Why ${ |z+w| }^{ 2 }=(z+w)(\overline { z+w } )$ in $\mathbb{C}$?

Question

I'm new into Analysis and I'm struggling to understand why ${ |z+w| }^{ 2 }=(z+w)(\overline { z+w } )$ in $\mathbb{C}$

Answer 1

This has nothing to do with sums. If $z\in\mathbb C$, then $\lvert z\rvert^2=z\overline z$. That's so because, if you write $z$ as $a+bi$, with $a,b\in\mathbb R$, then\begin{align}z\overline z&=(a+bi)(a-bi)\\&=a^2+b^2\\&=\lvert z\rvert^2.\end{align}

Answer 2

For simplicity's sake, I will just refer to $z+w$ as $\zeta$. Now, since $\zeta \in \Bbb{C}$, we know that $\zeta=a+bi$ for $a,b\in \Bbb R$. Hopefully, you are already familiar with the following formula from Pythagorean theorem:

$$|\zeta|=|a+bi|=\sqrt {a^2+b^2}\rightarrow |\zeta|^2=a^2+b^2$$

Now, for simplicity, we want to find a way to factor $a^2+b^2$ into two complex numbers. This formula looks very similar to $a^2-b^2$, which can be factored into real numbers:

$$a^2-b^2=(a+b)(a-b)$$

Now, we can re-express $a^2+b^2$ as $a^2-(-b^2)$. This is helpful because, according to complex arithmetic, $-b^2=(bi)^2$. Thus, $a^2+b^2=a^2-(bi)^2$. Since we have turned $a^2+b^2$ into the difference of two perfect squares, we can use the formula above:

$$a^2+b^2=(a+bi)(a-bi)$$

Now, notice that $a+bi=\zeta$ and $a-bi$ is the conjugate of $\zeta$, otherwise known as $\bar \zeta$. Thus, we have:

$$|\zeta|^2=a^2+b^2=(a+bi)(a-bi)=\zeta\bar \zeta$$