# Absolute Value of a Complex Number

Let $z=a+ib$ be a complex number. Then, the absolute value of a complex number $z$ denoted by $|z|$ or $|a+ib|$ is defined as the non negative real number $\sqrt{a^2+b^2}$. $|z|=\sqrt{a^2+b^2}$

The absolute value of a complex number is also known as the modulus of the complex number.

## Properties of Absolutes

### P1: $|z|=|\overline{z}|$

Let $z=a+ib$, then $\overline{z}=a-ib$. Now, $|z|=\sqrt{a^2+b^2}\;\;\text{and}\;\;|\overline{z}| =\sqrt{a^2+b^2}$ $\therefore |z|=|\overline{z}|$

### P2: $z\cdot\overline{z}=|\overline{z}|^2$

Let $z=a+ib$, then $\overline{z}=a-ib$ and $|z|=\sqrt{a^2+b^2}$. Now, $\begin{array}{l} z\cdot\overline{z} &=(a+ib)(a-ib) \\ &= a^2-i^2b^2 \\ &= a^2+b^2 \\ &= \left(\sqrt{a^2-b^2}\right)^2 \\ &= |z|^2 \end{array}$

### P3: $|z|=0 \Leftrightarrow z=0$

Let $z=a+ib$, then $|z|=\sqrt{a^2+b^2}$. Suppose $\begin{array}{l} & |z| &= 0 \\ \text{or,} & \sqrt{a^2+b^2} &= 0 \\ \text{or,} & a^2+b^2 &= 0 \end{array}$ $\therefore a=0\;\;\text{and}\;\; b=0$ $\therefore z=a+ib=0+i×0=0$

Conversely, suppose that, $\begin{array}{l} & z &=0 \\ \text{or,} & a+ib &= 0 \\ \text{or,} & a &= -ib \\ \text{or,} & a^2 &= i^2b^2 \\ \text{or,} & a^2 &= -b^2 \\ \text{or,} & a^2+b^2 &= 0 \\ \text{or,} & \sqrt{a^2+b^2} &= 0 \\ \therefore & |z| &= 0\end{array}$

### P4: $\text{Re}(z)≤|z|\;\;\text{and}\;\;\text{Im}(z)≤|z|$

Let $z=a+ib$, then $|z|=\sqrt{a^2+b^2}$. Now, $\begin{array}{l} & a^2 &≤ a^2+b^2 \\ \Rightarrow & a &≤ \sqrt{a^2+b^2} \\ \therefore & \text{Re}(z) &≤ |z| \\ \text{and,} & b^2 &≤ a^2+b^2 \\ \Rightarrow & b &≤ \sqrt{a^2+b^2} \\ \therefore & \text{Im}(z) &≤ |z| \end{array}$

### P5: $|z\cdot w|=|z|\cdot|w|$

Let $z=a+ib$ and $w=c+id$. Then, $|z|=\sqrt{a^2+b^2}\;\;\text{and}\;\;|w|=\sqrt{c^2+d^2}$ Now, $\begin{array}{l} z\cdot w &= (a+ib)(c+id) \\ &= ac+iad+ibc+i^2bd \\ &= (ac-bd)+(ad+bc)i \end{array}$

$\begin{array}{l} \therefore |z\cdot w| &= \sqrt{(ac-bd)^2+(ad+bc)^2} \\ &= \sqrt{a^2c^2+b^2d^2-2abcd+a^2d^2+b^2d^2+2abcd} \\ &= \sqrt{a^2(c^2+d^2)+b^2(c^2+d^2)} \\ &= \sqrt{(a^2+b^2)(c^2+d^2)} \\ &= \sqrt{a^2+b^2}\cdot\sqrt{c^2+d^2} \\ &=|z|\cdot|w| \end{array}$

### P6: $\left|\frac{z}{w}\right|=\frac{|z|}{|w|}$

Let $z=a+ib$ and $w=c+id$. Then, $|z|=\sqrt{a^2+b^2}\;\;\text{and}\;\;|w|=\sqrt{c^2+d^2}$

$\begin{array}{l} \frac{z}{w} &= \frac{a+ib}{c+id} \\ &= \frac{(a+ib)}{(c+id)}×\frac{(c-id)}{(c-id)} \\ &= \frac{ac-iad+ibc-i^2bd}{c^2-i^2d^2} \\ &= \frac{(ac+bd)+(bc-ad)i}{c^2+d^2} \\ &= \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i \end{array}$

$\begin{array}{l}\therefore \left|\frac{z}{w}\right| &= \sqrt{\left(\frac{ac+bd}{c^2+d^2}\right)^2 + \left(\frac{bc-ad}{c^2+d^2}\right)^2} \\ &= \frac{1}{c^2+d^2} \sqrt{a^2c^2+b^2c^2+2abcd+b^2c^2+a^2d^2-2abcd} \\ &= \frac{1}{c^2+d^2}\sqrt{(a^2+b^2)(c^2+d^2)} \\ &= \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} \\ &=\frac{|z|}{|w|} \end{array}$

### P7: (Triangle Inequality) $|z+w|≤|z|+|w|$

$\begin{array}{l} |z+w|^2 &=(z+w)(\overline{z+w}) &(\because |z|^2=z\cdot\overline{z}) \\ &=(z+w)(\overline{z}+\overline{w}) &(\because \overline{z+w}=\overline{z}+\overline{w}) \\ &= z\cdot\overline{z} + z\cdot\overline{w} + w\cdot\overline{z}+w\cdot\overline{w} \\ &= |z|^2+z\cdot\overline{w} +\overline{z}\cdot w+|w|^2 \\ &= |z|^2+z\cdot\overline{w}+\overline{z\cdot \overline{w}} + |w|^2 &(\because \overline{\overline{z}}=z) \\ &= |z|^2+2\text{Re}(z\cdot\overline{w})+|w|^2 &\left(\because\frac{1}{2}(z+\overline{z})=\text{Re}(z)\right) \\ &≤ |z|^2+2|z\cdot\overline{w}|+|w|^2 &(\because\text{Re}(z)≤|z|) \\ &\;\;\;\;= |z|^2+2|z||\overline{w}|+|w|^2 \\ &\;\;\;\; =|z|^2+2|z||w|+|w|^2 &(\because |z|=|\overline{z}|) \\ &\;\;\;\; =(|z|+|w|)^2 \\ \therefore |z+w|^2 &≤(|z|+|w|)^2 \\ \therefore |z+w| &≤ |z|+|w| \end{array}$

[See more: Triangle Inequality]

### P8: $|z+w|≥|z|-|w|$

Here, $\begin{array}{l} & z &= (z+w)-w \\ \text{or,} & z &= (z+w)-w \\ \therefore & |z| &= |(z+w)+(-w)| ≤ |z+w|+|-w| \\ \therefore & |z| &≤ |z+w|+|-w| \\ \text{or,} & |z| &≤ |z+w|+|w| \\ \text{or,} & |z|-|w| &≤ |z+w| \\ \therefore & |z+w| &≥ |z|-|w| \end{array}$

### P9: $|z-w|≤|z|+|w|$

Here, $\begin{array}{l} & z-w &= z+(-w) \\ \text{or,} & |z-w| &= |z+(-w)| ≤ |z|+|-w| \\ \text{or,} & |z-w| &≤ |z|+|-w| \\ \therefore & |z-w| &≤ |z|+|w| \end{array}$

### P10: $|z-w|≥|z|-|w|$

Here, $\begin{array}{l} & z &=(z-w)+w \\ \text{or,} & |z| &= |(z-w)+w| ≤ |z-w|+|w| \\ \therefore & |z| &≤ |z-w|+|w| \\ \text{or,} & |z|-|w| &≤ |z-w| \\ \therefore & |z-w| &≥ |z|-|w| \end{array}$