Complex Identities
Proposition W.4.1 Let $w,z\in\mathbb{C}$. Then $\overline{wz}=\overline{w}\;\overline{z}$.
Proof Let $w=a+bi$ and $z=c+di$. Then
$$\begin{align*} \overline{wz} &= \overline{(a+bi)(c+di)} \\ &= \overline{ac+adi+bci+bdi^2} \\ &= \overline{ac+adi+bci-bd} \\ &= \overline{ac+adi+bci+(-bd)} \\ &= \overline{ac}+\overline{adi}+\overline{bci}+\overline{-bd} \\ &= ac-adi-bci-bd \\ &= ac-adi-bci+bdi^2 \\ &= (a-bi)(c-di) \\ &= \overline{a+bi}\;\overline{c+di} \\ &= \overline{w}\;\overline{z}\quad\blacksquare \end{align*}$$
Proposition W.4.2 Let $w,z\in\mathbb{C}$. Then $\overline{w}z+w\overline{z}=2\Re{\big(\overline{w}z\big)}$.
Proof Let $w=a+bi$ and $z=c+di$. Then
$$\begin{align*} 2\Re{\big(\overline{w}z\big)} &= 2\Re{\Big(\overline{(a+bi)}(c+di)\Big)} \\ &= 2\Re{\Big((a-bi)(c+di)\Big)} \\ &= 2\Re{\Big(ac+adi-bci-bdi^2\Big)} \\ &= 2\Re{\Big(ac+(ad-bc)i+bd\Big)} \\ &= 2\Re{\Big((ac+bd)+(ad-bc)i\Big)} \\ &= 2(ac+bd) \\ &= ac+bd+ac+bd \\ &= ac-bdi^2+ac-bdi^2 \\ &= ac-bdi^2+ac-bdi^2+adi-bci-adi+bci \\ &= ac+adi-bci-bdi^2+ac-adi+bci-bdi^2 \\ &= (a-bi)(c+di)+(a+bi)(c-di) \\ &= \overline{(a+bi)}(c+di)+(a+bi)\overline{(c+di)} \\ &= \overline{w}z+w\overline{z}\quad\blacksquare \end{align*}$$
Proposition W.4.3 Let $w,z\in\wC$. Then $w\cj{z}=\cj{\cj{w}}\,\cj{z}=\cj{\cj{w}z}=\cj{z\cj{w}}$.
Proof Let $w=a+bi$ and $z=c+di$. For the first equality, note that $w=\cj{\cj{w}}$. Hence $w\cj{z}=\cj{\cj{w}}\,\cj{z}$. For the second equality, note that
$$\align{ \cj{\cj{w}}\,\cj{z} &= \cj{\cj{a+bi}}\,\cj{c+di} \\ &= \cj{a-bi}(c-di) \\ &= (a+bi)(c-di) \\ &= \cj{a-bi}\,\cj{c+di} \\ &= \cj{(a-bi)(c+di)} \\ &= \cj{\cj{a+bi}(c+di)} \\ &= \cj{\cj{w}z} }$$
For the third equality, note that $\cj{w}z=z\cj{w}$. $\blacksquare$
Proposition W.4.4 Let $z\in\wC$. Then $\norm{z}$ is nonnegative.
Proof Let $z=a+bi$. Recall that the square root of a nonnegative number is always nonnegative. Hence $\norm{z}=\sqrt{a^2+b^2}\geq0$. $\blacksquare$
Proposition W.4.5 Let $z\in\wC$. Then $\norm{z}=0$ if and only if $z=0$.
Proof Let $z=a+bi$. If $z=0$, then $0=a=b$ and $\norm{z}=\sqrt{a^2+b^2}=\sqrt{0}=0$. Conversely, suppose $\norm{z}=0$. If $a\neq0$, $\norm{z}=\sqrt{a^2+b^2}>\sqrt{b^2}\geq0$. Contradiction so $a=0$. Similarly $b=0$. Hence $z=0$. $\blacksquare$
Proposition W.4.6 Let $z\in\wC$. Then $z$ is nonnegative if and only if it has a nonnegative square root.
Proof Suppose $z$ is nonnegative. Then the principal square root, which is defined to be nonnegative, is $\sqrt{z}\geq0$. Conversely suppose $z$ has a nonnegative square root $a$. Hence $z=a^2$. But the square of a nonnegative number is nonnegative. Hence $z$ is nonnegative. $\wes$
Proposition W.4.7 Let $z\in\wC$. Then $z$ is nonnegative if and only if there exists a complex number $w$ such that $z=\cj{w}w$.
Proof Suppose $z$ is nonnegative and let $\sqrt{z}$ denote its principal square root. Then $\sqrt{z}\in\wR\subset\wC$ and
$$ z=\prn{\sqrt{z}}^2=\sqrt{z}\sqrt{z}=\sqrt{z}\cj{\sqrt{z}} $$
Conversely suppose there exists a complex number $w$ such that $z=\cj{w}w$. Then $z=\cj{w}w=\norm{w}^2\geq0$ by W.4.4. $\wes$