Question 1: Express the following complex numbers in the form a + ib

i) (1+i)(1+2i)           ii) \frac{3+2i}{-2+i}           iii) \frac{1}{(2+i)^2}           iv) \frac{1-i}{1+i}

v) \frac{(2+i)^3}{2+3i}           vi) \frac{(1+i)(1+\sqrt{3} i)}{1-i}           vii) \frac{2+3i}{4+5i}           viii) \frac{(1-i)^3}{1-i^3}

ix) (1+2i)^{-3}           x) \frac{3-4i}{(4+2i)(1+i)}

xi) \Big( \frac{1}{1-4i} - \frac{2}{1+i}  \Big)  \Big( \frac{3-4i}{5+i}  \Big)           xii) \frac{5+\sqrt{2} i}{1 - \sqrt{2} i}

Answer:

i) (1+i)(1+2i)

= 1 + 2i + i + 2i^2 = 1 + 3i - 2 = - 1 + 3i

ii) \frac{3+2i}{-2+i}

= \frac{3+2i}{-2+i} \times \frac{-2-i}{-2-i} = \frac{-6-3i-4i-2i^2}{4-i^2} = \frac{-6-7i+2}{4+1} = \frac{-4}{5} - i \big( \frac{7}{5} \big)

iii) \frac{1}{(2+i)^2}

= \frac{1}{4+i^2 + 4i} = \frac{1}{3+4i} \times \frac{3-4i}{3-4i} = \frac{3-4i}{9-16i^2} = \frac{3}{25} - i \big( \frac{4}{5} \big)

iv) \frac{1-i}{1+i}

= \frac{1-i}{1+i} \times \frac{1-i}{1-i} = \frac{1+i^2-2i}{1-i^2} = \frac{-2i}{2} = -i

v) \frac{(2+i)^3}{2+3i}

= \frac{( 4 + i^2 + 4 i )(2+i)}{2+3i} = \frac{8 + 2i^2 + 8i + 4i + i^3 + 4i^2}{2 + 3i} = \frac{2+11i}{2+3i} \times \frac{2-3i}{2-3i}

= \frac{4 - 6i + 22 i - 33i^2}{4 - 9i^2} = \frac{37+ 16i}{13} = \frac{37}{13} + i \big( \frac{16}{13} \big)

vi) \frac{(1+i)(1+\sqrt{3} i)}{1-i}

= \frac{(1+i)(1+\sqrt{3} i)}{1-i} \times \frac{1+i}{1-i} = \frac{(1+i^2+2i)(1+\sqrt{3}i}{1-i^2}

= \frac{2i ( 1 + \sqrt{3} i)}{2} = i(1 + \sqrt{3}i) = \sqrt{3} i^2 + i = - \sqrt{3}+ i

vii) \frac{2+3i}{4+5i}

= \frac{2+3i}{4+5i} \times \frac{4-5i}{4-5i} = \frac{8 - 10 i + 12 i - 15 i^2}{16 - 25 i^2} = \frac{23 + 2 i}{41} = \frac{23}{41} + i \big( \frac{2}{41} \big)

viii) \frac{(1-i)^3}{1-i^3}

= \frac{(1-i)^2 (1-i)}{1-i^3} = \frac{(1+i^2 - 2i)(1-i)}{1+i} = \frac{-2i(1-i)}{1+i} = \frac{-2(i - i^2)}{1+i} = \frac{-2(1+i)}{1+i} = - 2 

ix) (1+2i)^{-3}

= \frac{1}{(1+2i)^2(1+2i)} = \frac{1}{(1+4i^2+4i)(1+2i)} = \frac{1}{(-3+4i)(1+2i)} = \frac{1}{-3+4i-6i+8^2}

= \frac{1}{-11-2i} = \frac{1}{-11-2i} \times \frac{11-2i}{11-2i} = \frac{-11+2i}{121+4} = \frac{-11}{125} + i \big( \frac{2}{125} \big)

x) \frac{3-4i}{(4+2i)(1+i)}

= \frac{3-4i}{4 - 2i + 4i - 2i^2} = \frac{3-4i}{6+2i} \times \frac{6-2i}{6-2i} = \frac{18 - 24i - 6i + 8i^2}{36 + 4}

= \frac{10-30i}{40} = \frac{1-3i}{4} = \frac{1}{4} + i \big( \frac{-3}{4} \big)

xi) \Big( \frac{1}{1-4i} - \frac{2}{1+i}   \Big)  \Big( \frac{3-4i}{5+i} \Big)

= \Big( \frac{1+i-2+8i}{1-4i+i-4i^2}    \Big)  \Big( \frac{3-4i}{5+i}   \Big)   = \Big( \frac{-1+9i}{5-3i} \Big)  \Big( \frac{3-4i}{5+i}   \Big) = \frac{-3+27i+4i-36i^2}{25-15i+5i-3i^2}

= \frac{33+ 31i}{28-10i} \times \frac{28+10i}{28+10i} = \frac{924+330i+868i-310}{784+100} = \frac{614+1198i}{884} = \frac{614}{884} + i \big( \frac{1198}{884} \big)

xii) \frac{5+\sqrt{2} i}{1 - \sqrt{2} i}

= \frac{5+\sqrt{2} i}{1 - \sqrt{2} i} \times \frac{1 + \sqrt{2} i}{1 + \sqrt{2} i} = \frac{5 + \sqrt{2} i + 5 \sqrt{2} i + 2 i^2}{1+2}

= \frac{3 + 6 \sqrt{2} i}{3} = 1 + 2\sqrt{2} i = 1 + i ( 2\sqrt{2})

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Question 2: Find the real values of x and latex y, if

i) (x+iy)(2-3i) = 4+i           ii) (3x-2iy)(2+i)^2 = 10(1+i)

iii) \frac{(1+i)x- 2i}{3+i} - \frac{(2-3i)y+i}{3-i} = i           iv) (1+i)(x+iy)= 2- 5i

Answer:

i) (x+iy)(2-3i) = 4+i

\Rightarrow 2x + i 2y - 3x i - 3y i^2 = 4 + i

\Rightarrow ( 2x + 3y) + i ( -3x + 2y) = 4 + i

Therefore comparing, we get

2x + 3y = 4      … … … … … i)

-3x + 2y = 1      … … … … … ii)

Solving i) and ii), multiplying i) by 3 and ii) by 2 and adding we get

{ \hspace{1.0cm} 6x + 9y = 12} \\ \underline { (+) - 6x + 4 y = 2 } \\ { \hspace{2.0cm}13 y = 14}

\Rightarrow y = \frac{14}{13}

Now substituting in ii) we get

2x + 3 \big( \frac{14}{13} \big) = 4 \Rightarrow 2x = \frac{10}{13} \Rightarrow x = \frac{10}{26}

Hence x = \frac{10}{26} and y = \frac{14}{13}

ii) (3x-2iy)(2+i)^2 = 10(1+i)

\Rightarrow ( 3x - 2iy) ( 4 + i^2 + 4i) = 10 + 10 i

\Rightarrow  ( 3x - i 2y) ( 3 + 4i) = 10 + 10i

\Rightarrow 9x - i 6 y + i 12 x + 8 y = 10 + 10 i

\Rightarrow ( 9x + 8 y) + i ( 12 x - 6y) = 10 + 10 i

Therefore comparing, we get

9x + 8y = 10      … … … … … i)

12 x - 6y = 10      … … … … … ii)

Solving i) and ii), multiplying i) by 6 and ii) by 8 and adding we get

{ \hspace{1.0cm} 54x + 48 y = 60 } \\ \underline { (+) 96x - 48 y = 80 } \\ { \hspace{2.0cm} 150x  = 140}

\Rightarrow x = \frac{14}{15}

Now substituting in i) we get

9 \big( \frac{14}{15} \big) + 8y = 10 \Rightarrow 8y = \frac{8}{5} \Rightarrow y = \frac{1}{5}

Hence x = \frac{14}{15} and y = \frac{1}{5}

iii) \frac{(1+i)x- 2i}{3+i} - \frac{(2-3i)y+i}{3-i} = i

\Rightarrow \frac{[ (1+i) x - 2 i ] (3-i)+[(2-3i)y + 1](3+i)}{9+1} = i

\Rightarrow 3x + i 3x - 6i - i x - i^2 x + 2 i^2+ 6y - i 9y + 3 i + i 2y - i^2 3y + i^2 = 10i

\Rightarrow 4x + i ( 2x) - 6i - 2 + 9y + i ( -7y) + 3i - 1 = 10i

\Rightarrow ( 4x + 9y - 3) + i ( 2x - 7y - 3) = 10i

\Rightarrow 4x + 9y - 3 = 0 \Rightarrow 4x + 9y = 3     … … … … … i)

\Rightarrow 2x - 7y - 3 = 10 \Rightarrow 2x - 7y = 13     … … … … … ii)

Solving i) and ii), multiplying i) by 7 and ii) by 9 and adding we get

46 x = 117 + 21 \Rightarrow x = \frac{138}{46} = 3

Substituting in i) we get

4 ( 3) + 9y = 3 \Rightarrow y = -1

Hence x = 3   and y = -1

iv) (1+i)(x+iy)= 2- 5i

x + ix + iy + i^2 y = 2 - 5i

(x-y) + i( x+y) = 2 - 5i

Comapring we get

x-y = 2    … … … … … i)

x+y = -5    … … … … … ii)

Adding i) and ii) we get

2x = - 3 \Rightarrow x = \frac{-3}{2}

Substituting in i) we get

y = x - 2 = \frac{-3}{2} - 2 = \frac{-7}{2}

Hence x = \frac{-3}{2} and y = \frac{-7}{2}

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Question 3: Find the conjugates of the following complex numbers:

i) 4-5i           ii) \frac{1}{3+5i}           iii) \frac{1}{1+i}

iv) \frac{(3-i)^2}{2+i}           v) \frac{(1+i)(2+i)}{3+i}           vi) \frac{(3-2i)(2+3i)}{(1+2i)(2-i)}

Answer:

Note: if z = x + iy , then conjugate of z   is \overline{z} = x - iy

i) If z = 4-5i \Rightarrow \overline{z} = 4 + 5i

ii) If z = \frac{1}{3+5i} = \frac{1}{3+5i} \times \frac{3 - 5i}{3 - 5i} = \frac{3 - 5i}{34} = \frac{3}{34} -i \frac{5}{34}

Therefore \overline{z} = \frac{3}{34} + i \frac{5}{34}

iii) If z =  \frac{1}{1+i} = \frac{1}{1+i} \times \frac{1-i}{1-i} = \frac{1-i}{2} = \frac{1}{2} - i \big( \frac{1}{2} \big)

Therefore \overline{z} = \frac{1}{2} + i \big( \frac{1}{2} \big)

iv) If z =  \frac{(3-i)^2}{2+i} = \frac{9+i^2-6i}{2+i} = \frac{8-6i}{2+i} = \frac{8-6i}{2+i} \times \frac{2-i}{2-i}

= \frac{16-12i-8i-6}{4+1} = \frac{10-20i}{5} = 2-4i

Therefore \overline{z} = 2+4i

v) If z =  \frac{(1+i)(2+i)}{3+i} = \frac{2+2i+i+i^2}{3+i} = \frac{1+3i}{3+i} = \frac{1+3i}{3+i} \times \frac{3-i}{3-i}

= \frac{3+9i - i-3i^2}{9+1} = \frac{6+8i}{10} = \frac{3}{4} + i \big( \frac{4}{5} \big)

Therefore \overline{z} = \frac{3}{4} - i \big( \frac{4}{5} \big)

vi) If z = \frac{(3-2i)(2+3i)}{(1+2i)(2-i)} = \frac{6-4i+9i-6i^2}{2+4i-i-2i^2} = \frac{12+5i}{4+3i} \times \frac{4-3i}{4-3i}

= \frac{48+20i - 36i - 15i^2}{16+9} = \frac{63}{25} - i \big( \frac{16}{25} \big) 

Therefore \overline{z} = \frac{63}{25} + i \big( \frac{16}{25} \big)

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Question 4: Find the multiplicative inverse of the following complex numbers:

i) 1-i           ii) (1+i\sqrt{3})^2           iii) 4-3i           iv) \sqrt{5}+3i

Answer:

If z = x + iy is the complex number, then multiplication inverse of z is z^{-1} or \frac{1}{z}

i) z = 1-i

z^{-1} = \frac{1}{1-i} \times \frac{1+i}{1+i} = \frac{1+i}{1-i^2} = \frac{1}{2} + \frac{1}{2} i

ii) z = (1+i\sqrt{3})^2 = 1+3i^2+i (2\sqrt{3})= -2 +i (2\sqrt{3})

z^{-1} = \frac{1}{z} = \frac{1}{-2+i (2\sqrt{3})} \times \frac{-2-i (2\sqrt{3}) }{-2-i (2\sqrt{3}) } = \frac{-2-i (2\sqrt{3}) }{4 + 12} = \frac{-1}{8} -i \frac{\sqrt{3}}{6}

iii) z = 4-3i

z^{-1} = \frac{1}{z} = \frac{1}{4-3i} \times \frac{4+3i}{4+3i} = \frac{4+3i}{16+9} = \frac{4}{25} + i \big( \frac{3}{25} \big) 

iv) z = \sqrt{5}+3i

z^{-1} = \frac{1}{z} = \frac{1}{\sqrt{5}+ 3i} \times \frac{\sqrt{5}- 3i}{\sqrt{5}-3i} = \frac{\sqrt{5}-3i}{5+9} = \frac{\sqrt{5}}{14} - \frac{3}{14} i

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Question 5: If z_1 = 2 - i, z_2 = 1+i , find \Big| \frac{z_1+z_2+1}{z_1-z_2+1} \Big|

Answer:

If z = x + iy , then |z| = \sqrt{x^2+y^2}

z_1 = 2 - i       z_2 = 1 + i

\Rightarrow z_1 + z_2 + 1 = 2 - i + 1 + i + 1 = 4

\Rightarrow z_1 - z_2 + i = 2 - i - 1 - i + 1 = 1 - i 

\Rightarrow \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} = \frac{4}{1-i} \times \frac{1+i}{1+i} = \frac{4+4i}{2} = 2+2i

\Rightarrow \Big|  \frac{z_1 + z_2 + 1}{z_1 - z_2 + i}   \Big| = \sqrt{2^2+2^2} = \sqrt{8} =2\sqrt{2}

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Question 6: If z_1 = 2-i, z_2 = -2+i , find i) Re \Big( \frac{z_1z_2}{z_1} \Big)    ii) Im \Big( \frac{1}{z_1z_2} \Big)

Answer:

z_1 = 2-i, \hspace{1.0cm} z_2 = -2+i \hspace{1.0cm} \overline{z_1} =  2 + i

i) z_1z_2 = ( 2 - i) ( -2 + i) = -4 + 2i + 2i - i^2 = - 3 + 4i

\therefore \frac{z_1z_2}{\overline{z_1}} = \frac{-3+4i}{2+i} \times \frac{2-i}{2-i} = \frac{-6+8i+3i-4i^2}{4+1} = \frac{-2+5i}{5} = \frac{-2}{5} + i

\therefore Re \big( \frac{z_1z_2}{\overline{z_1}} \big) = \frac{-2}{5}

ii) z_1 = 2 - i    \hspace{1.0cm} \overline{z_1} = 2+i

\therefore z_1 \overline{z_1} = (2-i)(2+i) = 4 + 1 = 5

\therefore \frac{1}{z_1 \overline{z_1}} = \frac{1}{5}

\therefore Im \big( \frac{1}{z_1 \overline{z_1}} \big) = 0

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Question 7: Find the modulus of \frac{1+i}{1-i} - \frac{1-i}{1+i}

Answer:

\frac{1+i}{1-i} - \frac{1-i}{1+i} = \frac{(1+i^2+2i)-(1+i^2-2i)}{1-i^2} = \frac{2i+2i}{2} = 2i

|2i| = \sqrt{2^2} = 2

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Question 8: If x+iy = \frac{a+ib}{a-ib} , prove that x^2 + y^2 = 1

Answer:

x+iy = \frac{a+ib}{a-ib}

\overline{x+iy} = \Big( \overline{\frac{a+ib}{a-ib}} \Big) =  \frac{\overline{a+ib}}{\overline{a-ib}}

x-iy = \frac{a-ib}{a+ib}

\therefore ( x +iy)(x-iy) = \frac{a+ib}{a-ib} \times \frac{a-ib}{a+ib}

\Rightarrow x^2 + y^2 = \frac{a^2 +b^2}{a^2 +b^2} = 1 

Hence proved.

Therefore the general solution is given by \theta = 2 n \pi \pm \frac{\pi}{2} , n \in Z

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Question 9: Find the least positive integral value of n for which  \Big( \frac{1+i}{1-i} \Big)^n is real.

Answer:

For n = 1

\Big( \frac{1+i}{1-i} \Big)^1 = \frac{1+i^2+2i}{2} = i which is not real.

For n = 2

\Big( \frac{1+i}{1-i} \Big)^2 = \frac{1+i^2+2i}{1+i^2-2i} = \frac{2i}{-2i} = -1 which is real.

Therefore the least positive integral value of n = 2

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Question 10: Find the real values of \theta for which the complex number \frac{1+ i \cos \theta}{1 - 2 i \cos \theta} is purely real.

Answer:

Let z = \frac{1+ i \cos \theta}{1 - 2 i \cos \theta}

= \frac{1+ i \cos \theta}{1 - 2 i \cos \theta} \times \frac{1 + 2 i \cos \theta}{1 + 2 i \cos \theta}

= \frac{1+i \cos \theta + 2 i \cos \theta + 2 i \cos \theta + 2 i^2 \cos^2 \theta}{1+4 \cos^2 \theta}

= \frac{(1-2 \cos^2 \theta) + (i ( 3 \cos \theta)}{1+4 \cos^2 \theta}

For z to be purely Real, Im (z) = 0

\therefore \frac{3 \cos \theta}{1+4 \cos^2 \theta} = 0

\Rightarrow \cos \theta = 0

\Rightarrow \theta = \frac{\pi}{2}

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Question 11: Find the smallest positive integer value of n for which \frac{(1+i)^n}{(1-i)^{n-2}}   is a real number.

Answer:

\frac{(1+i)^n}{(1-i)^{n-2}}

= \frac{ (1+i)^n }{ (1-i)^{n-2} } \times (1-i)^2

= \Big(  \frac{1+i}{1-i} \times \frac{1+i}{1+i}   \Big)^m \times ( 1+i^2 - 2i)

= \Big(  \frac{1+i^2+ 2i}{1-i^2}    \Big)^m \times ( 1-1 - 2i)

= \Big(  \frac{1-1+ 2i}{1+1}    \Big)^m \times ( - 2i)

= -2i (i^m)

= - 2 i^{m+1}

For this to be real , the smallest positive value of n   will be 1 .

Thus i^{1+1} = i^2 = - 1 , which is real.

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Question 12:  If \Big( \frac{1+i}{1-i} \Big)^3 - \Big( \frac{1-i}{1+i} \Big)^3 = x+iy , find ( x, y)

Answer:

\Big( \frac{1+i}{1-i} \Big)^3 - \Big( \frac{1+i}{1-i} \Big)^3 = x+iy

\Rightarrow \Big( \frac{1+i}{1-i} \times \frac{1+i}{1+i} \Big)^3 - \Big( \frac{1+i}{1-i} \times \frac{1-i}{1-i} \Big)^3 = x+iy

\Rightarrow \Big( \frac{2i}{2} \Big)^3 - \Big( \frac{-2i}{2} \Big)^3 = x+iy

\Rightarrow i^3 - ( -i)^3 = x + i y

\Rightarrow -i - i = x + iy

\Rightarrow -2i = x + iy

\Rightarrow x = 0   and y = - 2

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Question 13: If \frac{(1+i)^2}{2-i} = x+iy , find x+y

Answer:

\frac{(1+i)^2}{2-i} = x + i y

\Rightarrow \frac{2i}{2-i} \times \frac{2+i}{2+i} = x+iy

\Rightarrow \frac{4i + 2i^2}{4+1} = x+iy

\Rightarrow \frac{-2+4i}{5} = x+iy

\Rightarrow x = \frac{-2}{5} , \ \ \ y = \frac{4}{5}

\therefore x + y = \frac{-2}{5} +  \frac{4}{5} = \frac{2}{5}

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Question 14: If \Big( \frac{1-i}{1+ i} \Big)^{100} = a+ib , find ( a, b)

Answer:

\Big( \frac{1-i}{1+ i} \Big)^{100} = a+ib 

\Rightarrow \Big( \frac{1-i}{1+ i} \times \frac{1-i}{1-i} \Big)^{100} = a+ib  

\Rightarrow \Big( \frac{-2i}{2} \Big)^{100} = a+ib  

\Rightarrow ( -i )^{100} = a+ib  

\Rightarrow 1 = a + ib 

\therefore a = 1 , b = 0  

\therefore (a, b) = ( 1, 0) 

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Question 15: If a = \cos \theta + i \sin \theta , find the value of \frac{1+a}{1-a}

Answer:

a = \cos \theta + i \sin \theta

\frac{1+a}{1-a} = \frac{(1+\cos \theta) + i (\sin \theta) }{(1-\cos \theta) - i (\sin \theta) }

= \frac{(1+\cos \theta) + i (\sin \theta) }{(1-\cos \theta) - i (\sin \theta) } \times \frac{(1-\cos \theta) + i (\sin \theta)}{(1-\cos \theta) + i (\sin \theta)}

= \frac{(1+i \sin \theta)^2 - \cos^2 \theta}{(1-\cos \theta)^2 + \sin^2 \theta}

= \frac{1 - \sin^2 \theta + 2 i \sin \theta- \cos^2 \theta}{1+ \cos^2 \theta - 2 \cos \theta + \sin^2 \theta}

= \frac{2 i \sin \theta }{2 ( 1 - \cos \theta)}

= \frac{i \sin \theta}{1 - \cos \theta}

= i \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}}

= i \cot^2 \frac{\theta}{2}

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Question 16:

i) 2x^3+2x^2-7x+72 , when x = \frac{3-5i}{2}

ii) x^4-4x^3+4x^2+8x+44 , when x= 3 + 2i

iii) x^4+4x^3+6x^2+4x+9 , when x = - 1 +i \sqrt{2}

iv) x^6 + x^4 + x^2+1 , when x = \frac{1+i}{\sqrt{2}}

v) 2x^4 + 5x^3 + 7x^2 - x +41 , when x = - 2 - \sqrt{3}i

Answer:

i) x = \frac{3-5i}{2}

\Rightarrow 2x = 3 - 5i

\Rightarrow ( 2x - 3)^2 = ( -5 i)^2

\Rightarrow 4x^2 + 9 - 12 x = - 25

\Rightarrow 4x^2 - 12 x + 34 = 0

\Rightarrow 2x^2- 6x + 17 = 0

\Rightarrow 2x^3 + 2 x^2 - 7x + 72 = 0

Now 2x^3+2x^2-7x+72 

=  x ( 2x^2 - 6x + 17) + 6x^2 - 17x + 2x^2 - 7x + 72

= x(0) + 8x^2 - 24x + 72

= 4 ( 2x^2 - 6x +17) + 4

= 4(0) + 4 = 4

ii) x = 3 + 2i

\Rightarrow (x-3)^2 = (2i)^2

\Rightarrow x^2 + 9 - 6x = - 4

\Rightarrow x^2 - 6x + 13 = 0

Now,  x^4-4x^3+4x^2+8x+44

= x^2( x^2 - 6x + 13) + 6x^2 - 13 x^2 - 4x^3 + 4x^2 + 8x + 144

= x^2 (0) + 2x^3 - 9x^2 + 8x + 44

= 2x ( x^2 - 6x + 13) + 12x^2 - 26 x - 9x^2 +8x+ 44

= 2x (0) + 3x^2 - 18x +44

= 3 ( x^2 - 6x + 13) + 5

= 3 ( 0) + 5 = 5

iii) x = - 1 + i \sqrt{2}

\Rightarrow  ( x + 1) ^2 = ( i \sqrt{2})^2

\Rightarrow x^2 + 1 + 2 x = - 2

\Rightarrow x^2 + 2x + 3 = 0

Now x^4+4x^3+6x^2+4x+9

= x^2 ( x^2 + 2x + 3) + 2x^3 + 3x^2 + 4x + 9

= x (0) + 2x ( x^2 + 2x + 3) - ( x^2 - 2x + 9)

= 2x (0) - ( x^2 + 2x + 3) + 3 + 9

= 12

iv)  x= \frac{1+i}{\sqrt{2}}

(\sqrt{2} x)^2 = ( 1 + i)^2

\Rightarrow 2x^2 = 2i

\Rightarrow x^2 = i

\Rightarrow x^4 = - 1

\Rightarrow x^4 + 1 = 0

Now,  x^6 + x^4 + x^2+1

= x^2 ( x^4 + 1)  + ( x^4 +1 )

= x^2 ( 0) + (0) = 0

v) x = - 2 - \sqrt{3} i

x^2 = ( -2 - \sqrt{3} i)^2 = 1 + 4 \sqrt{3} i

x^3 = ( 1+4\sqrt{3} i)( - 2 - \sqrt{3} i) = 10 - 9 \sqrt{3} i

x^4 = ( 10 - 9 \sqrt{3} i) ( - 2 - \sqrt{3} i) = - 20 + 18 \sqrt{3} i - 10 \sqrt{3} i + 27 i^2 = -47 + 8 \sqrt{3} i

Now,  2x^4 + 5x^3 + 7x^2 - x +41

= 2 ( -47 + 8\sqrt{3} i) + 5 ( 10 - 9 \sqrt{3} i) + 7 ( 1 + 4 \sqrt{3} i) - ( -2 - \sqrt{3} i) + 41

= - 94  + 16 \sqrt{3} i  + 50 - 45 \sqrt{3} i + 7  + 28 \sqrt{3} i + 2 + \sqrt{3} i + 41

= ( - 94 + 50 + 7 + 2 + 41) + ( 16 \sqrt{3} i - 45 \sqrt{3} i + 28 \sqrt{3} i + \sqrt{3} i)

= 6+0 =6

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Question 17: For a positive integer, find the value of  (1-i)^n \Big( 1 - \frac{1}{i} \Big)^n

Answer:

(1-i)^n \big( 1 - \frac{1}{i} \big)^n = \Big[ (1-i) \big( 1 - \frac{1}{i} \big) \Big]^n = \Big[ \frac{(1-i)(i-1)}{i} \Big]^n = \Big[ \frac{i+1-1+i}{i} \Big]^n = 2^n

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Question 18: If (1 +i) z = (1-i) \overline{z} , then show that z = -i \overline{z}

Answer:

(1+i) z = ( 1- i) \overline{z}

\Rightarrow z = \Big( \frac{1-i}{1+i} \Big) \overline{z}

\Rightarrow z = \Big( \frac{1-i}{1+i} \times \frac{1-i}{1-i} \Big) \overline{z}

\Rightarrow z = \Big( \frac{-2i}{2} \Big) \overline{z}

\Rightarrow z = -i \overline{z}

Hence proved.

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Question 19: Solve the system of equations Re(z^2) = 0 , |z|= 2 .

Answer:

Re(z^2) = 0 , |z| = 2

Let z = x + iy

z^2 = ( x + iy)^2 = ( x^2 - y^2)+ i ( 2xy)

Re(z^2) = 0 \Rightarrow x^2 - y^2 = 0    … … … … … i)

|z| = 2

\Rightarrow \sqrt{x^2+y^2} = 2

\Rightarrow x^2 + y^2 = 4    … … … … … ii)

Adding i) and ii) we get

2x^2 = 4 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}

\therefore y = \pm \sqrt{2}

Hence x + iy = \pm \sqrt{2} \pm i \sqrt{2}

For positive sign for x : z = \sqrt{2} \pm i \sqrt{2} = \sqrt{2} ( 1 \pm i)

For negative sign for x : z = - \sqrt{2} \pm i \sqrt{2} = \sqrt{2} ( - 1 \pm i )

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Question 20: If \frac{z-1}{z+1} is purely imaginary number (z \neq -1) , find the value of |z|

Answer:

Let z = x + iy

\frac{z-1}{z+1} = \frac{x + iy-1}{x + iy+1} = \frac{(x-1) + iy}{(x +1) + iy}

=  \frac{(x-1) + iy}{(x +1) + iy} \times \frac{(x+1) - iy}{(x +1) - iy}

= \frac{[(x-1)+iy][(x+1)-iy]}{(x+1)^2 + y^2}

= \frac{x^2 -1 + i ( yx+y) - i ( xy - y) + y^2}{(x+1)^2 + y^2}

= \frac{(x^2 + y^2 -1) + i ( yx + y - xy + y)}{(x+1)^2 + y^2}

= \frac{(x^2 + y^2 -1) }{(x+1)^2 + y^2} + i \frac{ 2y}{(x+1)^2 + y^2}

If it is purely imaginary,  Re(z) = 0

\Rightarrow x^2 + y^2 - 1 = 0

\Rightarrow x^2 + y^2 = 1

\Rightarrow \sqrt{x^2 + y^2} = 1

\Rightarrow |z| = 1

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Question 21: If z_1 is a complex number other than - 1 such that |z_1|= 1 and z_2 = \frac{z_1-1}{z_1+1} then show that the real parts of z_2 is zero.

Answer:

Let z_1 = x_1 + i y_2 and z_2 = x_ 2 + i y_2

|z| = 1 \Rightarrow {x_1}^2 + {y_1}^2  = 1    … … … … … i)

z_2 = \frac{z_1 - 1}{z_1 + 1}

\Rightarrow z_2 = \frac{(x_1-1)+iy_1}{(x_1+1)+iy_1}

= \frac{(x_1-1)+iy_1}{(x_1+1)+iy_1} \times \frac{(x_1+1)- iy_1}{(x_1+1)-iy_1}

= \frac{ ({x_1}^2-1) + i ( y_1x_1+y_1) - i ( y_1x_1-y_1)+{y_1}^2}{(x_1+1)^2+{y_1}^2}

= \frac{( {x_1}^2+{y_1}^2-1)+i ( 2y_1)}{(x_1+1)^2+{y_1}^2}

=0 + i  \frac{2y_1}{(x_1+1)^2+{y_1}^2}

Since there is no real part, z_2 is purely an imaginary  numbers

\Rightarrow x_2 = 0

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Question 22: If |z+1|= z + 2 (1+i) , find z

Answer:

Let z = x + i y

|z+1| = z + 2 ( 1  + i)

\Rightarrow |x+1+iy| = ( x+iy) + 2 ( 1+i)

\Rightarrow \sqrt{(x+1)^2 + y^2} = (x+2) + i ( y + 2) 

Comparing, y + 2 = 0 \Rightarrow y = - 2

\therefore \sqrt{ (x+1)^2+y^2 } = (x+2)

\Rightarrow (x+1)^2 + 4 = ( x+2)^2

\Rightarrow x^2 + 1 + 2x + 4 = ( x+2)^2

\Rightarrow 5-4 = 2x

\Rightarrow x = \frac{1}{2}

\therefore z = \frac{1}{2} + i (-2)

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Question 23: Solve the equation |z| = z+1+2i

Answer:

Let z = x + iy

|z| = z + 1 + 2i

|x+iy| = x + iy + 1 + 2i

\sqrt{x^2 + y^2} = (x+1) + i ( y+2)

Comparing, y+2 = 0 \Rightarrow y = -2

and \sqrt{x^2 + y^2} = (x+1)

\Rightarrow x^2 + y^2 = x^2 + 1 + 2x

\Rightarrow 2x = y^2 - 1 = ( -2)^2 - 1 = 3

\Rightarrow  x = \frac{3}{2}

\therefore z = \frac{3}{2} - 2i

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Question 24: What is the smallest positive integer n for which (1+i)^{2n}= ( 1- i)^{2n} ?

Answer:

(1+i)^{2n}= ( 1- i)^{2n}

\Rightarrow [(1+i)^{2}]^n= [( 1- i)^{2}]^n

\Rightarrow (2i)^n = ( -2i)^n

\Rightarrow (2i)^n = ( -1)^n (2i)^n

\Rightarrow ( -1 )^n = 1

\therefore n is a multiple of 2 .

Hence the smallest positive number n for which (1+i)^{2n} = (1-i)^{2n} is 2

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Question 25: If z_1, z_2, z_3 , are complex numbers such that,                 |z_1|= |z_2|=|z_3|= \Big| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \Big| = 1 , then find the value of |z_1+z_2+z_3| .

Answer:

|z_1+z_2+z_3| = \Big|  z_1 \frac{\overline{z_1}}{z_1} + z_2 \frac{\overline{z_2}}{z_2} + z_3 \frac{\overline{z_3}}{z_3}  \Big|

= \Big| \frac{| z_1|^2}{\overline{z_1}} + \frac{| z_2|^2}{\overline{z_2}} + \frac{| z_3|^2}{\overline{z_3}} \Big|

= \Big| \frac{1}{\overline{z_1}} + \frac{1}{\overline{z_2}} + \frac{1}{\overline{z_3}} \Big|

= 1

\therefore  |z_1+z_2+z_3| = 1

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Question 26: Find the number of solutions of z^2 + |z|^2=0

Answer:

Let z = x + iy   \Rightarrow  |z| = \sqrt{x^2 + y^2}

Now z^2 + |z|^2 = 0

\Rightarrow  (x + iy)^2 + (\sqrt{x^2 + y^2})^2 = 0

\Rightarrow (x^2 - y^2) + 2i xy + (x^2 + y^2) = 0

\Rightarrow 2x^2 + 2ixy = 0

\Rightarrow 2x ( x+iy) = 0

Therefore x = 0    or   x + iy = 0    \Rightarrow z = 0 $

For x = 0, z = iy

Thus there are infinite many solution of the form z = 0 + iy , y \in R