1. ## complex simplifying

Simplify $\sqrt{15+8i}$.

I got $4+i$ or $-4 - i$, but only one of these solutions is correct.

Obviously $\sqrt{15+8i}>0$, but with my answers, wouldn't the pesky $i$ term disable you from checking whether the expression is greater or less than 0? After all, $i$ is not on the number line... I dunno, it just feels risky dealing with that $i$ term.

thanks

2. Originally Posted by DivideBy0
Simplify $\sqrt{15+8i}$.

I got $4+i$ or $-4 - i$, but only one of these solutions is correct.

Obviously $\sqrt{15+8i}>0$, but with my answers, wouldn't the pesky $i$ term disable you from checking whether the expression is greater or less than 0? After all, $i$ is not on the number line... I dunno, it just feels risky dealing with that $i$ term.

thanks
Given any (non-zero) number $z$ its square root $\sqrt{z}$ is $\sqrt{|z|} \cdot e^{\theta i/2}$ where $\theta$ is the angle it makes with the x-axis measure $(-\pi,\pi]$.*

So if $z=15+8i$ then $|z| = \sqrt{15^2+8^2} = \sqrt{298}$.

The angle that $15+8i$ makes with the x-axis is $\theta = \tan^{-1} (15/8)$.

So,
$e^{i\cdot \left( \frac{\tan^{-1}(15/8)}{2} \right) } = \cos \left( \frac{\tan^{-1}(15/8)}{2} \right) + i \sin \left( \frac{\tan^{-1} (15/8)}{2} \right)$

Use half-angle formulas,
$\frac{1}{\sqrt{2}} \sqrt{ 1 + \cos \tan^{-1}(15/8) } + i\frac{1}{\sqrt{2}} \sqrt{ 1 - \cos \tan^{-1}(15/8)}$

Can you simplify that?

*)And this is computed with Euler formula: $e^{xi} = \cos x + i\sin x$.