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Write the following numbers in the polar form $π‘Ÿπ‘’^{π‘–πœƒ}, βˆ’πœ‹<πœƒβ‰€πœ‹$:

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

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

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

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  • $\begingroup$ why would you take tan(-170.265)? $\endgroup$ Nov 13, 2019 at 8:10
  • $\begingroup$ Put your software down and draw some pictures. If you sketch $-\sqrt{7} (1+i)$ you can see it makes an angle of 235 degrees anticlockwise from the positive real axis. Similarly, $\sqrt{2} + i$ in the first quadrant makes an angle of 35.26 degrees. If you divide two complex numbers you subtract their arguments which is 189.74 degrees or -170.26 degrees if you measure clockwise from the positive real axis. $\endgroup$
    – Paul
    Nov 13, 2019 at 8:45
  • $\begingroup$ Don't use degrees. $\endgroup$
    – user65203
    Nov 13, 2019 at 9:40

2 Answers 2

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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.)

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Firstly, you can rewrite $\frac{βˆ’\sqrt7(1+i)}{\sqrt2+i}$ 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 $$\begin{equation*} \begin{cases} r\cos(\theta) = -\frac{\sqrt7+\sqrt14}{3}\\ r\sin(\theta) = \frac{\sqrt7 - \sqrt14}{3} \end{cases} \end{equation*}$$ $$or$$ $$\begin{equation*} \begin{cases} \tan(\theta) = -\frac{\sqrt7-\sqrt14}{\sqrt7+\sqrt14}\\ r^2 = \frac{14}{3} \end{cases} \end{equation*}$$

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