=|z_1|+|z_2|` On squaring both sides ,we get `|z_1|^2 + |Z_2|^2+2|z_1||z_2|cos (arg z_1 - arg z_2)=|z_1|^2|z_2|^2+2|z_1||z_2|` ` rArr 2|z_1||z_2|cos (arg z_1-arg z_2)=2|z_1|z_2|` `rArr cos(arg z_1 - arg z_2)=1` `rArr arg(z_1)aargz_2)=0` |
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| 222. |
Prove that `z=i^i ,w h e r ei=sqrt(-1)`, is purely real. |
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Answer» `z = i^(i) = (cos.(pi)/(2) + i sin .(pi)/(2))^(i) = (e^(ipi//2))^(i) = e^(-(pi)/(2))` `rArr Re(z) = e^(-(pi)/(2))` |
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| 223. |
If `x=a+b i`is a complex number such that `x^2=3+4ia d nx^3=2+1i ,w h e r ei=sqrt(-1),t h e n(a+b)`equal to ______. |
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Answer» Correct Answer - 3 `x = (x^(3))/(x^(2)) = (2+11i)/(3+4i) xx (3-4i)/(3-4i) = (50+ 25i)/(2+i)` |
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| 224. |
If z is nonreal root of `[-1]^[1/7]` then,find the value of `z^86`+`z^175`+`z^289` |
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Answer» Correct Answer - `-1` Let `z = ""^(7)sqrt(-1)`. Then `Z^(7) = -1` `Therefore z^(86) + z^(175) + z^(289)` `= (z^(7))^(12) xx z^(2) + (z^(7))^(25) + (z^(7))^(41) xx z^(2)` `= (-1)^(12) xx z^(2) + (-1)^(25) + (-1)^(41) xx z^(2)` `= z ^(2) - 1-z^(2) = - 1` |
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| 225. |
Let `z and w` are two non zero complex number such that `|z|=|w|, and Arg(z)+Arg(w)=pi` then (a) `z=w` (b) `z=overlinew` (c) `overlinez=overlinew` (d) `overlinez=-overlinew`A. wB. `-w`C. `bar w`D. `- bar w` |
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Answer» Correct Answer - D Since |z|=|w| and arg (z)=`pi`-arg (w) Let `w=re^(I theta) , then bar w =re^(-ite)` `therefore z= re^i(pi - theta)= re ^(I pi).e^(-I theta)=-bar w` |
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| 226. |
Find the number of roots of the equation `z^(15) = 1` satisfying `|arg z| lt pi//2`. |
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Answer» Correct Answer - 7 We have `z^(15) = 1` `z=1^(1)/(15) = e^(i(2rpi)/(15)), r = 0, 1,2,.....,14` or `z =e^(pmi(2rpi)/(15)) , r=0,1,2,....7` The complex numbers having their arguments between `-(pi)/(2)` and `(pi)/(2)` are `e^(0), e^(pmi(2pi)/(15)),e^(pmi(4pi)/(15)),e^(pmi(6pi)/(15))`. Hence number of solutions is7. |
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| 227. |
If `z and w` are two complex numbers simultaneously satisfying te equations, `z^3+w^5=0 and z^2 +overlinew^4 = 1,` thenA. `z` and `w` both are purely realB. `z` is purely real and `w` is purely imagineryC. `w` is purely real and `z` is purely imagineryD. `z` and `w` both are imaginery |
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Answer» Correct Answer - A `(a)` `z^(3)=-omega^(5)implies|z|^(3)=|omega|^(5)implies|z|^(6)=|omega|^(10)` ………`(i)` and `z^(2)=(1)/(baromega^(4))implies|z|^(2)=(1)/(|omega|^(4))implies|z|^(6)=(1)/(|omega|^(12))`……..`(ii)` From `(i)` and `(ii)`, `|omega|=1` and `|z|=1implieszbarz=omegabaromega=1` Again `z^(6)=omega^(10)` and `z^(6)*baromega^(12)=1` `z^(6)=(1)/(baromega^(12))=omega^(10)` (using `(iii)`) `implies (omegabaromega)^(10)(baromega)^(2)=1` `implies(baromega)^(2)=1` `implies baromega=1` or `-1impliesomega=1` or `-1` If `omega=1`, then `z^(3)+1=0` and `z^(2)=1impliesz=-1` If `omega=-1`, then `z^(3)-1=0` and `z^(2)=1impliesz=1` Hence, `z=1` and `omega=-1` or `z=-1` and `omega=1` |
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| 228. |
If `alpha,beta` be the roots of the equation `u^2-2u+2=0` and if `cottheta=x+1,` then `((x+alpha)^n-(x+beta)^n)/(alpha-beta)` is equal to (a) `((sin n theta),(sin^n theta))` (b) `((cosn theta),(cos^n theta))` (c) `((sinn theta),cos^n theta))` (d) `((cosn theta),(sintheta^n theta))`A. `(sinn ntheta)/(sin^(n) theta)`B. `(cos ntheta)/( cos^(n)theta)`C. `(sin ntheta)/(cos^(n) theta)`D. `(cos ntheta)/(sin^(n) theta)` |
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Answer» Correct Answer - A `u^(2)-2u+2=0rArru=1pmi` `rArr ((x+alpha)^(n)-(x+beta)^(n))/(alpha-beta)` `=([(cottheta-1)+(1+i)]^(n)-[(cottheta-1)+(1-i)]^(n))/(2i)` `(thereforecottheta-1=x)` `=((costheta+isintheta)^(n)-(costheta-isintheta)^(n))/(sin^(n)theta 2i)` `=(2isinntheta)/(sin^(n)theta2i)` `(sin n theta)/(sin^(n)theta)` |
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| 229. |
If `alpha` is complex fifth root of unity and `(1+alpha +alpha^(2)+ alpha^(3))^(2005) = p + qalpha + ralpha^(2) + salpha^(3)` (where p,q,r,s are real), then find the value of `p+ q+r+s`. |
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Answer» Correct Answer - `-1` `alpha ` is a complex fifth roots of unity. `therefore 1+ alpha +alpha^(2) + alpha^(3)+ alpha^(4) = 0` Now, given that `(1+ alpha + alpha^(3) + alpha^(3))^(2005) = p + qalpha + ralpha^(2) + salpha^(3)` `rArr (-alpha^(4)) ^(2005) = p + qalpha + ralpha^(2) + salpha^(3)` `rArr - alpha^(8020) = p+ qalpha + ralpha^(2)+ salpha^(3)` `rArr -(alpha^(5))^(1640) = p +qalpha+ = ralpha^(2) + salpha^(3)` `rArr -1 = p + qalpha + ralpha^(2) + salpha^(3)` `rArr p = - 1, q = r = s = 0` `therefore p + q+ r+s = -1` |
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| 230. |
Let `z`be a complex number satisfying equation `z^p-z^(-q),w h e r ep ,q in N ,t h e n`if `p=q`, then number of solutions of equation will be infinite.if `p=q`, then number of solutions of equation will be finite.if `p!=q`, then number of solutions of equation will be `p+q+1.`if `p!=q`, then number of solutions of equation will be `p+qdot`A. if p=q, then number of solution of equation will infinte.B. if p=q, then number of solutions of equaiton will finiteC. if `pne q`, then number of solutions of equaiton will `p+q+1`.D. if `p ne q`, then number of solutions of equaiton will be `p+q` |
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Answer» Correct Answer - A::B If q=p, then equation becomes `z^(p)= z^(-q)` and it has infinte number of solutions because any z `in ` R will satisfly it. `p ne q`, let `p gt q`, then `z^(p) = z^(-q)` |
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| 231. |
The number of solutions of the equation `z^2+z=0` where z is a a complex number, isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - D Let `z = x + iy`, so that `barz = x -iy`. `therefore z ^(2) + barz = 0` `rArr (x^(2) -y^(2) + x) + i(2xy - y) = 0` Equating real and imaginary parts, we get `x ^(2)- y^(2) + x = 0` and `2xy - y = 0 rArr y = 0 or x = (1)/(2)` If `y 0`, then (1) given `x^(2) + x= 0` `rArr x = 0 or x = -1` If `x = 1//2`, then from Eq.(1) `y^(2) = (1)/(4) + (1)/(2) = (3)/(4) or y = pm (sqrt(3))/(2)` Hence, there are four solutions in all. |
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| 232. |
Is the following computation correct? If not give the correctcomputation: `[sqrt((-2))dotsqrt((-3))]=sqrt((-2)dot(-3))=sqrt(6)` |
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Answer» Correct Answer - Not correct The said computation is not correct, because -2 and -3 both are negative and `sqrt(ab) = sqrt(a) sqrt(b)` is ture when at least one of a and b is positive or zero. The correct computation is `(sqrt(-2))(sqrt(-3))= (isqrt(2)) (isqrt(3))=i^(2) sqrt(6) = -sqrt(6)` |
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| 233. |
If `sqrt(3)+i=(a+i b)//(c+i d)`, then find the value of `tan^(-1)(b//a)tan^(-1)(d//c)dot` |
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Answer» Correct Answer - `npi +(pi)/(6), n in Z` We have , `sqrt(3)+ i =(a+ ib) (c+id)` `rArr ac -bd = sqrt(3) and ad + bc = 1` Now, `tan^(-1)((b)/(a)) + tan^(-1)((d)/(c)) = tan^(-1)((b)/(a)+(d)/(c))/(1-(b)/(a)(d)/(c))` `=tan ^(-1).(bc+ad)/(ac-db)` `=tan^(-1)(1)/(sqrt(3)) = npi+(pi)/(6), n in Z` |
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| 234. |
Prove that the complex numbers `z_(1),z_(2)` and the origin form an equilateral triangle only if `z_(1)^(2) + z_(2)^(2) - z_(1)z_(2)=0`. |
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Answer» We know that if `z_(1),z_(2)` and `z_(3)` from an equilateral trinagle , then `z_(1)^(2) + z_(2)^(2) + z_(3)^(2) = z_(1)z_(2) + z_(2)z_(3) + z_(3)z_(1)` Putting `z_(3) = 0`, we get `z_(1)^(2) + z_(2)^(2) - z_(1)z_(2) =0` |
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| 235. |
Let `s , t , r`be non-zerocomplex numbers and `L`be the setof solutions `z=x+i y (x , y in RR, i=sqrt(-1))`of the equation`s z+t z +r=0`, where ` z =x-i y`. Then,which of the following statement(s) is (are) TRUE?If `L`has exactlyone element, then `|s|!=|t|`(b) If `|s|=|t|`, then `L`hasinfinitely many elements(c) Thenumber of elements in `Lnn{z :|z-1+i|=5}`is at most2(d) If `L`has morethan one element, then `L`hasinfinitely many elementsA. If L has exactly one element, then `|s| ne |t|`B. If `|s| = |t|` then L has infinitely many elementsC. The number of elements in `L nn {z :|z-1+i|=5}` is at most 2D. If L has most than one elements, then L has infinitely many elements. |
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Answer» Correct Answer - A::C::D Given `sz + bartz + z =0 , barz = x -iy" "(i)` Taking conjugate, we get `barz barz + bartz + barz = 0" "…(i)` Adding (i) and (ii), we get `(t + bars) barz + (bart + s)z+(r+barr) = 0` Clearly, this is the equations of a straight line . Now, eliminating `barz` form (i) and (ii) we,get `sbarsz + rbars + rbars -tbartz - barrt =0` `rArr (|s|^(2) - |t|^(2))z = barrt - rbars` If `|S| ne|t|` then L has unique solution. If `|s| = |t|` and `barrt = rbars`, then L has infinite solutions. If `|s| = |t|` and `barrt ne rbars`, then L has no solutions. Thus, L has either unique, infinite or no solution. Also, line can intersect circle in maximum two points. So, number of elements in L `nn{z:|z-1+i|=5}` is at most 2 |
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| 236. |
Show that `e^(2m itheta)((icottheta+1)/(i cottheta-1))^m=1.` |
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Answer» Let `cot^(-1) p = theta`. Then `cot theta = p` Now, L.H.S `= e^(2mi theta)((icot theta + 1)/(i cot theta -1))^(m)` `=e^(2mi theta)[(i(cot theta -i))/(i(cot theta +i))]^(m)` `= e^(2m i theta)((cot theta -i)/(cot theta + i))^(m)` `= e^(2 mitheta)((cos theta - i sintheta)/(cos theta + i sin theta))^(m)` `= e^(2 mitheta)((e^(-itheta))/(e^(itheta)))^(m)` `= e^(2mi theta)(e^(-2itheta))^(m)` `=e^(2mitheta)(e^(-2itheta))^(m) = e^(0) = 1= R.H.S` |
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| 237. |
Let `s , t , r`be non-zerocomplex numbers and `L`be the setof solutions `z=x+i y (x , y in RR, i=sqrt(-1))`of the equation`s z+t z +r=0`, where ` z =x-i y`. Then,which of the following statement(s) is (are) TRUE?If `L`has exactlyone element, then `|s|!=|t|`(b) If `|s|=|t|`, then `L`hasinfinitely many elements(c) Thenumber of elements in `Lnn{z :|z-1+i|=5}`is at most2(d) If `L`has morethan one element, then `L`hasinfinitely many elementsA. If L has exactly one element , then `|s| ne |t|`B. If `|s|=|t|` then L has infinitely many elementsC. The number of elements in `L cap {z:|z-1+i|=5}` is at most 2D. If L has more than one element , then L has infinitely many elements |
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Answer» Correct Answer - A::C::D We have `sz+t barz+r=0` On taking conjugate `bars bar z+ bari z+ bar r`=0 On solving Eqs (i) and (ii) we get `z=(barrt-rbars)/(|s^2-|t|^2` (a) For unique solitions of z `|s|^2-|t|^2 ne 0 rArr |s| ne |t|` It is true (b) If |s|=|t| then `bar r t - r bar s ` may or may not be zero so,z may have no soluiton L may be an empty set . It is false ( c) If elements ofset L repersents line then this line and given circle intersect at maximum two point Hence it is true . |
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| 238. |
Let `A={a in R}`the equation `(1+2i)x^3-2(3+i)x^2+(5-4i)x+a^2=0`has at least one real root. Then the value of `(suma^2)/2`is_______. |
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Answer» Correct Answer - 18 `A= (1+ 2i)x^(3) - 2(3 + i) x^(2) + (5-4i)x + 2a^(2) = 0` Let the real root of equations be `alpha`. Then `(1+2i) alpha^(3) -2(3+i) alpha^(2) + (5-4i) alpha + 2alpha^(2)= 0` Equating imaginary part zero, we get `2apha^(3) - 2alpha^(2) - 4alpha = 0` ` rArr or alpha(alpha^(2) - alpha -2) = 0` `rArr alpha = 0 or alpha = - 1,2` Now equating real part zero, we have `alpha^(3) - 6alpha^(2) + 5alpha+ 2alpha^(2) = 0` " `alpha = 0 rArr a =0` ` alpha= -1 rArr a = pm sqrt(6)` `alpha = 2 rArr alpha = pm sqrt(3)` `rArr sum alpha^(2) = (0)^(2) + (+sqrt(6))^(2) + (-sqrt(6))^(2) + (+sqrt(3))^(2)+ (-sqrt(3))^(2)` `= 18` |
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| 239. |
Prove that traingle by complex numbers `z_(1),z_(2)` and `z_(3)` is equilateral if `|z_(1)|=|z_(2)| = |z_(3)|` and `z_(1) + z_(2) + z_(3)=0` |
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Answer» Triangle is formed by complex numbers `z_(1),z_(2)` and `z_(3)` Let `|z_(1)| = |z_(2)| = |z_(3)|` and `z_(1) + z_(2)+z_(3)=0` If `|z_(1)| = |z_(2)| =|z_(3)|`, then `z_(1),z_(2)` and `z_(3)` are equidistant form origin. So, origin is circumcentre of the triangle. Also, centroid of the triangle is `(z_(1)+z_(2)+z_(3))/(3) = 0`. Thus, circumcentre and centroid coincide. Hence, triangle is equilateral. |
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| 240. |
Solve the equation `(x-1)^3+8=0`in the set C of all complex numbers. |
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Answer» We have, `(x-1)^(3) + 8 =0 pr (x-1)^(3) = -8 or x - 1 (-8)^(1//3)` or ` x -1= 2(-1)^(1//3)` or `x-1 = 2(-1) or x - 1 =2 (-omega) or x-1 = 2(-omega^(2)) " "[because (-1)^(1//3) = -1 or -omega or -omega^(2)]` `x -1 = -2 or x - 1 = -2omega or x - 1 = - 2omega^(2)` `or x = -1 or x =1 -2 omega or x = 1- 2omega^(2)` Hence , the solution of the equations `(x-1)^(3) + 8 = 0` are `-1,1,-2omega` and `1-2omega^(2)`. |
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| 241. |
Find the range of real number `alpha`for which the equation `z+alpha|z-1|+2i=0`has a solution. |
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Answer» Let `z = x + iy` We have, `z+ alpha|z-1| + 2i=0` `rArr x + i(y+2) + alpha sqrt((x-1)^(2) + y^(2))=0` Equating real and imaginary parts `therefore y = - 2 and x + alpha sqrt((x-1)^(2)+4)=0` `therefore x^(2) = alpha^(2) (x^(2)-2x +5)` `therefore (1-alpha^(2))x^(2) + 2alpha^(2)x-5alpha^(2) = 0` Since x is real, `therefore D =b^(2) - 4ac ge 0` `rArr 4alpha^(4) + 20alpha^(2)(1-alpha^(2)) ge 0` `rArr 4alpha^(4) + 5alpha^(2) ge 0` `rArr 4alpha^(2)(alpha^(2) - (5)/(4)) le 0` `rArr (-sqrt(5))/(2) le alpha le(sqrt(5))/(2)`. |
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| 242. |
Solve:`i x^2-3x-2i=0,` |
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Answer» `ix^(2) - 3x - 2i = 0,` or `x^(2) +3ix - 2i = 0" "("dividing by I")`, or `x = (-3ipmsqrt(-9 + 4.1.2))/(2.1) = (-3ipmsqrt(-1))/(2) = (-3i +i)/(2)` `rArr " " x= - "i" or x = -2i` |
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| 243. |
Solve the following for z: `z^2-(3-2i)z=(5i-5)` |
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Answer» `z^(2) - (3-2i)z = (5i -5)` `or z^(2)-(3-2i)z-(5i-5) =0` `or z=((3-2i)pmsqrt((3-2i)^(2)+4(5i-5)))/(2)` |
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| 244. |
Solve the equation `|z|=z+1+2i` |
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Answer» `|z|= z+1 + 2i` `rArr sqrt(x^(2) + y^(2)) = x + iy + 1+ 2i =x+1+(2+y)i` `rArr sqrt(x^(2) + y^(2)) = x + 1 and 0=2 +y or y = -2` `rArr sqrt(x^(2) + 4)) = x+1` `or x^(2) + 4 = x^(2) + 2x + 1` `or 2x = 3` `or x=(3)/(2)` `rArr x + iy = (3)/(2)-2i` |
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| 245. |
If `z_r=cos(pi/(3^r))+isin(pi/(3^r)),r=1,2,3, ,`prove that `z_1z_2z_3 z_oo=idot` |
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Answer» Here, `z_n = cos(pi/(3^r))+isin(pi/(3^r))` `:.L.H.S. = z_1z_2z_3...z_oo` `= (cos(pi/(3))+isin(pi/(3)))(cos(pi/(3^2))+isin(pi/(3^2)))(cos(pi/(3^3))+isin(pi/(3^3)))...oo` `=cos(pi/3+pi/3^2+pi/3^3+...oo)+isin(pi/3+pi/3^2+pi/3^3+...oo)` `= cos((pi/3)/(1-1/3))+isin((pi/3)/(1-1/3))` `=cos(pi/2)+isin(pi/2)` `= 0+i =i = R.H.S.` |
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| 246. |
if `|z^2-1|=|z|^2+1` then `z` lies onA. a circleB. a parabolaC. an ellipseD. none of these |
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Answer» Correct Answer - D Let `z=z+iy`.Then, `|z^(2)-1|=|z^(2)|+1` `rArr |(x^(2)-y^(2)-1)+2ixy|=x^(2)+y^(2)+1` `rArr (x^(2)-y^(2)-1)^(2)+4x^(2)y^(2)=(x^(2)+y^(2)+1)^(2)` `rArr x =0` Hence, z lies on imaginary axis. |
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| 247. |
If complex number z=x +iy satisfies the equation `Re (z+1) = |z-1|`, then prove that z lies on `y^(2) = 4x`. |
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Answer» We have `Re(z+1) = |z-1|` `rArr Re (x+iy+1) = |x + iy-1|` `rArr x + 1 = sqrt((x-1)^(2) + y^(2)))` `rArr (x+1)^(2) = (x+1)^(2) + y^(2)` `rArr y^(2) = 4x` Thus, z lies on `y^(2) = 4x`. |
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| 248. |
The modulus and amplitude of `(1+2i)/(1-(1-i)^(2))` areA. `sqrt(2)` and `(pi)/(6)`B. `1` and `(pi)/(4)`C. `1` and `0`D. `1` and `(pi)/(3)` |
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Answer» Correct Answer - C `(c )` `(1+2i)/(1-(1-i)^(2))=(1+2i)/(1-1-i^(2)+2i)=(1+2i)/(1+2i)=1+i.0` `:.` Modulus `=1` Amplitude `=tan^(-1)|(0)/(1)|=0` |
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| 249. |
Find the square root `9+40 idot` |
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Answer» Correct Answer - `(5+4i) or - (5+4i)` Let `sqrt(9+40i) = x +iy`. Then `(x + iy)^(2) = 9+40i` `or x^(2) - y^(2) = 9" "(1)` `and xy = 20" "(2)` Squaring (1) and adding with 4 times the square of (2), we get `x^(4) + y^(4) - 2x^(2)y^(2)+ 4x^(2)y^(2) = 81 + 1600` `or (x^(2) +y^(2))= 1681` or `x^(2) + y^(2) = 41" "(3)` From`(1) +(3)`, we get `x^(2)= 25 or x = pm 5` and `y =pm4` From Eq. (2) we can see that x and y are of same sign. Therefore, `x+iy= (5+4i) or - (5+41)` |
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| 250. |
Consider the equation `az + bar(bz) + c =0`, where a,b,c `in`Z If `|a| = |b| and barac ne b barc`, then z hasA. ifinite solutionsB. no solutionsC. finite solutionsD. cannot say anything |
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Answer» Correct Answer - B `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. |
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