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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.
| 1. |
Convert the following in the polar form : (i) `(1+7i)/((2-i)^2)`(ii) `(1+3i)/(1-2i)` |
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Answer» Let `z =((1+7i))/((2-i)^(2))=((1+7i))/((4+i^(2)-4i))=((1+7i))/((3-4i))xx((3+4i))/((3+4i))` `rArr" "z=((1+7i)(3+4i))/((9+16))=(-25+25i)/(25)=(-1+i)`. Let its polar form be `z = r(cos theta + i sin theta)`. Now, `r=|z|=sqrt((-1)^(2)+1^(2))=sqrt(2)`. Let `alpha` be the acute angle, given by `tan alpha=|(Im(z))/(Re(z))|=|(1)/(-1)|= 1 rArr alpha = (pi)/(4)`. Clearly, the point representing `z = (-1+i)` is P`(-1, 1)`, which lies in the second quadrant. `therefore" "arg(z) = theta = (pi-alpha)=(pi-(pi)/(4))=(3pi)/(4)`. Thus, `r=|z|=sqrt(2) and theta = (3pi)/(4)`. Hence, the required polar form is `z = sqrt(2)("cos"(3pi)/(4) + "i sin"(3pi)/(4))`. |
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| 2. |
Solve: `x^(2) + 3ix + 10 = 0` |
| Answer» Correct Answer - `{2i,- 5i}` | |
| 3. |
Express `(2 + 3i)^(3)` in the form `(a + ib).` |
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Answer» We know that `(z_(1) + z_(2))^(3) = z_(1)^(3) + z_(2)^(3) + 3z_(1)z_(2)(z_(1)+z_(2))`. `therefore" "(2+3i)^(3) = 2^(3)+(3i)^(3) + 3 xx 2 xx 3i xx (2 + 3i)` `= 8 + 27i^(3)+36i+54i^(2)` `= (8 - 27i + 36i - 54)" "[because i^(3) = -i and i^(2) = -1]` `= (-46 + 9i)`. Hence, `(2+3i)^(3) = (-46 + 9i)`. |
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| 4. |
Express `(2-3i)^(3)` in the form (a + ib). |
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Answer» Correct Answer - 46 - 9i `(2-3i)^(3)=2^(3)-(3i)^(3)-(3xx2xx3i)(2-3i)` `=(8-27i^(3)-36i-54)=(8+27i-36i-54)=(46-9i)`. |
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| 5. |
Solve the system of equations `R e(z^2)=0, |z|=2` |
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Answer» Given that, Re `(z^(2)) = 0, |z| = 2` Let `z = x + iy` `|z|=sqrt(x^(2) + y^(2))` `:. sqrt(x^(2) + y^(2)) = 2` `rArr x^(2) + y^(2) = 4 " " …(i)` and Re `(z) = x` Also, `z = x + iy` `rArr, z^(2) = x^(2) + 2ixy - y^(2)` `rArr z^(2) = (x^(2) - y^(2)) + 2ixy` `rArr Re(z^(2) = x^(2) - Y^(2)" " [:. Re (z^(2)) = 0 ]` `rArr x^(2) - Y^(2) = 0 " " ...(ii)` From Eqs. (i) and (ii), `x^(2) + x^(2) = 4` `rArr 2x^(2) = 4 rArr x^(2) = 2 ` `rArr x= pm sqrt (2)` `:. " " y =pmsqrt(2)` `z = x + iy` `rArr z = sqrt(2)pmisqrt(2),- sqrt(2)pm isqrt(2)` |
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| 6. |
Solve the equation, `z^(2) = bar(z)`, where z is a complex number. |
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Answer» Let z = x + iy. Then, `z^(2) = bar(z) rArr (x+iy)^(2)=x - iy` `rArr" "(x^(2)-y^(2))+2ixy = x - iy." "...(i)` Equating real parts and imaginary parts on both sides of (i) separately, we get `x^(2)-y^(2)=x" "...(ii) and 2xy = -y" "...(iii)`. From (iii), we get `2xy + y = 0 rArr y(2x + 1) = 0 rArr y = 0 or x = (-1)/(2)`. Case I When y = 0. Putting y = 0 in (ii), we get `x^(2)-x = 0 rArr x(x-1) = 0 rArr x = 0 or x = 1`. `therefore" "(x = 0, y = 0)or(x = 1, y = 0)` Thus, `z = (0+i0) or z = (1 + i0)`. Case II When `x = (-1)/(2)`. Putting `x = (-1)/(2)` in (ii), we get `((-1)/(2))^(2)-y^(2)=((-1)/(2)) rArr y^(2)=((1)/(4)+(1)/(2))=(3)/(4) rArr y = +-(sqrt(3))/(2)`. `therefore" "(x=(-1)/(2),y=(sqrt(3))/(2))or(x=(-1)/(2),y=(-sqrt(3))/(2))`. Thus, `z=((-1)/(2)+(sqrt(3))/(2)i)or z=((-1)/(2)-(sqrt(3))/(2)i)`. Hence, `z=0,1((-1)/(2)+(sqrt(3))/(2)i)and((-1)/(2)-(sqrt(3))/(2)i)` are the required roots of the given equation. |
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| 7. |
Convert each of the following complex numbers into polar form: `{:((i),3,(ii),-5,(iii),i,(iv),-2i):}` |
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Answer» (i) The given complex number is z = 3 + 0i. Let its polar form be `z = r(cos theta + i sin theta)`. Now, `r = |z| = sqrt(3^(2)+0^(2))=sqrt(9)=3`. Clearly, z = 3 + 0i is represented by the point P(3, 0), which lies on the positive side of the x-axis. `therefore" "arg(z) = theta = 0`. Thus, `r = 3 and theta = 0`. Hence, the required polar form of `z = 3 + 0i "is" 3(cos 0 + i sin 0)`. (ii) The given complex number is z = -5 + 0i. Let its polar form be `z = r(cos theta + i sin theta)`. Now, `r=|z|=sqrt((-5)^(2)+0^(2))=sqrt(25)=5`. Clearly, z = -5 + 0i is represented by the point P(-5, 0), which lies on the negative side of the x-axis. `therefore" "arg(z) = pi rArr theta = pi` Thus, r = 5 and `theta = pi`. Hence, the required polar form of z = -5 + 0i is `5(cos pi + i sin pi)`. (iii) The given complex number is z = 0 + i. Let its polar form be `z = r(cos theta + i sin theta)`. Now, `r = |z| = sqrt(0^(2)+1^(2))=1`. Clearly, z = 0 + i is respresented by the point P(0, 1), which lies on the y-axis and above the x-axis. `therefore" "arg(z)=(pi)/(2) rArr theta = (pi)/(2)`. Thus, `r = 1 and theta = (pi)/(2)`. Hence, the requird polar form of z = 0 + i is. `1.("cos"(pi)/(2)+"i sin"(pi)/(2)), i.e., ("cos"(pi)/(2)+"i sin"(pi)/(2))`. (iv) The given complex number is z = 0 -2i. Let its polar form be `z = r(cos theta + i sin theta)`. Now, `r = |z| = sqrt(0^(2) + (-2)^(2))=sqrt(4) =2`. Clearly, z = (0-2i) is represented by the point P(0, -2), which lies on the y-axis and below the x-axis. `therefore" "arg(z)=(-pi)/(2) rArr theta = (-pi)/(2)`. Thus, r = 2 and `theta = (-pi)/(2)`. Hence, the required polar form of z = 0 -2i is `z=2{cos((-pi)/(2))+i sin((-pi)/(2))}, i.e., 2("cos"(pi)/(2)-"i sin"(pi)/(2))`. |
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| 8. |
If the complex number `Z_(1)` and `Z_(2), arg (Z_(1))- arg(Z_(2)) =0`. then show that `|z_(1)-z_(2)| = |z_(1)|-|z_(2)|`. |
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Answer» Let `z_(1) = r_(1) (costheta_(1) + isin theta_(1))` and `z_(2) = r_(2) (costheta_(2) + isin theta_(2))` `rArr arg (z_(1)) = theta_(1)arg(z_(2)) = theta_(2)` Given that, arg `(z_(1)) - arg (z_(2)) = 0` `theta _(1)-theta _(2) -0 rArr theta _(1) = theta _(2)` ` z_(2) = r_(2) (costheta_(2) + isintheta_(2)) [:.theta_(1) = theta_(2)]` ` z_(1)-z_(2) = (r_(1) costheta_(1)-r_(2) costheta_(1)) + (r_(1) isintheta_(1)- r_(2)isintheta_(1)) ` ` |z_(1)-z_(2)| = sqrt((r_(1) costheta_(1)-r_(2) costheta_(1))^(2) + (r_(1) isintheta_(1)- r_(2)isintheta_(1))^(2)) ` `=sqrt(r_(1)^(2) +r_(1)^(2)-2 r_(1)r_(2)cos^(2)theta_(1) -2 r_(1)r_(2)sin^(2)theta_(1))` `=sqrt(r_(1)^(2) +r_(1)^(2)-2 r_(1)r_(2)(sin^(2)theta_(1) + cos^(2)theta_(1)))` `=sqrt(r_(1)^(2) +r_(1)^(2)-2 r_(1)r_(2))=sqrt((r_(1)-r_(2))^(2))` `rArr |z_(1) - z_(2)| =r_(1)-r_(2) [:. r = |z|]` `|z_(1) - z_(2)|` " "Hence proved |
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| 9. |
`Z_1 and Z_2` are two complex numbers represented by the plans A and B in the argand Plane `Z_! And Z_2` are the roots of `Z^2+pZ+q = 0`. `/_AOB = alpha` `alpha!= 0` and O is origin and OA = OB, then `p^2/q` is equal to |
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Answer» `Z_1=r*e^(itheta)` `Z_2=r*e^(1(theta+alpha))` `Z^2+PZ+q=0` `Z_1+Z_2=-P` `Z_1Z_2=q` `p^2/q=((Z_1+Z_2)^2/(z_1Z_2))` `(r^2*e^(120)(1+2e^(1alpha)+e^(12alpha)))/(r^2*e^(itheta)*e^(i(theta+alpha)` `=(1+2e^(1alpha)+e^(12alpha))/(e^(ialpha)` `=e^(-ialpha)+2+e^(ialpha)`. |
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| 10. |
For all complex numbers `z_(1) and z_(2)`, prove that `|z_(1)+z_(2)|^(2)+|z_(1)-z_(2)|^(2)=2(|z_(1)|^(2)+|z_(2)|^(2))`. |
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Answer» We have `|z_(1) + z_(2)|^(2)+|z_(1)-z_(2)|^(2)=2(|z_(1)|^(2)+|z_(2)|^(2))`. `=(z_(1)+z_(2))(bar(z)_(1)+bar(z)_(2))" "[because bar((z_(1)+z_(2)))=bar(z)_(1)+bar(z)_(2)]` `=z_(1)bar(z)_(1)+z_(2)bar(z)_(2)+z_(1)bar(z)_(2)+z_(2)bar(z)_(1)` `=|z_(1)|^(2)+|z_(2)|^(2)+z_(1) bar(z)_(2) + z_(2) bar(z)_(1)." "...(i)` `|z_(1)-z_(2)|^(2)=(z_(1)-z_(2))(bar(z_(1)-z_(2)))=(z_(1)-z_(2))(bar(z)_(1)-bar(z)_(2))` `=z_(1) bar(z)_(1)+z_(2)bar(z)_(2)-z_(1)bar(z)_(2)-z_(2)bar(z)_(1)` `=|z_(1)|^(2)+|z_(2)|^(2)-z_(1)bar(z)_(2)-z_(2)bar(z)_(1)." "...(ii)` On adding the corresponding sides of (i) and (ii), we get `|z_(1)+z_(2)|^(2)+|z_(1)-z_(2)|^(2)=2(|z_(1)|^(2)+|z_(2)|^(2))`. |
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| 11. |
Express each of the following in the form (a + ib) : `{:((i),(3+sqrt(-5))(3-sqrt(-5)),(ii),(-2+sqrt(-3))(-3 + 2 sqrt(-3))),((iii),(-2+3i)^(2),(iv),(sqrt(5)-7i)^(2)):}` |
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Answer» We have `(i)" "(3+sqrt(-5))(3-sqrt(-5))=(3+sqrt(5)i)(3-sqrt(5)i)={3^(2)-(sqrt(5)i)^(2)}` `=(9-5i^(2))=(9+5)=14 + 0i." "[because (z_(1) + z_(2))(z_(1)-z_(2))=(z_(1)^(2)-z_(2)^(2))]` `(ii)" "(-2+sqrt(-3))(-3+2 sqrt(-3))=(-2+sqrt(3)i)(-3+2sqrt(3)i)` `=(6-4 sqrt(3)i - 3 sqrt(3)i + 6i^(2))` `=-7 sqrt(3)i = 0 - 7 sqrt(3)i`. `(iii)" "(-2+3i)^(2)=(-2)^(2)+(3i)^(2)+2 xx (-2) xx 3i` `=(4+9i^(2)-12i)=(4-9-12i)=(-5-12i)`. `(iv)" "(sqrt(5)-7i)^(2)=(5)^(2)+(7i)^(2)-2 xx sqrt(5) xx 7i` `=(25+49i^(2)-14 sqrt(5)i)=(25-49-14 sqrt(5)i)` `=(-24-14 sqrt(5)i)`. |
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| 12. |
Reduce `((1)/(1+2i)+(3)/(1-i))((3-2i)/(1+3i))` to the form `(a + ib).` |
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Answer» We have `((1)/(1+2i)+(3)/(1-i))((3-2i)/(1+3i))` `={((1-i)+(3+6i))/((1+2i)(1-i))}((3-2i)/(1+3i))=((4+5i))/((3+i))xx((3-2i))/((1+3i))` `=((4+5i)(3-2i))/((3+i)(1+3i))=((12+10)+(15-8)i)/((3-3)+(9i+i))=((22+7i))/(10i)xx(i)/(i)` `=((7i^(2)+22i))/(10i^(2))=(-7+22i)/(-10)=((7)/(10)-(22)/(10)i)=((7)/(10)-(11)/(5)i)`. |
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| 13. |
`z_(1) "the"z_(2) "are two complex numbers such that" |z_(1)| = |z_(2)|`. "and" arg `(z_(1)) + arg (z_(2) = pi," then show that "z_(1) = - barz_(2).` |
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Answer» Let `z_(1) = r_(1) (costheta_(1) + i sintheta _(1))` and `z_(1) = r_(1)(costheta_(1) + i sintheta _(1)) are two complex numbers. Given that `|z_(1)| = |z_(2)|` and `arg (z_(1)) + arg (z_(2)) = pi` If `|z_(1)|= |z_(2)|` `rArr r_(1) = r_(2) ...(i) and if `arg (z_(1)) + arg (z_(2)) = pi` ` rArr theta _(1) + theta_(2) = pi` `rArr theta ^(1) = pi - theta_(2)` Now, `z_(1) = r_(1) (costheta_(1) + i sintheta _(1))` `rArr z_(1) = r_(2) [(cos(pi -theta_(2)) + i sin (pi- theta_(2))]" "[:.r_(1) = r_(2) and theta_(1) = (pi-theta_(2))]` `rArr `z_(1) = r_(2) (-costheta_(2) + i sin theta_(2))` ltbr. `rArr z_(1) = - r_(2) (costheta_(2) - i sin theta_(2))` `z_(1) = -[r_(2) (costheta_(2) - i sin theta_(2))]` `rArr z_(1) = - barz_(2)" "[:. barz = r_(2)(costheta_(2) -i sintheta _(2))]` |
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| 14. |
If ` (a^(2) + 1)^(2)/(2a-i)=x+iy," then when is the value of " x^(2)+y^(2)`? |
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Answer» Given that, `= (a^(2) + 1)^(2)/(2a-i)=x+iy rArr (a^(2) + 1)^(2)/((2a-i)) x+iy` `rArr = ((a^(2) + 1)^(2)(2a + i))/ ((2a + i) (2a + i))=x+iy` `rArr = ((a^(2) + 1)^(2)(2a + i))/(4a^(2)+1)=x+iy` `rArr = (2a(a^(2) + 1)^(2))/(4a^(2)+1) "and" y=(a^(2) + 1)^(2)/(4a^(2)+1) ` `:. x^(2)+y^(2)= 4a[((a^(2) + 1)^(2))/(4a^(2)+1)]^(2)+[((a^(2) + 1)^(2))/(4a^(2)+1)]^(2) ` `((4a^(2)+1)(a^(2)+1)^(4))/(4a^(2)+1)^(2)=(a^(2)+1)^(4)/(4a^(2)+1)^()` |
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| 15. |
Find the conjugate of each of the following : `{:((i),(-5-2i),(ii),(1)/((4+3i)),(iii),((1+i)^(2))/((3-i)),(iv),((1+i)(2+i))/((3+i))),((v),sqrt(-3),(vi),sqrt(2),(vii),-sqrt(-1),(viii),(2-5i)^(2)):}` |
| Answer» Correct Answer - `{:((i),(-5+2i),(ii),((4)/(25)+(3)/(25)i),(iii),((-1)/(5)-(3)/(5)i),"(iv)",((3)/(5)-(4)/(5)i)),((v),-sqrt(3)i,(vi),sqrt(2),(vii),i,"(viii)",(-21+20i)):}` | |
| 16. |
Write down the conjugate of each of the following : `(i)" "(-5+sqrt(-1))" "(ii)" "(-6-sqrt(-3))" "(iii)" "i^(3)" "(iv)" "(4+5i)^(2)` |
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Answer» We know that the conjugate of z = (a + ib) is given as `bar(z) = (a - ib)`. (i) Let `z = -5+sqrt(-1)=(-5+i)`. Then, `bar(z) = bar((-5+i))=(-5-i)`. (ii) Let `z = -6- sqrt(-3) = -6 - i sqrt(3)`. Then, `bar(z) = bar((-6-i sqrt(3))) = (-6 + i sqrt(3))`. (iii) Let `z = i^(3) = -i = 0 - i`. Then, `bar(z)=bar((0-i))=(0+i)=i`. (iv) Let `z = (4+5i)^(2)=(4)^(2) + (5i)^(2)+2 xx 4 xx 5i` `=(16 - 25 + 40i)=(-9+40i)`. `bar(z)=bar((-9+40i))=(-9-40i)`. |
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| 17. |
What is the conjugate of `(2-i)/((1-2i)^(2))`? |
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Answer» Given that, `z=(2-i)/((1-2i)^(2))= (2-i)/(-3-4i^(2)-4i)` `=(2-i)/(1-4-4i)= (2-i)/(-3-4i)` `=((2-i))/(-(3+4i))= [((2-i)(3-4i))/((3+4i)(3-4i))]` `-({6-8i-3i+4i)/(9+16))=-(-11i+2)/(25)` `=(-1)/(25)(-211i) rArr z=(1)/(25)-(-2+11i)` `barz=(-1)/(25)(-211i) =(-2)/(25)-(11)/(25)i` |
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| 18. |
Express each of the following in the form (a + ib) and find its conjugate: `{:((i),(1)/((4+3i)),(ii),(2+3i)^(2),(iii),((2-i))/((1-2i)^(2))),((iv),((1+i)(1+2i))/((1+3i)),(v),((1+2i)/(2+i))^(2),(vi),((2+i))/((3-i)(1+2i))):}` |
| Answer» Correct Answer - `{:((i),z=((4)/(25)-(3)/(25)i)","bar(z)=((4)/(25)+(3)/(25)i),(ii),z=(-5+12i)","bar(z)=(-5-12i)),((iii),z=((-2)/(25)+(11)/(25)i)","bar(z)=((-2)/(25)-(11)/(25)i),(iv),z=((4)/(5)+(3)/(5)i)","bar(z)=((4)/(5)-(3)/(5)i)),((v),z=((7)/(25)+(24)/(25)i)","bar(z)=((7)/(25)-(24)/(25)i),(vi),z=(-1+i)","z^(-1)=((-1)/(2)-(1)/(2)i)):}` | |
| 19. |
If `z^(1),z^(2)` and `z^(3),z^(4)` are two pairs of conjugate complex number, then find arg `2((z_(1))/(z_(4)))+ ((z_(2))/(z_(3))).` |
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Answer» Let `z_(1) = r_(1)(costheta_(1) + I sintheta_(1))`, Then, `z_(2) = barz_(1) = r_(1) (costheta_(1)- i sintheta_(1)) = r_(1)[cos(-theta_(1))+sin(-theta_(1))]` Also, let `z _(3) = r_(2)(costheta_(2) + isintheta_(2))`, Then, `z_(4) = barz_(3) = r_(2) (costheta _(2)-isin theta_(2))` arg `((z_(1))/(z_(4))) + arg ((z_(2))/(z_(3))) = arg (z_(1))-arg(z_(4)) +arg(z_(2))-arg(z_(3))` `=theta_(1)-(-theta_(2)+(-theta_(1))-theta_(2)" "[:.arg (z)=theta] ` `=theta_(1)+theta_(2)-theta_(1)-theta_(2) =0` |
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| 20. |
Find the value of `|(1+i)((2+i))/((3+i))|` |
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Answer» Given that, `|(1+i) ((2+i))/((3+i))|=|((2+i+2i+i^(2)))/((3+i))|=|(2+3i-1)/(3+i)|` `| ((1+3i))/((3+i))|=|((1+3i)(3+i))/((3+i)(3-i))|` `|(3+9i-i-3i^(2))/(9-i^(2))|=|(1+8i+3)/(9+1)|=|(6+8i)/(10)|` `=sqrt(6^(2)/(100)+8^(2)/(100))=sqrt((36+64)/(100))=sqrt((100)/(100))=1` |
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| 21. |
The real value of `alpha` for which the expression `(1-isinalpha)/(1+2isinalpha)` is purely real isA. `(n + 1)(pi)/(2)`B. `(2n + 1)(pi)/(2)`C. `npi`D. None of these |
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Answer» Correct Answer - C Given expression , `z = (1-isin alpha)/(1 +2isinalpha)` =`((1 - 2isin alpha)(1 - 2isin alpha))/((1 + 2isin alpha)( 1 - 2isin alpha))` `= (1-isinalpha - 2isinalpha + 2i^(2) sin^(2)alpha)/(1+4i^(2)sin^(2)alpha)` ` = (1-3 isinalpha - 2 sin^(2)alpha)/(1 + 4sin^(2)alpha` `(1-2sin^(2)alpha)/(1 + 4 sin ^(2)alpha)-(3isinalpha)/(1 +4 sin^(2)alpha` It is given that z is a purely real. `:. (-3sinalpha)/(1+ 4 sin^(2)alpha) = 0` `rarr -3sinalpha = 0 rArr sinalpha = 0 ` `alpha = npi` |
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| 22. |
sinx + i cos 2x and cos x - i sin2x are conjugate to each other for(A) x=nπ (B) x=(n+1/2)π/2 (C) x=0 (D) no value of xA. `x = npi`B. `x = ("n" +(1)/(2))(pi)/(2)`C. x=0D. No value of x |
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Answer» Correct Answer - D let `z= sinx +icos2x ` and `barz = sinx - 2x " "...(i)` Given that, `barz = cosx - isin 2x" "...(ii)` `:. sinx - icon 2x = cos x - isin 2x` `rArr sin x = cos x and cos2x = sin2 x ` `rArr tanx = 1 "and" 2x =1 ` `rArr tan x = "tan"(pi)/(4) "and" 2x = "tan"(pi)/(4)` `rArr x =npi + (pi)/(4) "and" 2x = 2pi + (pi)/(4) ` `rArr 2x - x = 0 rArr x = 0 ` |
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| 23. |
Find real q such that `(3+2isintheta)/(1-2isintheta)`is purelyreal. |
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Answer» We have `((3+2i sin theta)/(1-2i sin theta))=((3+2i sin theta))/((1-2i sin theta))xx((1+2i sin theta))/((1+2i sin theta))` `=((3+2i sin theta)(1+2i sin theta))/((1-4i^(2) sin^(2)theta))` `=((3-4sin^(2)theta)+i(6sin theta+2sin theta))/((1+4 sin^(2)theta))` `=((3-4sin^(2)theta)+i(8sin theta))/((1+4 sin^(2)theta))` Now, `((3+2i sin theta)/(1-2i sin theta))` will be purely real only when `(8 sin theta)/((1+4sin^(2) theta))=0`. This happens only when `8 sin theta = 0 iff sin theta = 0 iff theta = n pi, n in N`. Hence, the required value of `theta` is `n pi`, where n `in` N. |
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| 24. |
Find real `theta` such that `(3+2isintheta)/(1-2isintheta)`is purely real. |
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Answer» `(3+2isintheta)/(1-2isintheta) = (3+2isintheta)/(1-2isintheta)**(1+2isintheta)/(1+2isintheta)` `=((3+2isintheta)(1+2isintheta))/(1-4i^2sin^2theta))` `=(3+6isintheta+2isintheta+4i^2sin^2theta)/(1+4sin^2theta)` `=(3-4sin^2theta)/(1+4sin^2theta)+(8sintheta)/(1+4sin^2theta)i` For this number to be real, imaginary part should be `0`. So, `(8sintheta)/(1+4sin^2theta) =0` `=>sintheta =0=>theta = npi` So, when `theta = npi`, given number will be real. |
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| 25. |
A value of `theta` for which `(2+3isintheta)/(1-2isintheta)`purely imaginary, is :(1) `pi/3`(2) `pi/6`(3) `sin^(-1)((sqrt(3))/4)`(4) `sin^(-1)(1/(sqrt(3)))` |
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Answer» `(2+3isintheta)/(1-2isintheta) ** (1+2isintheta)/(1+2isintheta)` `=(2+4isintheta+3isintheta+6i^2sin^2theta)/(1-4i^2sin^2theta)` `=(2-6sin^2theta+7isintheta)/(1+4sin^2theta)` `=(2-6sin^2theta)/(1+4sin^2theta) +i(7sintheta)/(1+4sin^2theta)` For this number to be purely imaginary, `(2-6sin^2theta)/(1+4sin^2theta) = 0` `=> 2 - 6sin^2theta = 0` `=>sin^2theta = 1/3` `=> sintheta = 1/3` `=> theta = sin^-1(1/sqrt3)` So, option (4) is the correct option. |
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| 26. |
Find the principal argument of (-2i). |
| Answer» Correct Answer - `(-pi)/(2)` | |
| 27. |
Root of equation `x^2+1=0` areA. `i, -i`B. `i, i`C. `-i, - i`D. `1, -1` |
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Answer» Correct Answer - A |
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| 28. |
find `x` and`y` for which the inequalities hold `(x^4+2ix)-(3x^2+iy)=(3-5i)+(1+2iy)` |
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Answer» Correct Answer - `(x=2, y=3) or (x= -2, y=(1)/(3))` `(x^(4)-3x^(2))+i(2x-y)=4+(2y-5)i` `rArr" "(x^(4)-3x^(2)-4)+i(2x-y-2y+5)=0` `rArr" "x^(4)-3x^(2)-4=0 and 2x - 3y + 5 = 0`. Now, `x^(4)-3x^(2)-4=0 rArr (x^(2)-4)(x^(2)+1)=0 rArr x = 2 or x = -2`. `(x = 2 rArr y = 3) and (x = -2 rArr y = (1)/(3))`. |
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| 29. |
Find the complex number z for which |z| = z + 1 + 2i. |
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Answer» Correct Answer - `((2)/(2)-2i)` Let the required complex number be z = (x + iy). Then, |z| = z + 1 + 2i `rArr" "|x+iy|=(x+iy)+1+2i` `rArr" "sqrt(x^(2)+y^(2))=(x+1)+(y+2)i` `rArr" "sqrt(x^(2)+y^(2))=(x+1) and y + 2 = 0` `rArr" "y = -2 and sqrt(x^(2)+(-2)^(2))=(x+1)` `rArr" "y = -2 and x^(2) + 4 = (x + 1)^(2) rArr x = (3)/(2) and y = -2`. |
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| 30. |
If `(x+i y)^3=u+i v ,`then show that `u/x+v/y=4(x^2-y^2)`. |
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Answer» `(u+iv)=(x+iy)^(3)=x^(3)-iy^(3)+3ixy(x+iy)` `rArr" "(u+iv)=(x^(3)-3xy^(2))+i(3x^(2)y-y^(3))` `rArr" "u=x^(3)-3xy^(2))+3x^(2)y-y^(3)` `rArr" "((u)/(x)+(v)/(y))=(x^(2)-3y^(2))+(3x^(2)-y^(2))=4(x^(2)-y^(2))`. |
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| 31. |
`(x+iy)^(1/3) =(a+ib)` then prove that `(x/a+y/b)=4(a^2-b^2)` |
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Answer» `(x+iy)=(a+ib)^(3)=a^(3)-ib^(3)+3iab(a+ib)` `rArr" "(x+iy)=(a^(3)-3ab^(2))+i(3a^(2)b-b^(3))` `rArr" "x=a^(3)-3ab^(2) and y = 3a^(2)b-b^(3)` `rArr" "((x)/(a)+(y)/(b))=(a^(2)-3b^(2))+(3a^(2)-b^(2))=4(a^(2)-b^(2))`. |
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| 32. |
Express `(1-2i)^(-3)`in the standard form `a+i bdot` |
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Answer» Correct Answer - `((-11)/(125)-(2)/(125)i)` `(1-2i)^(-3)=(1)/((1-2i)^(3))=(1)/((1-8i^(3)-6i(1-2i)})` `=(1)/((-11+2i))xx((-11-2i))/((-11-2i))=((-11-2i))/(125)` |
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| 33. |
If `z^(2) + |z|^(2) = 0`, show that z is purely imaginary. |
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Answer» Let z = (x + iy). Then, `z^(2)+|z|^(2)=0 rArr x^(2)-y^(2)+2ixy + x^(2)+y^(2)=0` `rArr" "x^(2)+ixy=0 rArr x^(2)=0 and xy = 0 rArr x = 0`. `therefore" "z` is purely imaginary. |
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| 34. |
If `alpha, beta` are the roots of ` a x^2 + bx + c = 0` and `k in R` then the condition so that `alpha < k < beta` is : |
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Answer» a>0 c=x f(x)<0 af(x)<0 `a^2k^2+abk+ac<0` a<0 c=x f(k)>0 af(x)<0 option 4 is correct. |
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| 35. |
In Argand diagram all the complex numbers z satisfying `|z-4i|+ |z +4i|= 10` lie on a (A) straight line (B) circle (C) ellipse (D) parabola |
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Answer» here, `PS_1 + PS_2 = constant` P is an ellipse `|PS_1| - |PS_2| = `constant P is on hyperbola `|PS_1 - PS_2| = `C So option C is correct |
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| 36. |
The region of argand diagram defined by `|z-1|+|z+1| |
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Answer» `|z-1|+|z+1| le 4` shows that distance of `z` from a point `(-1,1)` is constant that is less than or equal to `4`. So, this is the case of ellipse as in case of ellipse distance of a point from two focii is always constant. As constant value is `4` or less than `4`, it will include the region interior of the ellipse also. So, option `(3)` is the correct option. |
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| 37. |
If A, G & H are respectively the A.M., G.M. & H.M. of three positive numbers a, b, & c, then equation whose roots are a, b, & c is given by |
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Answer» `A = (a+b+c)/3` `a+b+c = 3A `eqn(1) `G= (abc)^(1/3)` `abc = G^3` `H= 3/((1/a) + (1/b) + (1/c))` `H= 3/((bc+ab+ca)/(abc)) = (3abc)/(b+bc+ ca)` `ab+bc+ca = (3abc)/H = (3G^3)/H` eqn(3) `x^3 - (a+b+c)x^2 + (ab+bc+ca)x - (abc) = 0` `x^3 - 3Ax^2 +(3G^3/H)x - G^3 = 0` `x^3 - 3Ax^2 +(3G^3/H)x - G^3 = 0` answer |
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| 38. |
Solve: `3x^(2) + 5 = 7x` |
| Answer» Correct Answer - `{(7)/(6)+(sqrt(11))/(6)i,(7)/(6)-(sqrt(11))/(6)i}` | |
| 39. |
If `|z-4/z|=2`, then the maximum value of`|Z|`is equal to(1) `sqrt(3)+""1`(2) `sqrt(5)+""1`(3) 2(4) `2""+sqrt(2)` |
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Answer» `|z| = |z-4/z + 4/z| ` `<= |z-4/z| + |4/z|` `|z| <= 2 + 4/|z| ` `=> |z| - 2 - 4/|z| <= 0` `(|z|^2 - 2|z| -4)/|z| <= 0 ` roots are `|z = +- 2 + - (sqrt(4 +16))/2` `= 1 +- sqrt5 ` `0 <= |z| <= 1 + sqrt5 ` max `|z| = 1 + sqrt(5 )` option2 is correct |
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| 40. |
Let z1, z2, z3 be three vertices of an equilateral triangle circumscribing the circle `|z| = 1/2`. If `z1 = 1/2+ ((sqrt3)iota)/2` and z1, z2, z3 are in the anticlockwise sense then z2 is |
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Answer» `abs(z)=1/2`,it represents circle withcircle center `(0.0)`r=1/2 by drawing figure we get,since, z_2 is anticlockwise z_1=-1/2 `z_1=1/2+sqrt(3)i/2` andArgument (z)=`sqrt(3)` `tan theta=sqrt(3)``theta=60^@``Abs(z)=sqrt((1/2)^2+(sqrt(3)/2)^2)=1`similarly`abs(z_1)=abs(z_2)=abs(z_3)` So,`abs(z)`=1and `z` lies on the real axis of the positive direction`:.` Ans=D |
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| 41. |
`(32768)^(1/3)` |
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Answer» Here, `N = 32768` `=>N = 4096*8` `=>N = 512*8*8` `=>N = 64*8*8*8` `=>N = 4*4*4*8*8*8` `=>N = 4^3*8^3` `=>N^(1/3) = 4*8 = 32` `:. (32768)^(1/3) = 32.` |
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| 42. |
If z1 = -3 + 5`iota` and z2 = 5 - 3`iota` and z is a complex number lying on the line segment joining z1 & z2 then arg z can be |
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Answer» given that `z_1= -3+5i` `z_2= 5- 3i` `PQ = z_2 - z_1` `= (5-3i)- (-3+5i) ` `= 8-8i` `arg (PQ) = theta = tan^-1((img(PQ))/(real (PQ)))` `= tan^-1(-8/8)= tan^-1(-1)` `= -tan^-1 1 = -pi/4` `theta = pi/4` since z lie on PQ so, arg(z) is `pi/4` |
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| 43. |
The point represented by the complex number `(2 - i)` is rotated about origin through an angle` (pi)/(2)` in the clockwise direction, the new position of point is (A) 1+2i (B) -1-2i (C) 2+i (D) -1+2iA. `1 +2i`B. `- 1- 2i`C. `2 + i`D. `- 1 + 2i` |
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Answer» Correct Answer - B Givne that, `z = 2 - i` It is rotated about origin through an angle `(pi)/(2)` in the clockwise direction `:. "Now Position" = ze^(-ipi//2) = (2 - i )e^(-ipi//2)` `=(2 -i)["cos"((-pi)/(2))+ isin((-pi)/(2))] = (2 -i) [0 -i]` `= - 2i - 1 = -1 - 2i` . |
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| 44. |
Area of triangle formed by the vertices `z, omega z and z+omega z` is `4/sqrt(3)`, `omega` is complex cube roots of unity then `|z|` is (A) `1` (B) `4/3` (C) `3/4` (D) `4/(sqrt3)` |
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Answer» `1 + omega + omega^2 = 0` `z, z omega , z + z omega` `z , z omega , z(1 + omega)` `z , z omega , - z omega^2 ` area `/_ ABC= 1/2 xx b xxh ` `1/2 xx AB xx OD` `1/2 xx 1 xx sqrt3/2 = sqrt3/4` so `1, omega , -omega^2` `z, z omega , - z omega^2 ` area = `|z|^2 xx sqrt3/4 = 4/sqrt3` `|z|^2 = 16/3` `|z| = 4/ sqrt3` Answer |
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| 45. |
Let `alpha,beta`be real and z be a complex number. If `z^2+alphaz""+beta=""0`has two distinct roots on theline Re `z""=""1`, then it is necessary that :(1) `b"" in (0,""1)`(2) `b"" in (-1,""0)`(3) `|b|""=""1`(4) `b"" in (1,oo)` |
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Answer» `z^2 + alpha z + beta = 0` `z_1 + z_2 = - alpha` `z_1 z_2 = beta` `x= 1` `z_1 = 1 + i y` `z_2 = 1 - i y` now, `1^2 + y^2= beta` `beta in (1,oo)` option 4 is correct |
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| 46. |
If `a + ib = c + id`, thenA. `a^(2) + c^(2) = 0`B. `b^(2)+ c^(2) = 0`C. `b^(2) + d^(2) = 0`D. `a^(2) + b^(2) = c^(2) + d^(2)` |
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Answer» Correct Answer - D Given that, `a + ib = c + id` `rArr |a + ib| = | c + id|` `rArr sqrt(a^(2) + b^(2)) = sqrt(c^(2) + d^(2)) ` On squaring both sides, we get `a^(2) + b^(2)= c^(2) + d^(2)` |
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| 47. |
Let `x, y in R`. Then `x + iota y` is a non real complex number isA. x = 0B. y = 0C. `x ne 0`D. `y ne 0 ` |
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Answer» Correct Answer - D Given that , x, y, `in`R Then , x + iy is non- real complex number if and only if `y ne 0`. |
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| 48. |
If ` 2 cos alpha = x + 1/x , 2 cos beta = y + 1/ y , 2 cos gamma = z + 1/z` then `xyz + 1/(xyz)` = |
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Answer» ` 2 cos alpha = x + 1/x ` `2 x cos alpha = x^2 + 1` `x^2 - 2x cos alpha - 1= 0` `x = -((-2cos alpha) +- sqrt(4 cos^2 alpha- 4))/2` `= cos alpha +- i sin alpha` `= e^(i alpha) or e^(-i alpha)` `x = e^(i alpha) or e^(-i alpha)` `y = e^(i beta) or e^(- i beta)` `z = e^( i gamma) or e^(- i gamma)` `xyx = e^(i(+- alpha +- beta +- gamma))` `1/(xyz) = e^(-i(+- alpha +- beta +- gamma))` `xyz + 1/(xyz) = e^(i phi) + e^(- i phi)= 2 Re ( e^(i phi))` `= 2 cos phi ` `= 2 cos ( +- alpha +- beta +- gamma)` option 2 is correct |
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| 49. |
For real numbers x and y, if `x^2+xy-y^2+2x -y+1=0`, then (a) y can not be between 0 and `8/5` (b) y can not be between `-8/5 and 8/5` (c) y cannnot be between `-8/5 and 0`(d) y cannot be between `-16/5 and 0` |
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Answer» `x^2+x(y+2)-y^2-y+1=0` `D>=0` `(y+2)^2-4(-y^2-y+1)>=0` `y^2+4+4y+4y^2+4y-4>=0` `5y^2+8y>=0` `y(5y+8)>=0` `(-8/5,0)` Option C is correct. |
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| 50. |
If z is any complex number satisfying `|z-3-2i|lt=2` then the maximum value of `|2z-6+5i|` is |
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Answer» `Abs(z-3-2i=5/2i)<=2` `2*Abs[z-3+5/2i]=2*Abs[z-3-2i+2i+5/2i]` `=2Abs[z-3-2*i=(9/4)i` we know, `abs[z_1]+abs[z_2]>=abs[z_1+z_2]` `Abs[2-3-2i+(9/4)i]=Abs(abs(z-3-2i)-abs((9/4)i]``=Abs[2-9/4]``=5/2` `2Abs[z-3-2*i=(9/4)i=2*(5/2)` Ans=`5` |
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