Exercise 1.10

Question

Exercise 10: Suppose $z=a+ib$, $w=u+iv$, and $$a=\biggl(\frac{|w|+u}{2}\biggr)^{1/2},\quad\quad b=\biggl(\frac{|w|-u}{2}\biggr)^{1/2}.$$ Prove that $z^2=w$ if $v\ge 0$ and that $(\overline{z})^2=w$
if $v\le 0$. Conclude that every complex number (with one exception!) has two complex square roots.

analambanomenos · Accepted Answer

We have
$$(a^2-b^2) = \frac{|w|+u}{2}-\frac{|w|-u}{2} = u$$ $$2ab = (|w|+u)^{1/2}(|w|-u)^{1/2} = (|w|^2-u^2)^{1/2} = (v^2)^{1/2} = |v|.$$
    Hence $z^2=(a^2-b^2)+2abi=u+|v|i=w$ if $v\ge 0$, and $(\overline{z}^2)=(a^2-b^2)-2abi=u-|v|i=w$ if $v\le 0$. Hence every nonzero $w$ has two square roots $\pm z$ or $\pm\overline{z}$. Of course, 0 has
only one square root, itself.

Amit Awasthi · Answer

\begin{proof}
      Consider $z^2$ when $v \ge 0$:
      \begin{align*}
        z^2 &= (a+bi)^2 = a^2 + 2abi - b^2\
        &= u + i\left( |w| + u ight)^{1/2}\left( |w| - u ight)^{1/2}\
        &= u + i\left( |w|^2 - u^2 ight)^{1/2}\
        &= u + i\left( (u+iv)(u-iv) - u^2 ight)^{1/2}\
        &= u + i\left( v^2 ight)^{1/2} = u + iv = w.
        \intertext{Consider $(\overline{z})^2$ when $v \le 0$:}
        (\overline{z})^2 &= (a-bi)^2 = a^2 - 2abi - b^2\
        &= u - i\left( |w| + u ight)^{1/2}\left( |w| - u ight)^{1/2}\
        &= u - i\left( |w|^2 - u^2 ight)^{1/2}\
        &= u - i\left( (u+iv)(u-iv) - u^2 ight)^{1/2}\
        &= u - i\left( v^2 ight)^{1/2} = u + iv = w.
      \end{align*}
      Thus, every complex number $w = u + iv$ has two roots $z,-z$ when $v \ge 0$, or two roots $\overline{z},-\overline{z}$ when $v \le 0$, unless $z = -z$ or $\overline{z} = -\overline{z}$, that is, $u = v = 0$ which implies $w = 0$.
\end{proof}

Exercise 1.10

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Exercise 1.10

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