If a and b are complex numbers, prove: 1^(°) |ab|=|a||b|…

If $a$ and $b$ are complex numbers, prove: $\ 1^{\circ}\,\, |ab|=|a||b|,\ |a/b|=|a|/|b|\ (b\neq0);\quad 2^{\circ}\,\, \text{Re}(a)=\tfrac12(a+\bar a),\ \text{Im}(a)=\tfrac1{2i}(a-\bar a);\quad 3^{\circ}\,\, \overline{a\pm b}=\bar a\pm\bar b,\ \overline{ab}=\bar a\bar b,\ \overline{(a/b)}=\bar a/\bar b;\quad 4^{\circ}\,\, \text{Arg}(ab)=\text{Arg}\,a+\text{Arg}\,b,\ \text{Arg}(a/b)=\text{Arg}\,a-\text{Arg}\,b;\quad 5^{\circ}\,\, \big||a|-|b|\big|\leq|a+b|\leq|a|+|b|$.

Part $1^{\circ}$ · $|ab|=|a||b|$, $\ |a/b|=|a|/|b|$

Using $|z|^2=z\bar z$ and $\overline{ab}=\bar a\bar b$ (proved independently in $3^{\circ}$):

$$ |ab|^2=(ab)\overline{(ab)}=ab\bar a\bar b=(a\bar a)(b\bar b)=|a|^2|b|^2 \quad\Longrightarrow\quad \boxed{|ab|=|a||b|} $$

For the quotient, let $c=a/b$ (so $a=bc$); applying the product rule just proved: $|a|=|b||c|$, hence $\boxed{|a/b|=|c|=|a|/|b|}$.

Part $2^{\circ}$ · $\text{Re}(a)=\tfrac12(a+\bar a)$, $\ \text{Im}(a)=\tfrac1{2i}(a-\bar a)$

With $a=x+iy$, $\bar a=x-iy$:

$$ a+\bar a=2x \quad\Longrightarrow\quad \tfrac12(a+\bar a)=x=\text{Re}(a) $$

$$ a-\bar a=2iy \quad\Longrightarrow\quad \tfrac1{2i}(a-\bar a)=y=\text{Im}(a) $$

Part $3^{\circ}$ · Conjugate of sums, products, quotients

With $a=x_1+iy_1$, $b=x_2+iy_2$:

$$ \overline{a\pm b}=\overline{(x_1\pm x_2)+i(y_1\pm y_2)}=(x_1\pm x_2)-i(y_1\pm y_2)=(x_1-iy_1)\pm(x_2-iy_2)=\bar a\pm\bar b $$

$$ ab=(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1) \quad\Longrightarrow\quad \overline{ab}=(x_1x_2-y_1y_2)-i(x_1y_2+x_2y_1) = (x_1-iy_1)(x_2-iy_2)=\bar a\bar b $$

For the quotient, let $c=a/b$, so $a=bc$; taking conjugates via the product rule just shown: $\bar a=\bar b\bar c$, hence $\boxed{\overline{(a/b)}=\bar c=\bar a/\bar b}$.

Part $4^{\circ}$ · Argument of products and quotients

Write $a=r_1e^{i\theta_1}$, $b=r_2e^{i\theta_2}$. Then:

$$ ab=r_1r_2e^{i(\theta_1+\theta_2)} \quad\Longrightarrow\quad \text{Arg}(ab)=\theta_1+\theta_2=\text{Arg}\,a+\text{Arg}\,b \pmod{2\pi} $$

$$ \frac ab=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)} \quad\Longrightarrow\quad \text{Arg}(a/b)=\theta_1-\theta_2=\text{Arg}\,a-\text{Arg}\,b \pmod{2\pi} $$

Part $5^{\circ}$ · Triangle inequality and its reverse

Expand using conjugates:

$$ |a+b|^2=(a+b)(\bar a+\bar b)=|a|^2+|b|^2+a\bar b+\overline{a\bar b}=|a|^2+|b|^2+2\,\text{Re}(a\bar b) $$

Since $\text{Re}(z)\leq|z|$ for any $z$:

$$ |a+b|^2\leq|a|^2+|b|^2+2|a\bar b|=|a|^2+|b|^2+2|a||b|=(|a|+|b|)^2 \quad\Longrightarrow\quad \boxed{|a+b|\leq|a|+|b|} $$

For the reverse inequality, apply this to $a=(a+b)+(-b)$ and to $b=(a+b)+(-a)$:

$$ |a|=|(a+b)-b|\leq|a+b|+|b| \ \Rightarrow\ |a|-|b|\leq|a+b|, \qquad |b|=|(a+b)-a|\leq|a+b|+|a| \ \Rightarrow\ |b|-|a|\leq|a+b| $$

Combining both: $\boxed{\big||a|-|b|\big|\leq|a+b|}$.