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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.
| 101. |
Explain the fallacy : `-1 =(i xx i)=sqrt(-1)xx sqrt(-1) = sqrt((-1)xx(-1))=sqrt(1) = 1`. |
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Answer» We know that for any real numbers a and b, `sqrt(a) xx sqrt(b) = sqrt(ab)` is true only when at least one of a and b is either 0 or positive. `therefore" "sqrt(-1) xx sqrt(-1) ne sqrt((-1)xx(-1))`. |
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| 102. |
Show that the sum `(1 + i^(2) + i^(4)+....+i^(2n))` is 0 when n is odd and 1 when n is even. |
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Answer» Let `S = 1 + i^(2) + i^(4) +....+i^(2n)`. This is clearly a GP having (n + 1) terms with a = 1 and r = `i^(2)` = -1. `therefore" "S = (a(1-r^(n+1)))/((1+r))=(1xx{1-(i^(2))^(n+1)})/((1-i^(2)))` `" "= ({1-(-1)^(n+1)})/(1-(-1))=({q-(-1)^(n+1)})/(2)` `" "={{:((1)/(2)(1-1)=0",""when n is odd"),((1)/(2)(1+1)=1",""when n is even".):}` |
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| 103. |
Evaluate: `sqrt(3+4sqrt(-7))` |
| Answer» Correct Answer - `(sqrt(7)+2i)or (-sqrt(7)-2i)` | |
| 104. |
Evaluate `(i^(41)+(1)/(i^(71)))`. |
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Answer» Correct Answer - 2i `(i^(41)+(1)/(i^(71)))=(i+(1)/(i^(3)))=(i+(1)/(i^(3))xx(i)/(i))=(i+i)=2i`. |
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| 105. |
Show that `1+i^(10)+i^(20)+i^(30)`is a real number. |
| Answer» Given sum `=(1+i^(2)+1+i^(2))=0`. | |
| 106. |
Show that `(1-i)^(n)(1-(1)/(i))^(n)=2^(n)` for all `n in N` |
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Answer» `(1-(1)/(i))=((i-1))/(i)xx(i)/(i)=((i^(2)-i))/(i^(2))=((-1-i))/(-1)=(1+i)`. `therefore" ""given expression" = (1-i)^(n)xx(1+i)^(n)={(1-i)+(1+i)}^(2)=(1-i^(2))^(n)=2^(n)`. |
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| 107. |
If |z| = 6 and arg(z) `=(3pi)/(4)`, find z. |
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Answer» Correct Answer - `z = 3 sqrt(2) (-1 + i)` `z=6("cos"(3pi)/(4)+ "i sin"(3pi)/(4))=6{cos(pi-(pi)/(4))+i sin(pi-(pi)/(4))}` `rArr" "z=6("-cos"(pi)/(4)+"i sin"(pi)/(4))=6((-1)/(sqrt(2))+i.(1)/(sqrt(2)))=(6)/(sqrt(2))(-1+i)` `rArr" "z = 3 sqrt(2) (-1+i)`. |
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| 108. |
If `z_1=2-i , z_2=1+i ,`find `|(z_1+z_2+1)/(z_1-z_2+i)|` |
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Answer» Correct Answer - `2 sqrt(2)` `(z_(1)+z_(2)+1)/(z_(1)-z_(2)+i)=((2-i)+(1+i)+1)/((2-i)-(1+i)+i)=(4)/((1+i))xx((1-i))/((1-i))=2(1-i)` `rArr" "|(z_(1)+z_(2)+1)/(z_(1)-z_(2)+i)|=2 xx sqrt(1+(-1)^(2))=2sqrt(2)`. |
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| 109. |
Find the sum (`i + i^(2) + i^(3) + i^(4) +.....` up to 400 terms). |
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Answer» Given sum `=(i+i^(2)+i^(3)+i^(4))+(i^(5)+i^(6)_i^(7)+i^(8))+.......` `=(i+i^(2)+i^(3)+1)+i^(5)(1+i+i^(2)+i^(3))+....=0." "[because (1+i+i^(2)+i^(3))=0]` |
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| 110. |
Evaluate: `sqrt(4i)` |
| Answer» Correct Answer - `sqrt(2)(1+i)or sqrt(2)(-1-i)` | |
| 111. |
If `z = (sqrt(5)+3i)`, find `z^(-1)`. |
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Answer» Correct Answer - `((sqrt(5))/(14)-(3)/(14)i)` `z^(-1)=(bar(z))/(|z|^(2))=((sqrt(5)-3i))/((5+9))=((sqrt(5))/(14)-(3)/(14)i)`. |
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| 112. |
If `z = (1-i)`, find `z^(-1)`. |
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Answer» Correct Answer - `((1)/(2)+(1)/(2)i)` `z^(-1)=(bar(z))/(|z|^(2))=((1+i))/((1^(2)+1^(2)))=((1)/(2)+(1)/(2)i)`. |
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| 113. |
`" "(1+i)^(2)/(2-i)= x+ iy, "then find the value of x+y"`. |
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Answer» Given that , `" "(1+i)^(2)/(2-i)= x+ iy` `rArr ((1+i^(2)+2))/(2-i)= x + iy rArr (2i)/(2-i) = x +iy` `rArr (2i(2+i))/((2-i)(2+i))= x + iy rArr (4i+2-i^(2))/(4-i^(2)) = x +iy` `rArr (4i-)/(4+1) = iy rArr (-2)/(5)+(4i)/(5) = x+iy` On comparing both sides, we get `x = 2//2 rArr y = 4/5` `rArr" "x+y = (-2)/(5)+(4)/(5) = 2/5` |
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| 114. |
if `a = costheta +i sin theta `, prove that `(1+a)/(1-a) = cot(theta/2)i` |
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Answer» `(1+a)/(1-a)=((1+cos theta)+ i sin theta)/((1-cos theta)- i sin theta)xx((1-cos theta)+ i sin theta)/((1-cos theta)+ i sin theta)` `=((1-cos^(2)theta-sin^(2)theta)+2 i sin theta)/((1-cos theta)^(2)+ sin^(2) theta)=(1-(cos^(2)theta+sin^(2)theta)+2 i sin theta)/(1+(cos^(2)theta+sin^(2)theta)-2 cos theta)` `=(2 i sin theta)/(2(1-cos theta))=(i sin theta)/((1-cos theta))=(2 sin (theta//2)cos (theta//2))/(2 sin^(2)(theta//2)).i = ("cot"(theta)/(2))i`. |
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| 115. |
If `n in N ,`then find the value of `i^n+i^(n+1)+i^(n+2)+i^(n+3)dot` |
| Answer» Given sum `=i^(n) xx (1 + I + i^(2) + i^(3))=i^(n)(1+i-1-i)=0`. | |
| 116. |
Prove that: `x^4=4=(x+1+i)(x+1-i)(x-1+i(x-1-i)dot)` |
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Answer» Given expression `={(x+1)+i}" "{(x+1)-i}" "{(x-1)+i}" "{(x-1)-i}` `= {(x+1)^(2)-i^(2)}" "{(x-1)^(2)-i^(2)}={(x+1)^(2)+1}" "{(x-1)^(2)+1}` `={(x^(2)+2)+2x}" "{(x^(2)+2)-2x}` `=(x^(2)+2)^(2)-4x^(2)=(x^(4)+4)`. |
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| 117. |
Show that `(-sqrt(-1))^(4n+3) =i`, where n is a positive integer. |
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Answer» We have `(-sqrt(-1))^(4n+3)=(-i)^(4n+3)" "[because sqrt(-1)=i]` `=(-i)^(4n)xx(-i)^(3)` `={(-i)^(4)}^(n) xx (-i^(3))=(1 xx i) = i" "[because (-i)^(4) = 1 and -i^(3) = -(-i) = i]` Hence, `(-sqrt(-1))^(4n+3)=i`. |
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| 118. |
What is the smallest positive integer `n`for which `(1+i)^(2n)=(1-i)^(2n)?` |
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Answer» Correct Answer - n = 2 `((1+i))/((1-i))=((1+i))/((1-i))=((1+i))/((1+i))=((1+i)^(2))/(2)=((1+i^(2)+2i))/(2)=(2i)/(2)=i`. `therefore" "(1+i)^(2n)=(1-i)^(2n) rArr ((1+i)^(2n))/((1-i)^(2n))=1 rArr ((1+i)/(1-i))^(2n)=1 rArr i^(2n) = 1`. Least value of 2n is 4 and so the least value of n is 2. |
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| 119. |
`tan(ilog((a-ib)/(a+ib)))=`(i) `ab` (ii)`(2ab)/(a^2-b^2)` (iii) `(a^2-b^2)/(ab)` (iv) `(2ab)/(a^2+b^2)` |
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Answer» `a + ib = r e^(i theta)` `r = sqrt(a^2 + b^2)` `theta = tan^-1(b/a)` `a - ib = r e^(- i theta) ` `i log ( (a-ib)/(a + ib)) = i log( (r e^(_i theta))/(r e^(i theta))) ` `= i log_e e^(-2 i theta) ` `= i(-2 i theta) ` `= 2 theta` `2 tan^-1(b/a) ` `= tan^-1((2b/a)/(1- b^2/a^2)) = tan^-1((2ab)/(a^2- b^2))` `tan( tan^-1((2ab)/(a^2 - b^2))) = (2ab)/(a^2 - b^2)` option 2 is correct |
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| 120. |
Evaluate `(1)/(i^(78))`. |
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Answer» Correct Answer - -1 `(1)/(i^(78))=(1)/(i^(78))xx(i^(2))/(i^(2))=-1" "[because i^(80)=(i^(4))^(20)=1]` |
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| 121. |
Evaluate: `sqrt(1-i)` |
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Answer» Correct Answer - `{sqrt((sqrt(2)+1)/(2))}-{sqrt((sqrt(2)-1)/(2))}i or -{sqrt((sqrt(2)+1)/(2))}+{sqrt((sqrt(2)-1)/(2))}i` Let `sqrt(1-i) = x -iy`. On squaring both sides, we get `(1-i)=(x-iy)^(2)=(x^(2)-y^(2))-i(2xy)` `therefore" "(x^(2)-y^(2))=1 and 2xy = 1` `rArr" "(x^(2)-y^(2))=sqrt((x^(2)-y^(2))^(2)+4x^(2)y^(2))=sqrt(1^(2)+1^(2))=sqrt(2)` `rArr" "x^(2)-y^(2) = 1 and x^(2) + y^(2) = sqrt(2)` `rArr" "x^(2)=(sqrt(2)+1)/(2) and y^(2)=(sqrt(2)-1)/(2)` `rArr" "x = +- sqrt((sqrt(2)+1)/(2))and y = +- sqrt((sqrt(2)-1)/(2))`. Since xy `gt` 0, so x and y have the same sign. `therefore" "{x=sqrt((sqrt(2)+1)/(2)),y=sqrt((sqrt(2)-1)/(2))}or{x=-sqrt((sqrt(2)+1)/(2)),y=-sqrt((sqrt(2)-1)/(2))}` `therefore" "sqrt(1-i)={sqrt((sqrt(2)+1)/(2))-sqrt((sqrt(2)-1)/(2))}or{-sqrt((sqrt(2)+1)/(2))+ isqrt((sqrt(2)-1)/(2))}`. |
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| 122. |
Solve the following equation by factorizationmethod: `3x^2+7i x+6=0` |
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Answer» Correct Answer - `{(2)/(3)i, -3i}` a = 3, b = 7i and c = 6. `therefore" "(b^(2)-4ac)={(7i)^(2)-4xx3xx6}=(-49-72)=(-121)lt 0`. So, the given equation has complex roots, given by `(-b+-sqrt(b^(2)-4ac))/(2a)=(-7i+-sqrt(-121))/(2xx3)=(-7i+-11i)/(6)`. Solution set `={(-7i+11i)/(6),(-7i-11i)/(6)}={(2)/(3)i,-3i}` |
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| 123. |
lf `z(!=-1)` is a complex number such that `[z-1]/[z+1]` is purely imaginary, then `|z|` is equal to |
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Answer» Let z = (x + iy). Then, `(z-1)/(z+1)=((x+iy)-1)/((x+iy)+1)=((x-1)+iy)/((x+1)+iy)xx((x+1)-iy)/((x+1)-iy)=((x^(2)+y^(2)-1)+2iy)/((x+1)^(2)+y^(2))` `(z-1)/(z+1)` is purely imaginary `iff x^(2) - 1 = 0 iff x^(2) + y^(2) = 1 iff |z| = 1`. |
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| 124. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `((1-i))/(("cos"(pi)/(3)+"i sin"(pi)/(3)))` |
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Answer» Correct Answer - `sqrt(2),(7pi)/(12),sqrt(2)("cos"(7pi)/(12)-"i sin"(7pi)/(12))` Given number `=(2(1-i))/((1+sqrt(3)i))xx((1-sqrt(3)i))/((1-sqrt(3)i))={((1-sqrt(3)))/(2)-((1+sqrt(3)))/(2)i}` `therefore" "|z|^(2)=((1-sqrt(3))^(2))/(4)+((1+sqrt(3))^(2))/(4)=(2(1+3))/(4)=2 rArr |z| = sqrt(2)`. `tan alpha=|((-(1+sqrt(3)))/(2))/(((1+sqrt(3)))/(2))|=((sqrt(3)+1))/((sqrt(3)-1))=((1+(1)/(sqrt(3))))/((1-(1)/(sqrt(3))))=("tan"(pi)/(4)+"tan"(pi)/(6))/(1-"tan"(pi)/(4)*"tan"(pi)/(6))` `=tan((pi)/(4)+(pi)/(6))="tan"(5pi)/(12)`. The given number z is represented by the point `P((-(sqrt(3)-1))/(2),(-(sqrt(3)+1))/(2))` So, it lies in the third quadrant. `therefore" "arg(z) = theta = -(pi- alpha)=-(pi-(5pi)/(12))=(-7pi)/(12)`. Hence, `z=sqrt(2){cos((-7pi)/(12))+i sin((-7pi)/(12))}`. |
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| 125. |
If `z_1`is a complex number other than -1 such that `|z_1|=1 a n d z_2=(z_1-1)/(z_1+1)`, then show that the real parts of `z_2`is zero. |
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Answer» Let `z_(1) = x_(1) + iy_(1)`. Then, `|z|=1 rArr |z_(1)|= 1 rArr |z_(1)|^(2) = 2 rArr x_(1)^(2) + y_(1)^(2) = 1`. `z_(2)=(z_(1)-1)/(z_(1)+1)=((x_(1)+iy_(1))-1)/((x_(1)+iy_(1))+1)=((x_(1)-1)+iy_(1))/((x_(1)+1)+iy_(1))xx((x_(1)+1)-iy_(1))/((x_(1)+1)-iy_(1))` `=((x_(1)^(2)+y_(1)^(2)-1)+2iy_(1))/((x_(1)+1)^(2)+y_(1)^(2))=(2iy_(1))/((x_(1)+1)^(2)+y_(1)^(2))`, which is purely imaginary. `" "[because |z_(1)|^(2)=1]` |
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| 126. |
Find the modulus and argument of each of the following complex number: `-sqrt(3)-i` |
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Answer» Correct Answer - `2,(-5pi)/(6),2{cos((-5pi)/(6))+i sin((-5pi)/(6))}` Let `z = -sqrt(3) - i rArr r^(2) = |z|^(2) = (-sqrt(3))^(2)+(-1)^(2)=4 rArr r = |z| = 2`. `tan alpha=|(-1)/(-sqrt(3))|=(1)/(sqrt(3))rArr alpha = (pi)/(6)`. The given number represents the point `P(-sqrt(3), -1)` which lies in the third quadrant. `therefore" "arg(z)=theta=-(pi-alpha)=-(pi-(pi)/(6))=(-5pi)/(6)`. `therefore" "z=2[cos((-5pi)/(6))+i sin((-5pi)/(6))]`. |
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| 127. |
Show that a real value of `x`will satisfy hte equation `(1-i x)//(1+i x)=a-i b`if `a^2+b^2=1,w h e r ea ,b`real. |
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Answer» We have `(1-ix)/(1+ix)=(a-ib)/(1)rArr((1-ix)+(1+ix))/((1-ix)-(1+ix))=((a-ib)+1)/((a-ib)-1)" "["by componendo and dividendo"]` `rArr (2)/(-2ix)=((a+1)-ib)/((a-1)-ib)rArr -ix=((a-1)-ib)/((a+1)-ib)xx((a+1)+ib)/((a+1)+ib)` `rArr -ix = ((a^(2)-1+b^(2))+{(a-1)b-(a+1)b}i)/((a+1)^(2)+b^(2))` `=(-2bi)/((a+1)^(2)+b^(2))" "[because a^(2)+b^(2)=1]` `rArr x=(2b)/((a+1)^(2)+b^(2))`, which is purely real. |
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| 128. |
If `x+i y=sqrt((a+i b)/(c+i d))`prove that `(x^2+y^2)^2=(a^2+b^2)/(c^2+d^2)` |
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Answer» We have `(x+iy)=sqrt((a+ib)/(c+id))=(sqrt(a+ib))/(sqrt(c+id))` `rArr" "(x-iy)=(sqrt(a-ib))/(sqrt(c-id))` `rArr" "(x+iy)(x-iy)=(sqrt(a+ib))/(sqrt(c+id))xx(sqrt(a-ib))/(sqrt(c-id))=(sqrt((a+ib)(a-ib)))/(sqrt((c+id)(c-id)))` `rArr" "(x^(2)+y^(2))=(sqrt(a^(2)+b^(2)))/(sqrt(c^(2)+d^(2)))rArr (x^(2)+y^(2))^(2)=((a^(2)+b^(2)))/((c^(2)+d^(2)))`. Hence, `(x^(2)+y^(2))^(2)=((a^(2)+b^(2)))/((c^(2)+d^(2)))`. |
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| 129. |
If `(a + i b) (c + i d) (e + if) (g + i h) = A + i B`, then show that `(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2` |
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Answer» `(a+ib)(c+id)(e+if)(g+ih)=(A+iB)` `rArr" "|(a+ib)(c+id)(e+if)(g+ih)=|A+iB|` `rArr" "|a+ib|*|"c+id"|*|"e+if"|*|g+ih|=|A+iB|` `rArr" "|a+ib|^(2)*|"c+id"|^(2)*|"e+if"|^(2)*|g+ih|^(2)=|A+iB|^(2)` `rArr" "(a^(2)+b^(2))(c^(2)+d^(2))(e^(2)+f^(2))(g^(2)+h^(2))=(A^(2)+B^(2))`. Hence, `(a^(2)+b^(2))(c^(2)+d^(2))(e^(2)+f^(2))(g^(2)+h^(2))=(A^(2)+B^(2))`. |
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| 130. |
Which of the following is correct for any tow complex numbers `z_1a n dz_2?``|z_1z_2|=|z_1||z_2|`(b) `a r g(z_1z_2)=a r g(z_1)a r g(z_2)`(c) `|z_1+z_2|=|z_1|+|z_2|`(d) `|z_1+z_2|geq|z_1|+|z_2|`A. `|z_(1) z_(2)| = | z_(1) ||z_(2)|`B. arg` (z _(1) z_(2)) = arg (z_(1). Arg (z_(2))`C. `|z_(1) + z_(2)| = |z_(1)|+|z_(2)|`D. `|z_(1) + z_(2) ge |z_(1)|- | z_(2)|` |
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Answer» Correct Answer - A Let `z_(1) = r_(1) (costheta_(1) + isintheta_(1))` `rArr |z_(1)| = r-(1) …(i) ` and `z_(2) = r_(2)(costheta_(2) + isin theta_(2)) ` `rArr |z_(2)| =r_(2) ...(ii)` Now. ` z_(1) z_(2) = r_(1) r_(2) [costheta_(1) costheta_(2) + isin theta_(1) costheta_(2)+ icostheta_(1)isintheta_(2)+i^(2) sintheta_(1)sintheta_(2)]` `=r_(1)r_(2) [cos(theta_(1) + theta_(2))+ isin(theta_(1) + theta_(2))]` `rArr |z_(1)z_(2)| = r_(1)r_(2)` `:. |z_(1)z_(2)| = |z_(1)||z_(1)| [using Eqa. (i) and (ii)]` |
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| 131. |
If `((1 +i)/(1 -i))^(x) =1`, then (A) x=2n+1 (B) x=4n (C) x=2n (D) x=4n+1, n` in`N.A. `x = 2n + 1`B. `x = 4n`C. `x = 2n`D. `x = 4n + 1` |
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Answer» Correct Answer - B Given that, `((1 +i)/(1 -i))^(x) =1` `rArr [((1 + i)(1+ i))/((1 -i)(1 +i))]^(x) = 1 rArr [(1 + 2i+ i^(2))/(1-i^(2))]^x = 1 ` `[(2i)/(1+1)]^(x) = 1 rArr [(2i)/(2)]^(x) = 1` `rArr i^(x) = 1 rArr i^(x) = i^(4n) " " [:. i^(4n) = 1, n in N]` `rArr x = 4n` |
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| 132. |
The value of ` (z + 3) (barz + 3) ` is equivlent to(A) |z+3|^(2) (B) |z-3| (C) z^2+3 (D) none of theseA. `| z+ 3|^(2)`B. `|z -3|`C. `z^(2) +3`D. None of these |
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Answer» Correct Answer - A Given that, `(z + 3)( barz + 3)` Let z = x + iy `rArr ( z + 3) (barz + 3) = ( x +iy + 3) (x + 3 - iy)` `= (x + 3)^(2) - (iy) ^(2) = (x + 3)^(2) + y^(2)` `= | x + 3 + iy|^(2) = | z + 3|^(2)` |
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| 133. |
If `((z-1)/(z+1))` is purely an imaginary number and `z ne -1` then find the value of |z|. |
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Answer» Correct Answer - |z| = 1 Let z = x + iy. Then, `(z-1)/(z+1)=((x-1)+iy)/((x+1)+iy)xx((x+1)-iy)/((x+1)-iy)=((x^(2)+y^(2)-1)+(x+y)i)/((x+1)^(2)+y^(2))` Now, `(z-1)/(z+1)` is purely imaginary `iff x^(2) + y^(2) -1 = 0 iff x^(2) + y^(2) = 1 iff |z| = 1`. |
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| 134. |
if `(1+i)z=(1-i)barz` then `z` is |
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Answer» `(1+i)z=(1-i)bar(z) rArr (z)/(bar(z))=((1-i))/((1+i))xx((1-i))/((1-i))=((1-i)^(2))/(2)=((1+i^(2)-2i))/(2)= -i`. Hence, `z = -i bar(z)`. |
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| 135. |
Solve : `x^(2)+3=0`. |
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Answer» We have `x^(2)+3 = 0 = rArr x^(2) = -3 rArr x = +-sqrt(-3) = +- i sqrt(3)`. `therefore" ""solution set"= {i sqrt(3), -i sqrt(3)}`. |
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| 136. |
If a = `cos theta +theta i sin, "then find the value of" (1+a)/(1-a)`. |
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Answer» Givne that, `a = costheta + isintheta` `:. (1+a)/(1-a) = (1+costheta+isintheta)/(1-costheta -i sintheta)` = `(1+2cos^(2)theta//2 - 1+2isintheta//2.costheta/1-costheta //2)/(1-1+2sin^(2)theta//2-2isintheta//2.costheta//2)=(2costheta//2(costheta//2 + isintheta//2))/(2sintheta//2(sintheta//2-icostheta//2)) ` `=-(2costheta//2(costheta//2 + isintheta//2))/(2sintheta//2(sintheta//2-icostheta//2))= -(1)/(i)cot theta//2` `(+i^(2))/(i)cottheta//2=icottheta//2" "[:.(-1)/(i)=(i^(2))/(i)]` |
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| 137. |
Solve the equation:`x^2+3x+9=0` |
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Answer» The given equation is `x^(2)+3x+9=0`. This is of the form `ax^(2)+bx+c=0`, where a = 1, b = 3 and c = 9. `therefore" "(b^(2)-4ac)=(3^(2)-4xx1xx9)=(9-36)=27 lt 0`. So, the given equation has complex roots. These roots are given by `(-b+- sqrt(b^(2)-4ac))/(2a)=(-3 +- sqrt(-27))/(2xx1)" "[because b^(2)-4ac = -27]` `=(-3 +- isqrt(27))/(2)=(-3 +- i3 sqrt(3))/(2)` `therefore" ""solution set"={(-3+ i3 sqrt(3))/(2),(-3-i3 sqrt(3))/(2)}={(-3)/(2)+(3sqrt(3))/(2)i, (-3)/(2)-(3sqrt(3))/(2)i}`. |
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| 138. |
Solve : `9x^(2) + 10x + 3 = 0`. |
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Answer» The given equation is `9x^(2)+10x+3=0`. This is of the form `ax^(2)+bx+c=0`, Where a = 9, b = 10 and c = 3. `therefore" "(b^(2)-4ac)={(10)^(2)-4xx9xx3}=(100-108)=-8 lt 0`. So, the given equation has complex roots. These roots are given by `(-b+- sqrt(b^(2)-4ac))/(2a)=(-10 +- sqrt(-8))/(2xx9)" "[because (b^(2)-4ac)=-8]` `=(-10 +- i2 sqrt(2))/(18)=(-5+- i sqrt(2))/(9)`. `therefore" ""solution set"={(-5+ i sqrt(2))/(9),(-5- i sqrt(2))/(9)}={(-5)/(9)+(sqrt(2))/(9)i, (-5)/(9)-(sqrt(2))/(9)i}`. |
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| 139. |
Express of the complex number in the form `a + i b`.`(-2-1/3i)^3` |
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Answer» using the identity `(a+b)^3 = a^3 + b^3 + 3ab(a+b)` `(-2)^3 + (-1/3i)^2 + 3(-2)(-1/3i)(-2 - 1/3i)` =`-8 -1/27i^3 + 2i(-2-1/3i)` =`-8 + 1/27i - 4i -2/3i^2` =`-8 + 1/27i - 4*27/27i - 2/3(-1)` `= -8 + 2/3 + 1/29i -108/27i` `=-22/3 - 107i/27` `a=-22/3 & b= -107/27` answer |
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| 140. |
Find real values of `xa n dy`for which the complex numbers `-3+i x^2ya n dx^2+y+4i`are conjugate of each other. |
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Answer» Correct Answer - `(x =1, y = -4) or (x = -1, y = -4)` `(-3+iyx^(2))=(bar(x^(2)+y+4i))` `rArr" "(-3+iyx^(2))=(x^(2)+y-4i)` `rArr" "(x^(2)+y+3)-(4+yx^(2))i = 0` `rArr" "x^(2)+y+3=0 and 4 + yx^(2) = 0` `rArr" "x^(2)+y = -3" "...(i)" "and yx^(2) = -4" "...(ii)`. Putting `x^(2) = (-3-y)` from (i) in (ii), we get `y(-3-y) = -4 rArr y^(2)+3y - 4 = 0 rArr (y+4)(y-1)=0 rArr y = -4 or y = 1`. Now, `y = 1 rArr x^(2) = -4 rArr x` is imaginary. So, y `in` 1. When y = -4, we get `x^(2) = (-3+4)=1 rArr x = +- 1`. `therefore" "(x =1, y = -4) or (x = -1, y = -4)`. |
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| 141. |
Reduce `(1/(1-4i)-2/(1+i))((3-4i)/(5+i))`to the standard form. |
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Answer» `z = (1/(1-4i)-2/(1+i))((3-4i)/(5+i))` `=((1+i-2+8i)/((1-4i)(1+i)))((3-4i)/(5+i))` `=((-1+9i)/(1+i-4i+4))((3-4i)/(5+i))` `=((-1+9i)/(5-3i))((3-4i)/(5+i))` `=((-1+9i)(3-4i))/((5-3i)(5+i))` `=(-3+4i+27i-36i^2)/(25+5i-15i-3i^2)` `=(-3+4i+27i+36)/(25+5i-15i+3)` `=(33+31i)/(28-10i)**(28+10i)/(28+10i)` `=(924+868i+330+310i^2)/(784-100i^2)` `=(924+868i+330i-310)/(784+100)` `=(614+1198i)/884` `z = 307/442+599/442i` |
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| 142. |
Solve the equation:`sqrt(2)x^2+x+sqrt(2)=0` |
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Answer» The given equation is `sqrt(2)x^(2) + x + sqrt(2) = 0`. This is of the form `ax^(2)+bx+c=0`, where `a = sqrt(2), b = 1 and c = sqrt(2)`. `therefore" "(b^(2)-4ac)=(1^(2)-4xx sqrt(2)xx sqrt(2))=(1-8) =-7 lt 0`. So, the given equation has complex roots. These roots are given by `(-b +- sqrt(b^(2)-4ac))/(2a)=(-1 +- sqrt(-7))/(2 sqrt(2))" "[because b^(2)-4ac = -7]` `=(-1 +- i sqrt(7))/(2 sqrt(2))`. `therefore" ""solution set"={(-1+i sqrt(7))/(2 sqrt(2)),(-1-i sqrt(7))/(2 sqrt(2))}={(-1)/(2 sqrt(2))+(sqrt(7))/(2 sqrt(2))i,(-1)/(2 sqrt(2))-(sqrt(7))/(2 sqrt(2))i}`. |
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| 143. |
Solve : `3x^(2) + 8ix + 3 = 0`. |
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Answer» The given equation is `3x^(2)+8ix+3=0`. This is of the form `ax^(2)+bx+c=0`, where a = 3, b = 8i and c = 3. `therefore" "(b^(2)-4ac)={(8i)^(2)-4 xx 3 xx 3}=(-64-36)=-100 lt 0`. So, the given equation has complex roots. These roots are given by `(-b+- sqrt(b^(2)-4ac))/(2a)=(-8i+- sqrt(-100))/(2 xx 3)" "[because b^(2)-4ac=-100]` `=(-8i+-10i)/(6)=(-4i+-5i)/(3)`. Thus, the roots of the given equation are `(-4i+5i)/(3)=(i)/(3)and(-4i+5i)/(3)=(-9i)/(3)=-3i`. `therefore" ""solution set"={(i)/(3),-3i}`. |
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| 144. |
If `((1-i)/(1+i))^(100)=a+i b , fin d (a , b)` |
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Answer» Correct Answer - `a = 1, b = 0` `((1-i))/((1+i))=((1-i))/((1+i))xx((1-i))/((1-i))=((1-i)^(2))/(2)=((1+i^(2)-2i))/(2)= -i`. `therefore" "((1-i)/(1+i))^(100)=(-i)^(100) ={(-i)^(4)}^(25)=(1)^(25)=1" "[because (-i)^(4) = i^(4) = 1]`. Hence, a = 1 and b = 0. |
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| 145. |
If `((1+i)/(1-i))^3-((1-i)/(1+i))^3=x+i y , fin d (x , y)` |
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Answer» Correct Answer - `x = 0, y = -2` `((1+i)/(1-i))=((1+i))/((1-i))xx((1+i))/((1+i))=((1+i)^(2))/(2)=((1+i^(2)+2i))/(2)=(2i)/(2)=i`. Similarly, `((1-i)/(1+i)) = -i`. Given expression `= i^(3) -(-i)^(3) = 2i^(3) = -2i^(3) = -2i = 0 + (-2)i`. |
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| 146. |
Find the conjugate of `1/(3+4i)dot` |
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Answer» Correct Answer - `((3)/(25)+(4)/(25)i)` `z=(1)/((3+4i))xx((3-4i))/((3-4i))=(3-4i)/(25)=((3)/(25)-(4)/(25)i) rArr bar(z) = ((3)/(25)+(4)/(25)i)`. |
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| 147. |
Find the multiplicative inverse of the complex number. `4 - 3i` |
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Answer» let `4-3i= z` `^-1 = 1/z = 1/(4-3i)` `=1/(4-3i) * (4+3i)/(4+3i)` =`(4+3i)/(16-(-9)= (4+3i)/25` answer |
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| 148. |
Solve the equation:`sqrt(3)x^2-sqrt(2)x+3sqrt(3)= 0` |
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Answer» The given equation is `sqrt(3)x^(2)-sqrt(2)x +3 sqrt(3) = 0` This is of the form `ax^(2) + bx + c = 0`, where `a = sqrt(3), b = -sqrt(2) and c = 3 sqrt(3)`. `therefore" "(b^(2)-4ac)={(-sqrt(2))^(2)-4 xx sqrt(3) xx 3 sqrt(3)}=(2-36)=-34 lt 0`. So, the given equation has complex roots. These roots are given by `(-b +- sqrt(b^(2)-4ac))/(2a)=(sqrt(2)+- sqrt(-34))/(2 xx sqrt(3))" "[because b^(2)-4ac=-34]` `=(sqrt(2)+-i sqrt(34))/(2 sqrt(3))` `therefore" ""solution set"={(sqrt(2)+i sqrt(34))/(2 sqrt(3)),(sqrt(2)-i sqrt(34))/(2 sqrt(3))}={(1)/(sqrt(6))+(sqrt(34))/(2 sqrt(3))i,(1)/(sqrt(6))-(sqrt(34))/(2 sqrt(3))i}`. |
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| 149. |
If `x + i y =(a+i b)/(a-i b),`prove that `x^2+y^2=1`. |
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Answer» `(x+iy)=(a+ib)/(a-ib) rArr bar((x+iy))=((bar(a+ib)))/((bar(a-ib))) rArr (x-iy) = ((a-ib))/((a+ib))`. `therefore" "(x+iy)(x-iy)=((a+ib))/((a-ib))xx((a-ib))/((a+ib))=1 rArr x^(2) + y^(2) = 1`. |
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| 150. |
Evaluate: `sqrt(i)` |
| Answer» Correct Answer - `(1)/(sqrt(2))(1+i)or(1)/(sqrt(2))(-1-i)` | |