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) $-i$

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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