Exercise 3.2.2

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

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ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

In Section A.2, we use polar coordinates to construct square (and higher) roots of complex numbers. In this exercise, you will give an elementary argument that every complex number has a square root. The only fact you will use (besides standard algebra) is that every positive real number has a real square root.
\begin{enumerate}
\item[(a)] First explain why every real number has a square root in $\C$.
\item[(b)] Now fix $a+bi \in \C$ with $b 
e 0$. For $x,y \in \R$, show that the equation $(x+iy)^2 = a+bi$ is equivalent to the equations
$$x^2-y^2 = a, \qquad 2xy = b.$$
\item[(c)] Show that the equation of part (b) are equivalent to
$$x^2 =\frac{a\pm \sqrt{a^2+b^2}}{2}, \qquad y = \frac{b}{2x}.$$
Also show that $x
e 0$ and that $a\pm \sqrt{a^2+b^2}$ is positive when we choose the $+$ sign in the formula for $x^2$.
\item[(d)] Conclude that $a+bi$ has a square root in $\C$.
\end{enumerate}

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
\begin{enumerate}
\item[(a)] 
We know that the equation $x^2 = a$ has a real solution if $a\geq 0$ (see Ex. 3.2.3).
Therefore, if $a \in \mathbb{R}^-_*$, there exists $b\in \mathbb{R}^+$ such that $b^2 = -a = \vert a \vert$.
Thus $(ib)^2 = a$.

Conclusion: Every $a \in \mathbb{R}$ has a square root in $\mathbb{C}$.

\item[(b,c,d)] Let $z = a+ib,\ a,b \in \mathbb{R}$, and $Z = x+iy,\ x,y \in \mathbb{R}$ two complex numbers.

\begin{align*}
z^2 = Z &\iff (a+ ib)^2 = x+iy\
&\iff (a+ ib)^2 = x+iy \ \mathrm{and} \ \vert a+ ib\vert ^2 = \vert x+iy\vert \
&\iff a^2 - b^2 +2abi = x+iy\ \mathrm{and} \ a^2+b^2 = \sqrt{x^2+y^2}\
&\iff a^2 - b^2 = x, a^2 + b^2 =\sqrt{x^2+y^2}, 2ab = y.
\end{align*}

The system of equations
$
\left\{
\begin{array}{ccc}
a^2 - b^2 &=& x,   \
a^2 + b^2 &=&\sqrt{x^2+y^2}, 
\end{array}
ight.
$
is equivalent to
$$
\left\{
\begin{array}{ccc}
a^2  &=&\frac{1}{2}\left(  \sqrt{x^2+y^2}  + xight),  \
 b^2 &=&\frac{1}{2}\left( \sqrt{x^2+y^2} -xight).
\end{array}
ight.
$$
Therefore $$z^2 = Z \Rightarrow 
\left\{
\begin{array}{ccc}
a^2  &=&\frac{1}{2}\left(  \sqrt{x^2+y^2}  + xight),  \
 b^2 &=&\frac{1}{2}\left( \sqrt{x^2+y^2} -xight), \
 \mathrm{sgn}(ab) &= & \mathrm{sgn}(y).
\end{array}
ight.
$$

The converse is true, since these last equations imply
$$4a^2 b^2 = \left(  \sqrt{x^2+y^2}  + xight) \left( \sqrt{x^2+y^2} -xight) = x^2+y^2-x^2 = y^2,$$
and since $\mathrm{sgn}(ab) =  \mathrm{sgn}(y)$, we conclude $2ab=y$. So we have proved the equivalence

$$z^2 = Z \iff
\left\{
\begin{array}{ccc}
a^2  &=&\frac{1}{2}\left(  \sqrt{x^2+y^2}  + xight),  \
 b^2 &=&\frac{1}{2}\left( \sqrt{x^2+y^2} -xight), \
 \mathrm{sgn}(ab) &= & \mathrm{sgn}(y).
\end{array}
ight. 
$$

As $x^2+y^2 \geq x^2, \sqrt{x^2+y^2} \geq \vert x \vert $, and $\vert x \vert \geq x, \vert x \vert \geq -x$, so
\begin{align*}z^2 = Z &\iff
\left\{
\begin{array}{ccl}
a  &=&\varepsilon \sqrt{\frac{1}{2}\left ( \sqrt{x^2+y^2}  + xight)}  \
 b &=&\varepsilon\ \mathrm{sgn}(y)\sqrt{\frac{1}{2}\left( \sqrt{x^2+y^2} -xight)},\qquad  \varepsilon \in \{-1,1\} \
\end{array}
ight.\
&\iff z \in \{ z_0, - z_0   \},
\end{align*}
where $$z_0 = \sqrt{\frac{1}{2}\left ( \sqrt{x^2+y^2}  + xight)} + i \ \mathrm{sgn}(y)\sqrt{\frac{1}{2}\left( \sqrt{x^2+y^2} -xight)}.$$
Conclusion: Every $z \in \mathbb{C}$ has a square root in $\C$.
\end{enumerate}
\end{proof}

Exercise 3.2.2

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Exercise 3.2.2

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