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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.
| 51. |
If `omega`is a complex nth root of unity, then `sum_(r=1)^n(ar+b)omega^(r-1)`is equal toA..`(n(n+1)a)/2`B. `(n b)/(1+n)`C. `(n a)/(omega-1)`D. none of these |
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Answer» `E= sum_(r=1)^n(a+b) omega^(r-1)` `= (a+b) +((2a+b) omega + (3a+b)omega^2 + (4a+b) omega^3 + .... + (na+b)omega^(n-1)` `=b (1+ omega + omega^2 + .....+ omega^(n-1)) + a(1+ 2omega + 3omega^2 + 4 omega^3 + .... + n omega^(n-1)` S`= 1 + 2omega + 3omega^2 + 4omega^3 + ..... + nomega^(n-2)` S`omega= omega + 2omega^2 + 3omega^3 + .... + nomega^n` subtracting these equations we get `S(1-omega)= 1+ omega + omega^2 + omega^3 + .... + omega^(n-1) - n omega^n` `= 0- n omega^n = -n` `S= -n/(1-omega)` E`= 0 + aS= (-an)/(1- omega)` `E= (-na)/(1- omega) = (na)/(omega-1)` option C is corect |
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| 52. |
For a positive integer `n`, find the value of (`1-i)^n(1-1/i)^ndot` |
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Answer» Given expression = `(1-i)^(n)(1-(1)/(i))^(n)` `=(1 -i)^(n)(i-1)^(n).i^(-n)= (1-i)^(n)(-1)^(n). i^(-n)` `=[(1 -i)^(2)]^(n)(-1)^(n).i^(-n)= (1-i^(2)-2i)^(n)(-n)^(n) i^(-n)" "[:.i^(2)=-1]` `=(1-1-2i)^(n)i^(-n)=(-2)^(n). i^(n)(-1)^(n)i^(-n) ` `= (-1)^(2n).2^(n) = 2^(n)` |
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| 53. |
Solve the equation:`x^2-x+2=0` |
| Answer» Correct Answer - `{(1)/(2)+(sqrt(7))/(2)i,(1)/(2)-(sqrt(7))/(2)i}` | |
| 54. |
Express of the complex number in the form `a + i b`.`(1/3+3i)^3` |
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Answer» Here, `z = (1/3+3i)^3` `z = (1/3)^3 + (3i)^3+3(1/3)(3i)(1/3+3i)` `=1/27+27i^3+i+9i^2` `=1/27-27i+i-9` `z=-242/7-26i` Here, `a = -242/7 and b = -26` |
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| 55. |
Solve the following quadratic: `x^2+2x+2=0` |
| Answer» Correct Answer - `{-1 + i, -1 -i}` | |
| 56. |
Express of the complex number in the form `a + i b`.`(1-i)^4` |
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Answer» `(1-i)^4` `((1-i)^2)^2` `= (1^2 + i^2 - 2i)^2` `= (cancel(1) cancel(- 1) - 2i)^2` = `(-2i)^2` `=4i^2` `= 4(-1)` `=-4` so, a= -4 & b=0 |
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| 57. |
Solve: `sqrt(5)x^(2) + x + sqrt(5) = 0` |
| Answer» Correct Answer - `{(-1)/(2sqrt(5))+(sqrt(19))/(2sqrt(5))i,(-1)/(2sqrt(5))-(sqrt(19))/(2sqrt(5))i}` | |
| 58. |
If `a + i b =((x+i)^2)/(2x^2+1),`prove that `a^2+b^2=((x^2+1)^2)/((2x^2+1)^2)` |
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Answer» We have `(a+ib)=((x+i)^(2))/((2x^(2)+1))=((x^(2)+i^(2)+2ix))/((2x^(2)+1))=((x^(2)-1)+i(2x))/((2x^(2)+1))` `rArr" "(a+ib)=((x^(2)-1))/((2x^(2)+1))+i.(2x)/((2x^(2)+1))` `rArr" "|a+ib|^(2)=|((x^(2)-1))/((2x^(2)+1))+i.(2x)/((2x^(2)+1))|^(2)` `rArr" "(a^(2)+b^(2))={((x^(2)-1)^(2))/((2x^(2)+1)^(2))+(4x^(2))/((2x^(2)+1)^(2))}` `" "=((x^(2)-1)^(2)+4x^(2))/((2x^(2)+1)^(2))=((x^(2)+1)^(2))/((2x^(2)+1)^(2))`. Hence, `(a^(2)+b^(2))=((x^(2)+1)^(2))/((2x^(2)+1)^(2))` |
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| 59. |
Number of solutions of the equation `z^3+[3(barz)^2]/|z|=0` where z is a complex number is |
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Answer» `Z^3+(3(overlineZ)^2)/|Z|=0` `r^3e^(i3theta)+(3r^2e^(i(-2theta)))/r=0` `r^3e^(i3theta)+3re^(i(-2theta)=0` `r/e^(i2theta)(r^2e^(i5theta)+3)=0` `r!=0` `|r^2e^(i5theta)|=|-3|` `r^2=3` `r=sqrt3` `e^(i5theta)=-1` `cos5theta+isintheta=-1` `cos5theta=-1` `sin5theta=0` `theta in(-pi,pi]` `-5pi<5theta<=5pi` `sin5theta=0` the correct answer is 5. |
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| 60. |
Evaluate: `sqrt(-4-3i)` |
| Answer» Correct Answer - `(1)/(sqrt(2))(1-3i)or(1)/(sqrt(2))(-1+3i)` | |
| 61. |
where does z lie , if `|(z-5i)/(z+5i)|=1?` |
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Answer» Let z = (x + iy). Then, `|(z-5i)/(z+5i)|=1 rArr (|z-5i|)/(|z+5i|)=1" "[because |(z_(1))/(z_(2))|=(|z_(1)|)/(|z_(2)|)]` `rArr" "|z-5i|=|z+5i| rArr |z-5i|^(2)=|z+5i|^(2)` `rArr" "|(x+iy)-5i|^(2)=|(x+iy)+5i|^(2)" "[because z = (x+iy)]` `rArr" "|x+i(y-5)|^(2)=|x+i(y+5)|^(2)` `rArr" "x^(2)+(y-5)^(2)=x^(2)+(y+5)^(2)" "[because |x+iy|^(2)=(x^(2)+y^(2))]` `rArr" "(y+5)^(2)-(y-5)^(2)= 0 rArr 4 xx y xx 5 = 0 rArr y = 0`. `therefore" "z=x + i0 rArr z = x`, where x is real. Hence, z is a real number. |
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| 62. |
In `Delta ABC , A(z_1) B(z_2) C(z_3)` is inscribed in a circle |z| = 5. If `H(z_H)` be the orthocentre of triangle ABC `z_H` |
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Answer» Cirumcentre=O From number line `OC:CH=1:2` `C=(Z_1+Z_2+Z_3)/3` `OH=3x=3OC=Z_1+Z_2+Z_3` `Z_H=Z_1+Z_2+Z_3`. |
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| 63. |
Express `sin""pi/5+i(1-cos"" pi/5)`in polar form. |
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Answer» Let `z = "sin"(pi)/(5)+ i(1-"cos"(pi)/(5))`. Let its polar form be `z = r(cos theta + i sin theta)`. Now, `r^(2)=|z|^(2)="sin"^(2)(pi)/(5)+(1-"cos"(pi)/(5))^(2)=("sin"^(2)(pi)/(5)+"cos"^(2)(pi)/(5))+1 -2 "cos"(pi)/(5)` `rArr" "r^(2)=2(1-"cos"(pi)/(5))=4"sin"^(2)(pi)/(10) rArr r=2"sin(pi)/(10)`. Let `alpha` be the acute angle, given by `tan alpha=|(Im(z))/(Re(z))|=|(1-"cos"(pi)/(5))/("sin"(pi)/(5))|=(2"sin"^(2)(pi)/(10))/(2"sin"(pi)/(10)."cos"(pi)/(10))= "tan"(pi)/(10)rArr alpha = (pi)/(10)`. Clearly, the point representing z lies in the first quadrant as `x gt 0 and y gt 0`. `therefore" "arg(z)= theta = alpha = (pi)/(10)`. Thus, `r=2"sin"(pi)/(10)and theta=(pi)/(10)`. Hence, the required polar form is `2"sin"(pi)/(10)("cos"(pi)/(10)+"i sin"(pi)/(10))`. |
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| 64. |
If `alpha, beta, gamma` are the roots of the equation `x^3+ qx +r =0`then find the equation whose roots are (a) `alpha+beta, beta+gamma, gamma+alpha` (b) `alpha beta, beta gamma, gamma alpha` |
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Answer» a)`alpha+beta+gamma=0` `alphabeta+betagamma+gammaalpha=q` `alphabetagamma=-r` `2(alpha+beta+gamma)=0=b` `x^3-bx^2+cx-d=0` `C=alphabeta+alphagamma+beta^2+betagamma+betaalpha+gamma^2+gammaalpha+alphabeta+alpha^2+alphabeta` `c=(alpha+beta+gamma)^2+(alphabeta+gammabeta+alphagamma)` `c=q` `d=(alpha+beta)(beta+gamma)(gamma+alpha)` `d=alphabeta+alphagamma+beta^2+betagamma)(alpha+gamma)` `d=alphabetagamma+alphagamma^2+beta^2gamma+betagamma^2+alpha^2beta+alpha^2gamma+beta^2gamma+alphabetagamma` `d=r` `x^3+2x+r=0` `b)b=alphabeta+betagamma+gammaalpha=q` `c=alphabeta^2gamma+betagammaa^2alpha+alpha^2betagamma` `=alphabetagamma(alpha+beta+gamma)=0` `d=alpha^2beta^2gamma^2=r^2` `x^3-qx^2+0x-r^2=0` `x^3-qx^2-r^2=0`. |
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| 65. |
Given that `alpha` and `beta` are the roots of the equation `x^2=x+7` Prove that `(a) 1/alpha=(alpha-1)/7` `(b) alpha^3=8 alpha+7` Find the numerical value of `alpha/beta+beta/alpha` |
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Answer» Given equation is, `x^2 = x+7` `=>x^2-x-7 = 0` (a) As `alpha ` is root of this equation, `x = alpha` should satisfy this equation. `alpha^2-alpha-7 = 0` `=>alpha(alpha-1) = 7` `=>1/alpha = (alpha-1)/7` (b) `alpha^2-alpha-7 = 0` `=>alpha^2 = alpha+7` `=>alpha^3 = alpha^2+7alpha ` `=>alpha^3 = (alpha+7)+7alpha` `=>alpha^3 = 8alpha+7` As, `alpha and beta` are the roots of this equation, `:. alpha+beta = -b/a = 1` `alphabeta = c/a = -7` Now, `alpha/beta+beta/alpha = (alpha^2+beta^2)/(alphabeta)` `=((alpha+beta)^2-2alphabeta)/(alphabeta)` `=(1^2-2(-7))/(-7) = -15/7` `:. alpha/beta+beta/alpha = -15/7` |
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| 66. |
`x=2^(1/3)-2^(-1/3)` then find `2x^3+6x` |
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Answer» `x=2^(1/3)-2^(-1/3)` `2x^3+6x=2(2^(1/3)-2^(-1/3))^3+6(2^(1/3)-2^(-1/3))` `=2(2-2^(-1)-3(2^(1/3)-2^(-1/3)))+6(2^(1/3)-2^(-1/3))` `=4-1-6(2^(1/3)-2^(-1/3))+6(2^(1/3)-2^(-1/3))` `=3`. |
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| 67. |
Convert `4(cos 300^(@) + i sin 300^(@))` into Cartesian form. |
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Answer» `4(cos 300^(@)+isin 300^(@))=4[cos(360^(@)-60^(@))+i sin(36^(@)-60^(@))]` `" "=4[cos(-60^(@))+i sin(-60^(@))]` `" "=4[cos60^(@)-i sin60^(@)]4((1)/(2)-i(sqrt(3))/(2))` `" "=(2-2i sqrt(3))` |
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| 68. |
`1+i^2+i^4+i^6+i^8++i^(20)` |
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Answer» This is a GP in which a = 1, `r = i^(2) = -1` and n = 11. `therefore" "S=(a(1-r^(n)))/((1-r))=(1xx{1-(-1)^(11)})/({1-(-1)})=(2)/(2)=1`. |
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| 69. |
If `z_1,z_2.........z_n=z`, then `argz_1+arg z_2+.......+arg z_n and arg z` differ by a |
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Answer» (i) Let `z=r(cos theta + i sin theta)`. Then, `|z| = r and arg(z) = theta`. Now, `z = r(cos theta +i sin theta)` `rArr" "z=r cos theta + i(r sin theta)` `rArr" "bar(z)=r cos theta-i(r sin theta)=r(cos theta -i sin theta)` `" "=r{cos(-theta)+ i sin(-theta)}` `rArr" "|bar(z)|=r and arg(bar(z))=-theta =-arg(z)`. Hence, `arg(bar(z))=-arg(z)`. (ii) Let `z_(1)=r_(1)(cos theta_(1)+ i sin theta_(1))and z_(2)=r_(2)(cos theta_(2)+ i sin theta_(2))`. Then, `|z_(1)|=r_(1), arg(z_(1))=theta and |z_(2)|=r_(2), arg(z_(2))=theta_(2)`. `therefore" "z_(1)z_(2)=r_(1)(cos theta_(1)+ i sin theta_(1)).r_(2)(cos theta_(2)+ i sin theta_(2))` `=r_(1)r_(2){(cos theta_(1)cos theta_(2)-sin theta_(1)sin theta_(2))+i(sin theta_(1)cos theta_(2)+cos theta_(1)sin theta_(2))}` `=r_(1)r_(2){cos(theta_(1)+theta_(2))+ i sin (theta_(1) + theta_(2))}` `rArr" "arg(z_(1)z_(2))=(theta_(1)theta_(2))=arg(z_(1))+arg(z_(2))`. REMARKS (I) Note here that `|z_(1)z_(2)|=r_(1)r_(2)=|z_(1)||z_(2)|`. (II) In general, we have `|z_(1)z_(2)....z_(n)|=|z_(1)|.|z_(2)|....|z_(n)| and arg(z_(1)z_(2)....z_(n))=arg(z_(1)) + arg(z_(2))+....+arg(z_(n))`. (iii) Let `z_(1)=r_(1)(cos theta_(1)+i sin theta_(1))and z_(2)=r_(2)(cos theta_(2)+ i sin theta_(2))`. Then, `bar(z)_(2)=bar(r_(2)cos theta_(2)+i(r_(2)sin theta_(2)))=r_(2)cos theta_(2)-i(r_(2)sin theta_(2))` `rArr" "bar(z)_(2)=r_(2){cos(-theta_(2))+ i sin (-theta_(2))}`. `therefore" "z_(1)bar(z)_(2)=r_(1)(cos theta_(1)+ i sin theta_(1)).r_(2){cos(-theta_(2))+ i sin (-theta_(2))}` `=r_(1)r_(2)(cos theta_(1)+ i sin theta_(1)){cos(-theta_(2))+ i sin(-theta_(2))}` `=r_(1)r_(2)[cos{theta_(1)+(-theta_(2))}+ i sin {theta_(1)+(-theta_(2)}]` `=r_(1)r_(2){cos(theta_(1)-theta_(2))+i sin(theta_(1)-theta_(2))}` Hence, `arg(z_(1)bar(z)_(2))=(theta_(1)-theta_(2))=arg(z_(1))-arg(z_(2))`. Let `z_(1)=r_(1)(cos theta_(1)+i sin theta_(1))and z_(2)=r_(2)(cos theta_(2)+i sin theta_(2))`. Then, `|z_(1)|=r_(1),|z_(2)|=r_(2), arg(z_(1))=theta_(1)and arg(z_(2))=theta_(2)`. `therefore" "(z_(1))/(z_(2))=(r_(1)(cos theta_(1)+i sin theta_(1)))/(r_(2)(cos theta_(2)+i sin theta_(2)))xx((cos theta_(2)-i sin theta_(2)))/((cos theta_(2)-i sin theta_(2)))` `=(r_(1))/(r_(2)).{((cos theta_(1).cos theta_(2)-sin theta_(1).sin theta_(2))+i(sin theta_(1).cos theta_(2)-cos theta_(1).sin theta_(2)))/((cos^(2) theta_(2)+sin^(2)theta_(2)))}` `=(r_(1))/(r_(2)).{cos(theta_(1)-theta_(2))+i sin(theta_(1)-theta_(2))}` `rArr" "arg((z_(1))/(z_(2)))=(theta_(1)-theta_(2))=arg(z_(1))-arg(z_(2))`. Hence, `arg((z_(1))/(z_(2)))=arg(z_(1))-arg(z_(2))`. |
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| 70. |
If `f(z)=(7-z)/(1-z^2)`, where `z=1+2i ,`then `|f(z)|`is`(|z|)/2`(b) `|z|`(c) `2|z|`(d) none of theseA. `(|z|)/(2)`B. `|z|`C. `2|z|` None of theseD. |
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Answer» Correct Answer - A Let z = 1 + 2i `rArr |z|= sqrt( 1+ 4 ) = sqrt(5)` Now. `f(z) = (7-z)/(1-z^(2))=(7-1-2i)/(1-(1+2i)^(2))` `(6-2i)/(1-1-4i^(2)-4i)=(6-2i)/(4-4i)` `((3-i)(2+2i))/((2-2i)(2+2i))` `(6-2i+6i-2i^(2))/(4-4i^(2))=(6+4i+2)/(4+4)` `(8+4i)/(8)=1+(1)/(2)i` `f(z) = 1 + (1)/(2)i` ` :. |f(z)|=sqrt( 1+(1)/(4))=sqrt((4+1)/4)=sqrt(5)/(2) = (|z|)/(2)` |
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| 71. |
If `alpha and beta` are the roots of `x^2-2 x +4=0` then the value of `alpha^6+beta^6` is |
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Answer» `alpha*beta=4` `alpha+beta=2` `alpha^2+beta^2=(alpha+beta)^2-2alphabeta` `=4-8=-4` `(a+b)^3=a^3+b^3+3ab(a+b)` `(alpha^2+beta^2)^3=alpha^6+beta^6+3alpha^2beta^2(alpha^2+beta^2)` `-64=alpha^6+beta^6+3*16(-4)` `alpha^6+beta^6=192-64=128` option3 is correct. |
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| 72. |
If `(a+i b)/(c+i d)=x+i y ,`prove that `(a-i b)/(c-i d)=x-i ya n d(a^2+b^2)/(c^2+d^2)=x^2+y^2dot` |
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Answer» `(i) ((a+ib)/(c+id))=(x+iy) rArr (bar((a+ib)/(c+id)))=bar((x+iy))` `" " rArr bar((a+ib))/(bar(c+id))=(x+iy)rArr ((a-ib))/((c-id))=(x-iy)`. (ii) We have `((a+ib))/((c+id))=(x+iy) and ((a-ib))/((c-id))=(x-iy)` `rArr" "((a+ib))/((c+id))xx((a-ib))/((c-id))=(x+iy)(x-iy)` `rArr" "((a+ib)(a-ib))/((c+id)(c-id))=(x+iy)(x-iy)` `rArr" "((a^(2)+b^(2)))/((c^(2)+d^(2)))=(x^(2)+y^(2))`. |
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| 73. |
If `|z_1|=|z_2|=dot=|z_n|=1,`prove that `|z_1+z_2+z_3+...+z_n|=1/(z_1)+1/(z_2)+1/(z_3)+...+1/(z_n).` |
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Answer» We have `|z_(1)|=|z_(2)|=|z_(3)|=...=|z_(n)|=1` `rArr" "|z_(1)|^(2)=|z_(2)|^(2)=|z_(3)|^(2)=...=|z_(n)|^(2)=1` `rArr" "z_(1)bar(z)_(1)=1, z_(2)bar(z)_(2)=1, z_(3)bar(z)_(3)=1,...,z_(n)bar(z)_(n)=1` `rArr" "(1)/(z_(1))=bar(z)_(1), (1)/(z_(2))=bar(z)_(2), (1)/(z_(3))=bar(z)_(3),...,(1)/(z_(n))=bar(z)_(n)` `rArr" "|(1)/(z_(1))+(1)/(z_(2))+(1)/(z_(3))+...+(1)/(z_(n))|=|bar(z)_(1)+bar(z)_(2)+bar(z)_(3)+...+bar(z)_(n)|` `" "=|bar(z_(1)+z_(2)+z_(3)+...+z_(n))|` `" "=|z_(1)+z_(2)+z_(3)+...+z_(n)|" "because |bar(z)|=|z|]` `therefore" "|(1)/(z_(1))+(1)/(z_(2))+(1)/(z_(3))+...+(1)/(z_(n))|=|z_(1)+z_(2)+z_(3)+...+z_(n)|`. |
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| 74. |
A real value of x satisfies the equation `(3-4ix)/(3+4ix)=alpha-ibeta(alpha,beta in R)`, if `alpha^2+beta^2=`A. ` x = 2n +1`B. `x = 4n`C. `x = 2n`D. `x = 4n + 1` |
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Answer» Correct Answer - A Given equation, ` ((3 - 4 ix)/(3 + 4 ix )) = alpha - i beta ( alpha , beta in R ) ` `rArr [(3 - 4 ix)/(3 + 4 ix )] = alpha - i beta` Now, ` (alpha -ibeta ) =((3 -4ix)(3 - 4ix))/((3 + 4ix)(3 - 4 ix) )=(9+ 26i^(2) x^(2) - 24ix)/(9 - 16i^(2)x^(2))` `rArr (alpha -ibeta ) =(9+ 26i^(2) x^(2) - 24ix)/(9 - 16i^(2)x^(2))` `rArr alpha - ibeta = (9- 16x^(2))/(9 + 16x^(2) - i24x) /(9 + 16x^(2)` `rArr alpha - ibeta = (9 - 16x^(2))/(9 + 16x^(2)) - (i24x)/(9 + 16x^(2)) ...(i)` `:." " alpha + ibeta = (9 - 16x^(2))/(9 + 16x^(2)) - (i24x)/(9 + 16x^(2)) ...(ii)` So, `(alpha - ibeta)(alpha + i beta)= ((9 - 16x^(2))/(9 + 16x^(2))^(2)) - ((i24x)/(9 + 16x^(2)))^(2)` `alpha^(2) + beta^(2) = (81+ 256x^(4) - 288x^(2) + 576x^(2))/((9 + 16x^(2))^(2))` `(( 9 + 16x^(2))^(2))/(9 + 16x^(2) )^(2) = 1` |
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| 75. |
If `alpha` is non real and `alpha="^5sqrt(1)` then value of `2^|1+alpha+alpha^2+alpha^-2-alpha^-1|` is equal to |
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Answer» `|1+alpha+alpha^2+1/alpha^2-1/alpha|` `|(alpha^2+alpha^3+alpha^4+1+alpha-2alpha)/alpha^2|` `|-2alpha/alpha^2|` `=2^|-2/alpha|` `=2^(2/|alpha|` `2^2` `4` |
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| 76. |
Evaluate `sqrt(6 + 8i)`. |
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Answer» Let `sqrt(6+8i)=(x+iy)." "...(i)` On squaring both sides of (i), we get `6+8i=(x+iy)^(2) rArr 6+8i=(x^(2)-y^(2))+i(2xy)." "...(ii)` On comparing real parts and imaginary parts on both sides of (ii), we get `x^(2)-y^(2)=6 and 2xy = 8` `rArr" "x^(2)-y^(2)=6 and xy = 4` `rArr" "(x^(2)+y^(2))=sqrt((x^(2)-y^(2))^(2)+4x^(2)y^(2))=sqrt(6^(2)+4 xx 16)=sqrt(100)=10` `rArr" "x^(2)-y^(2)=6 and x^(2)+y^(2)=10` `rArr" "2x^(2)=16 and 2y^(2)=4` `x^(2)=8 and y^(2)=2` `rArr" "x = +- 2 sqrt(2) and y = +- sqrt(2)`. Since `xy gt 0`, so x and y are of the same sign. `therefore" "(x=2 sqrt(2) and y = sqrt(2)) or (x = -2 sqrt(2) and y = -sqrt(2))`. Hence, `sqrt(6+8i) = (2 sqrt(2) + sqrt(2)i) or (-2 sqrt(2)-sqrt(2)i)`. |
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| 77. |
If `(x+i y)^(1//3)=a+i b ,x ,y ,a b in R.`Show that(i) `x/a+y/b=4(a^2-b^2)`(ii) `x/a-y/b=2(a^2+b^2)` |
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Answer» We have `(x+iy)^(1//3)=(a+ib)` `rArr" "(x+iy)=(a+ib)^(3)" "["on cubing both sides"]` `rArr" "(x+iy)=a^(3)+i^(3)b^(3)+3iab(a+ib)` `=a^(3)-ib^(3)+3a^(2)bi-3ab^(2)=(a^(3)-3ab^(2))+i(3a^(2)b-b^(3))` `rArr" "x = a^(3) - 3ab^(2) and y = 3a^(2)b-b^(3)" "["on equating real and imaginary parts separetely"]` `rArr" "(x)/(a)=(a^(2)-3b^(2))and(y)/(b)=(3a^(2)-b^(2))` `rArr" "((x)/(a)+(y)/(b))=4(a^(2)-b^(2))and ((x)/(a)-(y)/(b))= -2(a^(2)+b^(2))`. Hence, (i) `(x)/(a)+(y)/(b)=4(a^(2)-b^(2)) and (ii) (x)/(a)-(y)/(b) = -2(a^(2)+b^(2))`. |
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| 78. |
Express the following in the form of `a + b i`:(i) `(-5i)(1/8i)` (ii) `(-i)(2i)(-1/8i)^3` |
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Answer» (i) `(-5i)(1/8i)` `= -5/8i^2` `=-5/8(-1)` `=5/8 + 0i` answer (ii) `(-i)(2i)(-1/8i)^3` `(-2i^2)(-i^3/512)` `(-2(-1)) (-i(-1))/512` `= 0 + i*1/256` |
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| 79. |
Show that if `iz^3+z^2-z+i=0`, then `|z|=1` |
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Answer» We have `iz^(3) + z^(2) - z + i = 0` `rArr" "z^(3)-iz^(2)+iz + 1 = 0" "["on dividing both sides by I"]` `rArr" "z^(2)(z-i) + i(z-i) = 0` `rArr" "(z-i)(z^(2)+i)=0` `rArr" "z = i or z^(2) = -i`. Now, `z = i rArr |z| = |i| rArr |z| = 1` And, `z^(2) =-i rArr |z^(2)|=|-i|=1 rArr |z| = 1`. |
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| 80. |
Convert the complex number `(1+isqrt(3))` into polar form. |
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Answer» The given complex number is `z = (1 + i sqrt(3))`. Let its polar form be `z = r(cos theta + i sin theta)`. Now, `r = |z| = sqrt(1^(2)(sqrt(3))^(2)) = sqrt(4) = 2`. Let `alpha` be the acute angle, given by `tan alpha = |(Im(z))/(Re(z))|=|(sqrt(3))/(1)|=sqrt(3) rArr alpha = (pi)/(3)`. Clearly, the point representing `z = (1 + i sqrt(3))` is P`(1, sqrt(3))`, which lies in the firest quadrant. `therefore" "arg(z) = theta = alpha = (pi)/(3)`. Thus, `r = |z| = 2 and theta arg(z) = (pi)/(3)`. Hence, the required polar form of `z = (1 + i sqrt(3))` is `2("cos"(pi)/(3) + "i sin"(pi)/(3))`. |
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| 81. |
Express of the complex number in the form `a + i b`. `i + i` |
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Answer» given that `i + i` =`2i` comparing with the form `0 + 2i` `a=0 & b=2` answer |
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| 82. |
If `|z^2=1|=|z|^2+1`, then show that `z`lies on the imaginary axis. |
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Answer» Let z = (x + iy). Then, `z^(2) = (x^(2)-y^(2)) and |z|^(2) = (x^(2) + y^(2))`. Now, `|z^(2)-1|= |z|^(2) + 1` `rArr" "|(x^(2)-y^(2)-1)+i(2xy)|=(x^(2)+y^(2)+1)` `rArr" "|(x^(2)-y^(2)-1)+i(2xy)|^(2)=(x^(2)+y^(2)+1)^(2)` `rArr" "(x^(2)-y^(2)-1)^(2)+4x^(2)y^(2)=(x^(2)+y^(2)+1)^(2)` `rArr" "[x^(2)+(y^(2)+1)]^(2)-[x^(2)-(y^(2)+1)]^(2)=4x^(2)y^(2)` `rArr" "4x^(2)(y^(2)+1)=4x^(2)y^(2) rArr 4x^(2){(y^(2)+1)-y^(2)}=0` `rArr" "4x^(2)= 0 rArr x = 0` Hence, z = 0 + iy, which shows that z is imaginary. |
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| 83. |
Show that the equation `e^(sinx)-e^(-sinx)-4=0`has no real solution. |
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Answer» `e^(sinx)-e^(-sinx)-4=0` `e^(sinx)-1/(e^(sinx))-4=0` `((e^(sinx))^2-1)/(e^(sinx))=4` `(e^(sinx))^2-4e^(sinx)-1=0` `e^(sinx)=(4pmsqrt20)/2=(4pm2sqrt5)/2` `e^(sinx)=2pmsqrt5=2pm2.73` `=-0.236` `0lte^(sinx)lte`. |
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| 84. |
Convert of the complex number in the polar form: `1-i` |
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Answer» Polar form of a complex number is given by, `r(costheta+isintheta)` So, `1-i = r(costheta+isintheta)` Comparing, real and unreal part, `rcostheta = 1 and rsintheta = -1` Squaring and adding both expression, `r^2cos^2theta + r^2sin^2theta = 1+1` `r^2(cos^2theta+sin^2theta) = 2` `r = sqrt(2)` So, `sintheta = -1/sqrt2 and costheta = 1/sqrt2` So, polar form will be `sqrt2(1/sqrt2-1/sqrt2i)` |
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| 85. |
Solve for `x: (1-i)x + (1+i)y=1 - 3i`. |
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Answer» Correct Answer - x = 2, y = -1 `(1-i)x+(1+i)y=1-3i iff (x+y)+(y-x)i=1-3i` `therefore" "x+y=1 and y-x=-3 rArr x = 2, y = -1`. |
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| 86. |
Express the following in the form `a + i b`(i) `(5+sqrt(2i))/(1-sqrt(2i))` (ii) `i^(-35)` |
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Answer» (i)`(5+sqrt2i)/(1-sqrt2i) = (5+sqrt2i)/(1-sqrt2i)**(1+sqrt2i)/(1+sqrt2i)` `=((5+sqrt2i)(1+sqrt2i))/(1^2+sqrt2^2)` `=(5+5sqrt2i+sqrt2i+2i^2)/3 = (3+6sqrt2i)/3` `=1+2sqrt2i` (ii)`i^-35 = i^(-36+1) = i/i^36 = i/(i^4)^9` We know, `i^4 = 1`,So, `i^-35 = i = 0+1i` |
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| 87. |
Simplify and express each of the following in the form (a + ib) : `(i)" "((5)/(-3+2i)+(2)/(1-i))((4-5i)/(3+2i))" "(ii)" "((1)/(1-4i)-(2)/(1+i))((1-i)/(5+3i))` |
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Answer» Correct Answer - `{:((i),((5)/(13)+(4)/(13)i),"(ii)",((4)/(17)+(5)/(17)i)):}` `(i) ((5)/(-3+2i)+(2)/(1-i))((4-5i)/(3+2i))=(5(1-i)+2(-3+2i))/((-3+2i)(1-i))xx((4-5i))/((3+2i))` `=((-1-i)(4-5i))/((-1+5i)(3+2i))=((1+i)(4-5i))/((1-5i)(3+2i))` `=((9-i))/(13(1-i))xx((1+i))/((1+i))=((10+8i))/(26)=((5)/(13)+(4)/(13)i)`. |
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| 88. |
Simplify : `(i)" "(-2i)((1)/(6)i)" "(ii)" "(-i)(3i)((-1)/(6)i)^(3)" "(iii)" "4 sqrt(-4)+5 sqrt(-9)-3 sqrt(-16)` |
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Answer» We have `(i)" "(-2i)((1)/(6)i)+(-2xx(1)/(6))xxi^(2)=(-1)/(3)xx(-1)=(1)/(3)`. `(ii)" "(-i)(3i)((-1)/(6)i)^(3)=(-3i^(2))((-1)/(216)i^(3))` `=(-3)xx(-1)[(-1)/(216)xx(-i)]" "[because i^(3) = -i]` `=(3xx(1)/(216)xxi)=(1)/(72)i`. `(iii)" "4 sqrt(-4)+5 sqrt(-9)-3 sqrt(-16)` `=(4xx2i)+(5xx3i)-(3xx4i)" "[because sqrt(-4)=2i, sqrt(-9)=3i, sqrt(-16) = 4i]` `=(8i + 15i - 12i)=(11)i`. |
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| 89. |
If `|z+1|=z+2(1+i),`find `zdot` |
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Answer» Given that, `|z +1| = z + 2 ( + i) ... (i)` `z = x + iy` then, `|x + iy + 1| = x + iy + 2 (1 + i)` `rArr |x + 1 + iy| = (x + 2) + i (y + 2)` `rArr" sqrt((x + 1)^(2) + y^(2) = (x + 2) + i (y + 2))` On squaring both sides, we get `(x + 1)^(2) + y^(2) = (x + 2)^(2) +i ^(2)(y + 2)^(2) +2i (x + 2)(y +2)` `rArr x^(2) + 2x + 1 + y^(2) = x^(2) + 4x + 4 - y^(2) - 4y - 4 +2i(x + 2)(y + 2)` `rArr x^(2) + y^(2)+ 2x + 1 = x^(2) - y^(2) + 4x - 4y + 2i(x + 2)(y + 2)` On comparing real and imaginary parts, we get ` x^(2) + y^(2) + 2x + 1 = x^(2) - y^(2) + 4x -4y` `rArr 2y^(2) - 2x + 4y + 1 = 0 ...(ii)` and `2(x + 2) (y + 2) =0` `rArr x + 2 = 0 or y + 2 = 0 ` `x = - 2 or y = - 2 ...(iii)` For `x = - 2, 2y^(2) + 4 + 4y + 1 = 0 [ using Eq. (ii)]` `rArr 2y^(2) + 4y + 5 =` `rArr 16 - 4 xx 2 xx 5 lt0 ` `:. "Discriminant" , D = b^(2) - 4ac lt 0 ` `rArr 2y^(2) + 4y + 5 "has no real roots"`. For y = -2, `2(-)6(2) - 2x + 4 (-2) + 1 = 0 [ using Eq. (ii)]` `rArr 8 - 2x - 8 + 1 = 0` `rArr x = 1//2` `:. " " z = x + iy = (1)/(2) - 2i` |
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| 90. |
e that both the roots Prove that both the roots ofthe equation (x (x-b) t (x-b)(x-cy t (x- c) (x-a 0 are alw ind the interval in which lies if (a2 t a-2) (a |
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Answer» `-2<(x^2+4x+1)/(x^2+x+1)<2` `-1<(x^2+kx+1)/(x^2+x+1)` `0<2+(x^2+kx+1)/(x^2+x+1)` `0<(3x^2+(2+k)x+3)/(x^2+x+1)` `3x^2+(2+k)x+3>0` `(2+k)^2-4*3*2<0` `(k+z)^2<36` `-6ltk+2<6` `-8`(x^2+kx+1)/(x^2+x+1)-2<0` `(-x^2+(k-1)x-1)/(x^2+x+1)<0` `-x^2+(k-2)x-1<0` `x^2-(k-2)x+1>0` `D<0` `(k-2)^2-4<0` `(k-2)^2<4` `-2ltk-2<2` `0ltk<4-(2)` `k in (0,4)`. |
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| 91. |
Perform the indicated operation and find the result in the form `a+i b :(3-sqrt(-16))/(1-sqrt(-9))` |
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Answer» Correct Answer - `((3)/(2)+(1)/(2)i)` Given expression `=(3-4i)/(1-3i)xx(1+3i)/(1+3i)=((3-4i)(1+3i))/((1+9))=(15+5i)/(10)=(1)/(2)(3+i)`. |
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| 92. |
If the real part of `(barz +2)/(barz-1)` is 4, then show that the locus of the point representing z in the complex plane is a circle. |
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Answer» Let z = x +`i`y Now, `(barz + 2)/(barz -1) =(x - `i`y +2)/(x-`i`y-1)` `=([(x+2)-i y][(x -1)+ i y])/([(x -1) - i y ][(x -1)+ i y])` =`((x -1)(x+2)-iy(x -1)+i y(x + 2) + y^(2))/((x -1)^(2) +y^(2))` =`((x -1)(x+2)+y^(2)+i[(x + 2)y-(x - 1) y])/((x -1)^(2) +y^(2)) " "[:. -i^(2) = 1]` Taking real part, `((x-1)(x +2)+ y^(2))/((x - 1)^(2) +Y^(2)) = 4` `rArr x^(2) - x + 2x - 2 + y^(2) = 4 (x^(2) - 2x + 1 + Y^(2))` `rArr 3x^(2) + 3y^(2) - 9x + 6 = 0. which reperesents a circle`. Hence, z lines on the circle. |
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| 93. |
Solve the equation `|z|=z+1+2idot` |
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Answer» The given equation is `|z| = z + 1 + 2i " " ...(i) ` Let `z = x + iy` From Eq. (i). |x + iy| = x + iy + 1 + 2i` `rArr sqrt(x^(2)+Y^(2) = x + iy + 1 + 2i)" " [:. |z| + sqrt (x^(2)+Y^(2))]` `rArr sqrt(x^(2)+Y^(2)) = (x + 1) + i (y + 2)` On squaring both sides, we get `x^(2) + y^(2) = (x+ 1)^(2) + i^(2) (y +2)^(2) + 2i(x + 1)(y +2)` `rArr x^(2) + y^(2) = x^(2) +2x + 1- y ^(2) - 4y -4 + 2i(x + 1)(y +2)` On comparing real and imaginary parts , `x^(2) + y^(2) = x^(2) +2x + 1- y ^(2) - 4y -4` i. e., `2y^(2) = 2x - 4y - 3` ....(ii) and `2(x + 1)(y + 2) 0` `(x + 1) = 0 or (y + 2) = 0` `rArr x = - 1 or y = - 2` For ` x = - 1` we get, `2y^(2) = -2- 4 y 3` `2y^(2) = + 4 + 5 = 0 [using Eq. (ii)]` `rArr y = (- 4pmsqrt(16 - 2 xx 4 xx 5))/(4)` `rArr y = (- 4pm(sqrt(-24)))/(4)notinR` Now, for y = - 2, Then, `2(-2)^(2) = 2x - 4 (-2) -3 [using Eq. (ii)]` `8 = 2x + 8 - 3` `2x = 3 rArr x = 3//2` `z = x + iy = 3/2 - 2i` |
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| 94. |
Express the complex number `(-sqrt(3)-i)` in polar form. |
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Answer» The given complex number is `z = (-sqrt(3)-i)`. Let its polar form be `z=r(cos theta + i sin theta)`. Now, `r=|z|=sqrt((-sqrt(3))^(2)+(-1)^(2))=sqrt(4)=2`. Let `alpha` be the acute angle, given by `tan alpha=|(Im(z))/(Re(z))|=|(-1)/(-sqrt(3))|=(1)/(sqrt(3)) rArr alpha = (pi)/(6)`. Clearly, the point representing the complex number `z = (-sqrt(3)-i)` is `P(-sqrt(3), -1)`, which lies in the third quadrant. `therefore" "arg(z) = theta = -(pi-alpha)=-(pi-(pi)/(6))=(-5pi)/(6)`. Thus, `r=|z|=2 and theta = (-5pi)/(6)`. Hence, the polar form of `z = (-sqrt(3)-i)` is given by `z=2{cos((-5pi)/(6))+i sin((-5pi)/(6))}, i.e., 2("cos"(5pi)/(6)-"i sin"(5pi)/(6))`. |
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| 95. |
Show that the complex number `z ,`satisfying are `(z-1)/(z+1)=pi/4`lies on a circle. |
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Answer» Let z = x +`i`y Given that, arg` ((z - 1)/(z + 1)) = pi//4` `rArr arg(z -1) -arg (z + 1) = pi//4` `rArr arg(x +iy - 1)- arg (x + iy + 1) = pi //4` `rArr arg(x - 1 + iy)- arg ( x + 1 + iy) = pi/4` `rArr tan^(-1) (y)/(x - 1) - tan^(-1) (y)/(x + 1) = pi//4` `rArr tan^(-1)[((y)/(x -1)-(y)/(x +1))/ (1 +((y)/(x -1))((y)/(x+1)))]=pi//4` `rArr [y[(x + 1- x +1)/(x^(2) - 1)]]/((x ^(2)-1 + Y^(2))/(x^(2)-1))= tan pi //4` `rArr (2y)/(x^(2) + Y^(2) - 1) = 1` `rArr x^(2) + y^(2) - 1 = 2 y` `rArr x^(2) + Y^(2) - 1 = 0` which represents a circle . |
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| 96. |
Show that : `(1)/(i)+(1)/(i^(2))+(1)/(i^(3))+(1)/(i^(4))=0` |
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Answer» `(1)/(i)=(1)/(i)xx(i^(3))/(i^(3))=i^(3) = -i, (1)/(i^(2))= -1. (1)/(i^(3))=(1)/(i^(3))xx(i)/(i)=i and (1)/(i^(4)) = 1`. `therefore" "{(1)/(i)-(1)/(i^(2))+(1)/(i^(3))-(1)/(i^(4))}=(-i+1+i-1)=0`. |
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| 97. |
Evaluate: `sqrt(16-30i)` |
| Answer» Correct Answer - `(5-3i)or(-5+3i)` | |
| 98. |
Evaluate `(sqrt(-36) xx sqrt(-25))`. |
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Answer» Correct Answer - -30 `(sqrt(-36)xx sqrt(-25))=(6ixx5i)=30i^(2) = -30`. |
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| 99. |
Convert of the complex number in the polar form:`1" "" "i` |
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Answer» The given complex number is z = -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 the complex number z = -1 -i is P(-1, -1), which lies in the third 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 of z = (-1 -i) is given by `z=sqrt)2){cos((-3pi)/(4))+i sin((-3pi)/(4))}, i.e., sqrt(2)("cos"(3pi)/(4)+"i sin"(3pi)/(4))` |
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| 100. |
Express `(-sqrt(3)+sqrt(-2))(2sqrt(3)-i)`in the form of `a + i b` |
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Answer» `(-sqrt3 + sqrt(-2))(2sqrt3 - i)` `(-sqrt2 + sqrt2 i )(2 sqrt3 - i)` `= - sqrt3(2sqrt3 - i) + sqrt2i(2 sqrt3 - i)` `=-6 + sqrt3i + 6i - sqrt2(-1)` `= (sqrt2- 6) + i( sqrt3 + 6)` answer |
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