# Chapter 6 - Section 6.5 - Complex Numbers in Polar Form; DeMoivre's Theorem - Exercise Set - Page 767: 25

$$z=\sqrt{7} \left ( \cos 5.57 +i \sin 5.57 \right )$$

#### Work Step by Step

We plot the complex number $z=2-\sqrt{3}i$ the same way we plot $(2, -\sqrt{3})$ in the rectangular coordinate system. Please note that the complex number $z=a+bi$ is written in polar form as$$z=r(\cos \theta +i \sin \theta ),$$where $a=r\cos \theta$, $b=r\sin \theta$, $r=\sqrt{a^2+b^2}$, and $\tan \theta = \frac{b}{a}$, $0 \le \theta \lt 2\pi$. So, for the complex number $z=2-\sqrt{3}i$, we have$$r=\sqrt{2^2+(\sqrt{3})^2}=\sqrt{7}, \\ \tan \theta =\frac{-\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}.$$Since $\cos \theta =\frac{a}{r} = \frac{2}{\sqrt{7}} \gt 0$, the argument, $\theta$, must lie in quadrant $IV$. So$$\theta =\tan^{-1} (-\frac{\sqrt{3}}{2})+\pi \approx 5.57$$Thus, the complex number has the polar form$$z=\sqrt{7} \left ( \cos 5.57 +i \sin 5.57 \right ).$$

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