Toronto Math Forum
MAT334-2018F => MAT334--Tests => Quiz-1 => Topic started by: Victor Ivrii on September 28, 2018, 04:17:06 PM
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$\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}$
Find all solutions of the given equation.
\begin{equation*}
(z+1)^2 =1-i.
\end{equation*}
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Solution is scanned
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Still, typed is better
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$(z+1)^2=1-i$ which is $(1, -1)$ on the axis with angle $-\pi /4$ and length $2^{1/2}$ so $1-i = 2^{1/2}e^{i(-\pi /4+2k\pi )}$
$$z+1=(1-i)^{1/2} =\bigl(2^{1/2}e^{i(-\pi /4+2k\pi )}\bigr)^{1/2}$$ $$z=2^{1/4}e^{i(-\pi /8 + k\pi )}-1$$
when $k=0$ $z=2^{1/4}e^{-i\pi /8}-1=2^{1/4}(\cos \pi /8 - i\sin \pi /8)-1$
when $k=1$ $z=2^{1/4}e^{i7\pi /8}-1=-2^{1/4}(\cos \pi /8 - i\sin \pi /8)-1$
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