Author Topic: Q1: TUT 5101  (Read 1511 times)

Victor Ivrii

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Q1: TUT 5101
« on: September 28, 2018, 04:17:06 PM »
$\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}$
Find all solutions of the given equation.
\begin{equation*}
(z+1)^2 =1-i.
\end{equation*}

Mengyang Li

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Re: Q1: TUT 5101
« Reply #1 on: September 28, 2018, 08:16:23 PM »
Solution is scanned

Victor Ivrii

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Re: Q1: TUT 5101
« Reply #2 on: September 29, 2018, 03:38:51 PM »
Still, typed is better

Yatong Yu

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Re: Q1: TUT 5101
« Reply #3 on: September 29, 2018, 05:58:01 PM »
$(z+1)^2=1-i$ 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 $k=1$   $z=2^{1/4}e^{i7\pi  /8}-1=-2^{1/4}(\cos \pi  /8 - i\sin   \pi  /8)-1$

Do not use pitiful html options to display mathematical snippets! I fixed it
« Last Edit: September 29, 2018, 08:05:17 PM by Victor Ivrii »