Question 1: Find the square root of the following complex numbers: $\displaystyle \text{i) } -5 + 12 i \hspace{1.0cm} \text{ii) } -7-24i \hspace{1.0cm} \text{iii) } 1-i \hspace{1.0cm} \text{iv) } -8-6i$ $\displaystyle \text{v) } 8-15i \hspace{1.0cm} \text{vi) } -11-60\sqrt{-1} \hspace{1.0cm} \text{vii) } 1+4 \sqrt{-3} \hspace{1.0cm} \text{viii) } 4i \hspace{1.0cm} \text{ix) }-i$

There are two ways by which we can solve the problems. We have a few solved by each method. $\displaystyle \text{i) } \text{Let } \sqrt{-5+12i} = x+iy$

Squaring both sides $\displaystyle \Rightarrow -5 + 12 i = ( x + iy)^2$ $\displaystyle \Rightarrow -5 + 12 i = ( x^2 - y^2) + i ( 2xy)$ $\displaystyle \Rightarrow x^2 - y^2 = - 5$ … … … … … i) $\displaystyle \Rightarrow 2xy = 12$ … … … … … ii) $\displaystyle \text{Now } (x^2 + y^2)^2 = ( x^2 - y^2)^2 + 4 x^2 y^2$ $\displaystyle \Rightarrow (x^2 + y^2)^2 = ( - 5)^2 + (12)^2$ $\displaystyle \Rightarrow (x^2 + y^2)^2 = 169$ $\displaystyle \Rightarrow x^2 + y^2 = 13$ … … … … … iii) [ since $\displaystyle x^2 + y^2 > 0$ ]

Solving i) and iii) $\displaystyle 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$ $\displaystyle \text{From } \text{i) } y^2 = x^2 + 5 \Rightarrow y^2 = 4 + 5 \Rightarrow y = \pm 3$

Since $\displaystyle 2xy$ is positive, therefore $\displaystyle x$ and $\displaystyle y$ are of same signs. $\displaystyle \therefore \sqrt{-5+12i} = \pm ( 2 + 3i)$ $\displaystyle \text{ii) } \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) > 0$ $\displaystyle \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) < 0$ $\displaystyle \text{Let } z = -7 - 24 i \therefore Re(z) = - 7 |z| = \sqrt{(-7)^2+(-24)^2} = 25$

Here $\displaystyle Im(z) < 0$, therefore $\displaystyle \sqrt{-7 - 24 i} = \pm \Bigg\{ \sqrt{\frac{25+(-7)}{2}} - i \sqrt{\frac{25-(-7)}{2}} \Bigg\}$ $\displaystyle \Rightarrow \sqrt{-7 - 24 i} = \pm ( 3-4i )$ $\displaystyle \text{iii) } \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) > 0$ $\displaystyle \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) < 0$ $\displaystyle \text{Let } z = 1-i \therefore Re(z) = 1 |z| = \sqrt{(1)^2+(-1)^2} = \sqrt{2}$

Here $\displaystyle Im(z) < 0$, therefore $\displaystyle \sqrt{1-i} = \pm \Bigg\{ \sqrt{\frac{\sqrt{2}+1}{2}} - i \sqrt{\frac{\sqrt{2} - 1}{2}} \Bigg\},$ $\displaystyle \text{iv) } \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) > 0$ $\displaystyle \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) < 0$ $\displaystyle \text{Let } z = -8-6i \therefore Re(z) = -8 |z| = \sqrt{(-8)^2+(-6)^2} = 10$

Here $\displaystyle Im(z) < 0$, therefore $\displaystyle \sqrt{-8-6i} = \pm \Bigg\{ \sqrt{\frac{10-8}{2}} - i \sqrt{\frac{10+8}{2}} \Bigg\}$ $\displaystyle \Rightarrow \sqrt{-7 - 24 i} = \pm ( 1-3i )$ $\displaystyle \text{v) } \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) > 0$ $\displaystyle \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) < 0$ $\displaystyle \text{Let } z = 8-15i \therefore Re(z) = 8 |z| = \sqrt{(8)^2+(-15)^2} = 17$

Here $\displaystyle Im(z) < 0$, therefore $\displaystyle \sqrt{8-15i} = \pm \Bigg\{ \sqrt{\frac{17+8}{2}} - i \sqrt{\frac{17-8}{2}} \Bigg\}$ $\displaystyle \Rightarrow \sqrt{8-15i} = \pm \frac{1}{\sqrt{2}} ( 5-3i )$ $\displaystyle \text{vi) } \text{Let } \sqrt{-11-60i} = x+iy$

Squaring both sides $\displaystyle \Rightarrow -11-60i = ( x + iy)^2$ $\displaystyle \Rightarrow -11-60i = ( x^2 - y^2) + i ( 2xy)$ $\displaystyle \Rightarrow x^2 - y^2 = -11$ … … … … … i) $\displaystyle \Rightarrow 2xy = -60$ … … … … … ii) $\displaystyle \text{Now } (x^2 + y^2)^2 = ( x^2 - y^2)^2 + 4 x^2 y^2$ $\displaystyle \Rightarrow (x^2 + y^2)^2 = ( - 11)^2 + (-60)^2$ $\displaystyle \Rightarrow (x^2 + y^2)^2 = 3121$ $\displaystyle \Rightarrow x^2 + y^2 = 61$ … … … … … iii) [ since $\displaystyle x^2 + y^2 > 0$ ]

Solving i) and iii) $\displaystyle 2x^2 = 50 \Rightarrow x^2 = 25 \Rightarrow x = \pm 5$ $\displaystyle \text{From } \text{i) } y^2 = x^2 + 11 \Rightarrow y^2 = 25 + 11 \Rightarrow y = \pm 6$

Since $\displaystyle 2xy$ is negative, therefore $\displaystyle x$ and $\displaystyle y$ are of opposite signs. $\displaystyle \therefore \sqrt{-11-60i} = \pm ( 5-6i)$ $\displaystyle \text{vii) } \text{Let } \sqrt{1 + 4\sqrt{3} i} = x+iy$

Squaring both sides $\displaystyle \Rightarrow 1 + 4\sqrt{3} i = ( x + iy)^2$ $\displaystyle \Rightarrow 1 + 4\sqrt{3} i = ( x^2 - y^2) + i ( 2xy)$ $\displaystyle \Rightarrow x^2 - y^2 = 1$ … … … … … i) $\displaystyle \Rightarrow 2xy = 4\sqrt{3}$ … … … … … ii) $\displaystyle \text{Now } (x^2 + y^2)^2 = ( x^2 - y^2)^2 + 4 x^2 y^2$ $\displaystyle \Rightarrow (x^2 + y^2)^2 = ( 1)^2 + (4\sqrt{3})^2$ $\displaystyle \Rightarrow (x^2 + y^2)^2 = 49$ $\displaystyle \Rightarrow x^2 + y^2 = 7$ … … … … … iii) [ since $\displaystyle x^2 + y^2 > 0$ ]

Solving i) and iii) $\displaystyle 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$ $\displaystyle \text{From } \text{i) } y^2 = x^2 -1 \Rightarrow y^2 = 3 \Rightarrow y = \pm \sqrt{3}$

Since $\displaystyle 2xy$ is positive, therefore $\displaystyle x$ and $\displaystyle y$ are of same signs. $\displaystyle \therefore \sqrt{1 + 4\sqrt{3} i} = \pm ( 2 + \sqrt{3} i)$ $\displaystyle \text{viii) } \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) > 0$ $\displaystyle \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) < 0$ $\displaystyle \text{Let } z = 4i \therefore Re(z) = 0 |z| = \sqrt{(0)^2+(4)^2} = 4$

Here $\displaystyle Im(z) > 0$, therefore $\displaystyle \sqrt{4i} = \pm \Bigg\{ \sqrt{\frac{4}{2}} + i \sqrt{\frac{4}{2}} \Bigg\}$ $\displaystyle \Rightarrow \sqrt{4 i} = \pm ( \sqrt{2} + i \sqrt{2} ) = \pm \sqrt{2} ( 1 + i)$ $\displaystyle \text{ix) } \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} + i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) > 0$ $\displaystyle \sqrt{z} = \pm \Bigg\{ \sqrt{\frac{|z|+Re(z)}{2}} - i \sqrt{\frac{|z|-Re(z)}{2}} \Bigg\}, \ \text{if} \ Im(z) < 0$ $\displaystyle \text{Let } z = -i \therefore Re(z) = 0 |z| = \sqrt{(0)^2+(-1)^2} = 1$

Here $\displaystyle Im(z) < 0$, therefore $\displaystyle \sqrt{-i} = \pm \Bigg\{ \sqrt{\frac{1+0}{2}} - i \sqrt{\frac{1-0}{2}} \Bigg\}$ $\displaystyle \Rightarrow \sqrt{-i} = \pm \frac{1}{\sqrt{2}} ( 1-i )$