Convert complex to polar cord

Question

Write the following numbers in the polar form $^{}, −<≤$:

$r$ = ?, $\theta$ =?

$$-\dfrac{\sqrt{7}(1+i)}{\sqrt{2}+ i} $$

When i used wolfram I got $r = 2.16025$ and $\theta = -170.264^{°}$. However, when I input this in WebAssign I got the right answer for $r$ but the wrong answer for $\theta$. I then tried doing $tan(-170.264^{°}) =.171579$ (in degree) and $tan(-170.264) = -.710889017$ (in radians), and they were both wrong. I'm lost what should be $\theta$ ?

Answer 1

Firstly, you can rewrite $\frac{−\sqrt7(1+i)}{\sqrt2+i}$ as $-\frac{\sqrt7+\sqrt14}{3} + i(\frac{\sqrt7 - \sqrt14}{3})$. Then use that $re^{i\theta} = r(cos(\theta) + isin(\theta))$ and you get that $$\begin{equation*} \begin{cases} r\cos(\theta) = -\frac{\sqrt7+\sqrt14}{3}\\ r\sin(\theta) = \frac{\sqrt7 - \sqrt14}{3} \end{cases} \end{equation*}$$ $$or$$ $$\begin{equation*} \begin{cases} \tan(\theta) = -\frac{\sqrt7-\sqrt14}{\sqrt7+\sqrt14}\\ r^2 = \frac{14}{3} \end{cases} \end{equation*}$$

Answer 2

Considering all factors in turn, the modulus will be

$$\frac{\sqrt7\sqrt2}{\sqrt3}$$

and the argument

$$-\pi+\frac{\pi}4-\arctan\frac1{\sqrt2}.$$

($-\pi$ is chosen rather than $\pi$ so that the sum be in the allowed range.)