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Express the following in the form $x+i y$, with $x, y \in \mathbb{R}$ :
(a) $\frac{i}{1-i}+\frac{1-i}{i}$;
(b) all the 3 rd roots of $-8 i$;
(c) $\left(\frac{i+1}{\sqrt{2}}\right)^{1337}$
in Mathematics by Gold Status (10,261 points) | 14 views

1 Answer

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Solution. (a)
$$
\begin{aligned}
\frac{i}{1-i}+\frac{1-i}{i} &=\frac{i^{2}+(1-i)^{2}}{(1-i) i} \\
&=\frac{-1-2 i}{1-i} \cdot \frac{1-i}{1-i} \\
&=\frac{-1+i-2 i-2}{2} \\
&=\frac{-3-i}{2}=-\frac{3}{2}-\frac{i}{2}
\end{aligned}
$$

(b) We have that
$$
-8 i=2^{3} \exp \left(\frac{-i \pi}{2}\right)
$$
Thus the cube roots are
$$
2 \exp \left(\frac{-i \pi}{6}\right), \quad 2 \exp \left(\frac{i \pi}{2}\right) \text { and } 2 \exp \left(\frac{7 i \pi}{6}\right)
$$
That is
$$
\sqrt{3}-i, \quad 2, \quad-\sqrt{3}-i
$$

\begin{array}{l}
\text { (c) }\\
\begin{aligned}
\left(\frac{i+1}{\sqrt{2}}\right)^{1337} &=\left(\exp \frac{i \pi}{4}\right)^{1337} \\
&=\exp \frac{1337 \pi i}{4} \\
&=\exp \left(167 \cdot 2 \pi i+\frac{\pi}{4} i\right) \\
&=\exp \frac{\pi}{4} i=\frac{1+i}{\sqrt{2}}
\end{aligned}
\end{array}
by Gold Status (10,261 points)

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