【归档】What is the square root of i?

Note: 旧的wordpress博客弃用，于是将以前的笔记搬运回来。

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Solution:
Let $z=a+bi$ be the complex number which is a square root of i, that is
$z^2=(a+bi{)}^2=a^2-b^2+2abi=i$
Equating real and imaginary parts, we have
$\{\begin{array}{ll}{a}^2-b^2& =0,\\ 2ab& =1.\end{array}$
The two real solution to this pair of equations are
$\{\begin{array}{l}a=\frac{1}{\sqrt{2}},\\ b=\frac{1}{\sqrt{2}},\end{array}$
and
$\{\begin{array}{l}a=-\frac{1}{\sqrt{2}},\\ b=-\frac{1}{\sqrt{2}}.\end{array}$
The two square root of i therefor are $\pm \frac{1}{\sqrt{2}}(i+1)$.