Question 1: Find the square root of the following complex numbers:

i) -5 + 12 i           ii) -7-24i           iii) 1-i          iv) -8-6i

v) 8-15i           vi) -11-60\sqrt{-1}           vii) 1+4 \sqrt{-3}          viii) 4i           ix) -i

Answer:

There are two ways by which we can solve the problems. We have a few solved by each method.

i) Let \sqrt{-5+12i} = x+iy

Squaring both sides

\Rightarrow -5 + 12 i = ( x + iy)^2

\Rightarrow -5 + 12 i = ( x^2 - y^2) + i ( 2xy)

\Rightarrow x^2 - y^2 = - 5     … … … … … i)

\Rightarrow 2xy = 12     … … … … … ii)

Now (x^2 + y^2)^2 = ( x^2 - y^2)^2 + 4 x^2 y^2

\Rightarrow (x^2 + y^2)^2 = ( - 5)^2 + (12)^2

\Rightarrow (x^2 + y^2)^2 = 169

\Rightarrow x^2 + y^2 = 13     … … … … … iii)  [ since x^2 + y^2 > 0 ]

Solving i) and iii)

2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2

From i) y^2 = x^2 + 5  \Rightarrow y^2 = 4 + 5  \Rightarrow y = \pm 3

Since 2xy is positive, therefore x and y are of same signs.

\therefore \sqrt{-5+12i} = \pm ( 2 + 3i)

ii)      We know

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) > 0 

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) < 0 

Let z = -7 - 24 i      \therefore Re(z) = - 7      |z| = \sqrt{(-7)^2+(-24)^2} = 25

Here Im(z) < 0 ,  therefore

\sqrt{-7 - 24 i} = \pm \Bigg\{ \sqrt{\frac{25+(-7)}{2}} - i \sqrt{\frac{25-(-7)}{2}} \Bigg\} 

\Rightarrow \sqrt{-7 - 24 i} = \pm (  3-4i )

iii)    We know

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) > 0 

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) < 0 

Let z = 1-i      \therefore Re(z) = 1      |z| = \sqrt{(1)^2+(-1)^2} = \sqrt{2}

Here Im(z) < 0 ,  therefore

\sqrt{1-i} = \pm \Bigg\{ \sqrt{\frac{\sqrt{2}+1}{2}} - i \sqrt{\frac{\sqrt{2} - 1}{2}} \Bigg\}, 

iv)     We know

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) > 0 

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) < 0 

Let z = -8-6i      \therefore Re(z) = -8      |z| = \sqrt{(-8)^2+(-6)^2} = 10

Here Im(z) < 0 ,  therefore

\sqrt{-8-6i} = \pm \Bigg\{ \sqrt{\frac{10-8}{2}} - i \sqrt{\frac{10+8}{2}} \Bigg\} 

\Rightarrow \sqrt{-7 - 24 i} = \pm (  1-3i )

v)      We know

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) > 0 

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) < 0 

Let z = 8-15i      \therefore Re(z) = 8      |z| = \sqrt{(8)^2+(-15)^2} = 17

Here Im(z) < 0 ,  therefore

\sqrt{8-15i} = \pm \Bigg\{ \sqrt{\frac{17+8}{2}} - i \sqrt{\frac{17-8}{2}} \Bigg\} 

\Rightarrow \sqrt{8-15i} = \pm \frac{1}{\sqrt{2}} (  5-3i )

vi)     Let \sqrt{-11-60i} = x+iy

Squaring both sides

\Rightarrow -11-60i = ( x + iy)^2

\Rightarrow -11-60i = ( x^2 - y^2) + i ( 2xy)

\Rightarrow x^2 - y^2 = -11     … … … … … i)

\Rightarrow 2xy = -60     … … … … … ii)

Now (x^2 + y^2)^2 = ( x^2 - y^2)^2 + 4 x^2 y^2

\Rightarrow (x^2 + y^2)^2 = ( - 11)^2 + (-60)^2

\Rightarrow (x^2 + y^2)^2 = 3121

\Rightarrow x^2 + y^2 = 61     … … … … … iii)  [ since x^2 + y^2 > 0 ]

Solving i) and iii)

2x^2 = 50 \Rightarrow x^2 = 25 \Rightarrow x = \pm 5

From i) y^2 = x^2 + 11  \Rightarrow y^2 = 25 + 11  \Rightarrow y = \pm 6

Since 2xy is negative, therefore x and y are of opposite signs.

\therefore \sqrt{-11-60i} = \pm ( 5-6i)

vii)    Let \sqrt{1 + 4\sqrt{3} i} = x+iy

Squaring both sides

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

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

\Rightarrow x^2 - y^2 = 1     … … … … … i)

\Rightarrow 2xy = 4\sqrt{3}     … … … … … ii)

Now (x^2 + y^2)^2 = ( x^2 - y^2)^2 + 4 x^2 y^2

\Rightarrow (x^2 + y^2)^2 = ( 1)^2 + (4\sqrt{3})^2

\Rightarrow (x^2 + y^2)^2 = 49

\Rightarrow x^2 + y^2 = 7     … … … … … iii)  [ since x^2 + y^2 > 0 ]

Solving i) and iii)

2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2

From i) y^2 = x^2 -1  \Rightarrow y^2 = 3  \Rightarrow y = \pm \sqrt{3}

Since 2xy is positive, therefore x and y are of same signs.

\therefore \sqrt{1 + 4\sqrt{3} i} = \pm ( 2 + \sqrt{3} i)

viii)   We know

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) > 0 

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) < 0 

Let z = 4i      \therefore Re(z) = 0      |z| = \sqrt{(0)^2+(4)^2} = 4

Here Im(z) > 0 ,  therefore

\sqrt{4i} = \pm \Bigg\{ \sqrt{\frac{4}{2}} + i \sqrt{\frac{4}{2}} \Bigg\} 

\Rightarrow \sqrt{4 i} = \pm ( \sqrt{2} + i \sqrt{2} ) = \pm \sqrt{2}  ( 1 + i)

ix)     We know

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) > 0 

\sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \  Im(z) < 0 

Let z = -i      \therefore Re(z) = 0      |z| = \sqrt{(0)^2+(-1)^2} = 1

Here Im(z) < 0 ,  therefore

\sqrt{-i} = \pm \Bigg\{ \sqrt{\frac{1+0}{2}} - i \sqrt{\frac{1-0}{2}} \Bigg\} 

\Rightarrow \sqrt{-i} = \pm \frac{1}{\sqrt{2}} ( 1-i )

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