### Author Topic: LEC0101 - Quiz 1 C  (Read 318 times)

#### Kuba Wernerowski

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##### LEC0101 - Quiz 1 C
« on: September 25, 2020, 12:37:22 PM »
$\textit{Find all solutions of the given equation:}$ $$(z+1)^4 = 1 - i.$$

$1 - i$ on the xy plane is equivalent to $(1, -1)$, which has an angle $\theta = \frac{-\pi}{4}$ and length $\sqrt{2} = 2^{1/2}$.

In polar representation, $$1-i = 2^{1/2} e^{i(- \pi/4 + 2k \pi)} \quad k \in \mathbb{Z}.$$

Back to the original question,

\begin{align*}

(z+1)^4 &= 2^{1/2} e^{i(- \pi/4 + 2k \pi)} \\
z+1 &= (2^{1/2} e^{i(- \pi/4 + 2k \pi)})^{1/4} \\
z &= 2^{1/8} e^{i(- \frac{\pi}{16} + \frac{1}{2}k \pi)} - 1
\end{align*}

We only need to solve for $k = 0, 1, 2, 3$ since $(z+1)^4$ has 4 roots.

$$z = \begin{cases} 2^{1/8} e^{i \left(- \frac{\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{-\pi}{16} - i \sin\frac{-\pi}{16}) & \text{for } k=0, \\ 2^{1/8} e^{i \left(\frac{7\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{7\pi}{16} - i \sin\frac{7\pi}{16}) & \text{for } k=1, \\ 2^{1/8} e^{i \left(\frac{15\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{15\pi}{16} - i \sin\frac{15\pi}{16}) & \text{for } k=2, \\ 2^{1/8} e^{i \left(\frac{23\pi}{16} \right)} - 1 = 2^{1/8} (\cos\frac{23\pi}{16} - i \sin\frac{23\pi}{16}) & \text{for } k=3 \end{cases}$$
« Last Edit: September 25, 2020, 12:39:13 PM by Kuba Wernerowski »