1. ## Complex number equation

Hi;

Find the square roots of 5 + 12i. Hence find the roots of the equation
z^2-(1+6i)z-10=0

Thanks for helping up.

2. Hello, eureka!

Didn't I do one of these for you already?

Find the square roots of $\displaystyle 5 + 12i$.
Hence find the roots of the equation: $\displaystyle \,z^2 - (1+6i)z - 10\:=\:0$

Let $\displaystyle z \:=\:5 + 12i$

In polar form: .$\displaystyle z\:=\:13(\cos\theta + i\sin\theta)$ . . . where $\displaystyle \tan\theta = \frac{12}{5}$

Then: .$\displaystyle z^{\frac{1}{2}}\;=\;13^{\frac{1}{2}}\left[\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right]$ [1]

Since $\displaystyle \tan\theta = \frac{12}{5}$, then: $\displaystyle \cos\theta = \frac{5}{13}$

. . $\displaystyle \cos\frac{\theta}{2}\:=\:\sqrt{\frac{1 + \cos\theta}{2}} \:=\:\sqrt{\frac{1 + \frac{5}{13}}{2}} \:=\: \sqrt{\frac{18}{26}}\:=\:\frac{3}{\sqrt{13}}$

. . $\displaystyle \sin\frac{\theta}{2}\:=\:\sqrt{\frac{1 - \cos\theta}{2}} \:=\:\sqrt{\frac{1 - \frac{5}{13}}{2}} \:=\:\sqrt{\frac{8}{26}} \:=\:\frac{2}{\sqrt{13}}$

Substitute into [1]: .$\displaystyle z^{\frac{1}{2}} \;=\;\sqrt{13}\left(\frac{3}{\sqrt{13}} + i\frac{2}{\sqrt{13}}\right) \;=\;3 + 2i$

Therefore: .$\displaystyle \sqrt{5 + 12i} \;=\;\boxed{\pm(3 + 2i)}$