Solution: To solve this we will write $z = r e^{\iota \theta }$ ot $z = x + \iota y$.
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We want to find $z \in \mathbb{C} $ such that $z^3 = 1$. So,
\begin{align*}
z^3 = 1 & \implies \left( r e^{\iota \theta } \right) ^3 = 1 = e^{0}\\
& \implies r^3 e^{3\iota \theta } = e^0 \\
& \implies r^3 = 1 \text{ and } 3\theta = 2k\pi , k \in \mathbb{Z} \\
& \implies r = 1 \text{ and } \theta = \frac{2k\pi}{3}, \quad k = 0,1,2.
\end{align*}
Note that we only took three values of $k$, because for other values, the angles will repeat. Thus, the solution will be
\begin{align*}
z_1 & = e^0 = 1 \\
z_2 & = e^{\iota \frac{2\pi}{3} } = -\frac{1}{2} + \iota \frac{\sqrt{3} }{2}, \text{ and } \\
z_3 & = e^{\iota \frac{4\pi}{3} } = -\frac{1}{2} - \iota \frac{\sqrt{3}}{2}.
\end{align*}
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We want to solve $\left( \frac{z - 1}{z + 1} \right)^2 = 2 \iota $. Let $w = \frac{z - 1}{z + 1}$, and write $w = \rho e^{\iota \phi }$. Then we want to solve
\begin{align*}
w^2 = 2\iota & \implies \rho ^2 e^{2\iota \phi } = 2 e^{\iota \pi /2} \\
& \implies \rho = \sqrt{2} \text{ and } 2\phi = 2k\pi + \frac{\pi}{2}, \quad k = 0,1 \\
& \implies \rho = \sqrt{2} \text{ and } \phi = k\pi + \frac{\pi }{4}, \quad k = 0,1. \\
& \implies w_0 = \sqrt{2} e^{\iota \pi /4} \text{ and } w_1 = \sqrt{2} e^{\iota 5\pi /4} \\
& \implies w_0 = 1 + \iota \text{ and } w_1 = -1 - \iota .
\end{align*}
Now let $z = x + \iota y$. Then
\begin{align*}
\frac{z - 1}{z + 1} = w_0 & \implies \frac{x + \iota y - 1}{ x + \iota y + 1} = 1 + \iota \\
& \implies x + \iota y - 1 = x + \iota y + 1 + \iota x - y + \iota \\
& \implies 2 - y + \iota (x + 1) = 0 \\
& \implies x = -1 \text{ and } y = 2.
\end{align*}
Thus,
\[
z_0 = -1 + 2\iota .
\]
Similarly,
\begin{align*}
\frac{z - 1}{z + 1} = -1 - \iota & \implies x + \iota y - 1 = -x - \iota y - 1 - \iota x + y - \iota \\
& \implies x - 1 = -x + y - 1 \text{ and } y = -y -x - 1\\
& \implies 2x - y = 0 \text{ and } 2y + x - 1 = 0 \\
& \implies x = -\frac{1}{5} \text{ and } y = -\frac{2}{5}.
\end{align*}
Thus,
\[
z_1 = \frac{-1 - 2 \iota }{5}.
\]
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We want to solve $z^2 = 24 \iota - 7$. Take $z = x + \iota y$. Then
\begin{align*}
(x + \iota y)^2 = 24 \iota - 7 & \implies x^2 - y^2 + \iota 2xy = 24 \iota - 7 \\
& \implies \left\{
\begin{array}{c}
x^2 - y^2 = -7 \\
2xy = 24 \\
\end{array}
\right. \\
& \implies \left\{
\begin{array}{c}
x^2 - y^2 = -7 \\[1ex]
y = \dfrac{12}{x}
\end{array}
\right.
\end{align*}
Then substituting $y = \frac{12}{x}$ in the first equation yields,
\begin{align*}
x^2 - \left( \frac{12}{x} \right) ^2 + 7 = 0 & \implies x^4 - 144 + 7x^2 = 0 \\
& \implies \left( x^2 - 9 \right) \left( x^2 + 16 \right) = 0 \\
& \implies x = \pm 3 \quad (\text{ since } x \in \mathbb{R} ).
\end{align*}
Therefore,
\[
y = \pm 4 \text{ and } z = \pm 3 \pm 4\iota .
\]
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We want to solve $z^3 = \iota - 1$. Take $z = r e^{\iota \theta }$. Note that $\iota - 1 = \sqrt{2} e^{\iota \frac{3\pi}{4}}$. Thus,
\begin{align*}
z^3 = \iota - 1 & \implies r^3 e^{\iota 3\theta } = \sqrt{2} e^{\iota \frac{3\pi}{4}} \\
& \implies r^3 = \sqrt{2} \text{ and } e^{\iota 3\theta } = e^{\iota 3\pi /4} \\
& \implies r = \sqrt[6]{2} \text{ and } 3\theta = 2k\pi + 3\pi /4, \quad k = 0,1,2 \\
& \implies r = \sqrt[6]{2} \text{ and } \theta = \frac{2k\pi}{3} + \frac{\pi}{4}, \quad k = 0,1,2.
\end{align*}
Therefore,
\begin{align*}
z_0 & = \sqrt[6]{2} e^{\iota \frac{\pi}{4}} , \quad z_1 = \sqrt[6]{2} e^{\iota \left(\frac{11\pi}{12}\right)} \text{ and } z_2 = \sqrt[6]{2} e^{\iota \left(\frac{19\pi}{12}\right)}.
\end{align*}
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We want to solve $z^4 = -1$. Take $z = r e^{\iota \theta }$. Then,
\begin{align*}
z^4 = -1 & \implies r^4 e^{\iota 4 \theta } = e^{\iota \pi} \\
& \implies r^4 = 1 \text{ and } 4\theta = \pi + 2k\pi, \quad k \in \mathbb{Z} \\
& \implies r = 1 \text{ and } \theta = \frac{\pi}{4} + \frac{k\pi}{2}, \quad k = 0, 1, 2, 3.
\end{align*}
Thus,
\begin{align*}
z_0 & = e^{\iota \frac{\pi}{4}} = \frac{1 + \iota }{\sqrt{2} } \\
z_1 & = e^{\iota \frac{3\pi}{4}} = \frac{-1 + \iota }{\sqrt{2} } \\
z_2 & = e^{\iota \frac{5\pi}{4}} = \frac{-1 - \iota }{\sqrt{2} } \\
z_3 & = e^{\iota \frac{7\pi}{4}} = \frac{-1 + \iota }{\sqrt{2} }.
\end{align*}