Author Topic: Quiz 1 (Quiz-5101)  (Read 362 times)

kaye

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Quiz 1 (Quiz-5101)
« on: September 25, 2020, 01:09:15 PM »
Problem(3pt). Find all solutions of the given equation:
$$z^5=-i .$$

Let $z=r(cos \theta + isin \theta)$, then $z^5 = r^5[cos(5\theta)+isin(5\theta)]$:
$$-i=1\left[cos\left(\frac{3\pi}{2}+2k\pi\right) +isin\left(\frac{3\pi}{2}+2k\pi\right)\right].$$

Therefore we get
$$r=1^{\frac{1}{5}}=1 \text{ and } \theta=\frac{1}{5}\left(\frac{3\pi}{2}+2k\pi\right) \text{ for } k=0,1,2,3,4$$

The five solutions of the given equation:
\begin{align*}
    k&=0: \theta = \frac{3\pi}{10} \Rightarrow z=cos\left(\frac{3\pi}{10}\right) + isin\left(\frac{3\pi}{10}\right)\\
    k&=1: \theta = \frac{7\pi}{10} \Rightarrow z=cos\left(\frac{7\pi}{10}\right) + isin\left(\frac{7\pi}{10}\right)\\
    k&=2: \theta = \frac{11\pi}{10} \Rightarrow z=cos\left(\frac{11\pi}{10}\right) + isin\left(\frac{11\pi}{10}\right)\\
    k&=3: \theta = \frac{3\pi}{2} \Rightarrow z=cos\left(\frac{3\pi}{2}\right) + isin\left(\frac{3\pi}{2}\right)=-i\\
    k&=4: \theta = \frac{19\pi}{10} \Rightarrow z=cos\left(\frac{19\pi}{10}\right) + isin\left(\frac{19\pi}{10}\right)
\end{align*}