 # Operations on Complex Number Modulus of Complex Number

Let $z$ = $x~+~iy$ be a complex number, modulus of a complex number $z$ is denoted as $|z|$ which is equal to $\sqrt{x^{2}+y^{2}}$.

Geometrically, modulus of a complex number $z$ = $x ~+ ~iy$ is the distance between the corresponding point of $z$ which is $(x,y)$ and the origin $(0,0)$ in the argand plane. In the above figure, $OP$ is equal to the distance between the point $(x,y)$ and origin $(0,0)$ in argand plane.

Therefore,

$|z|$ = $OP$ = $\sqrt{x^{2}+y^{2}}$

Example: Find the value of b if the modulus of the complex number, $z$ = $3~+~ib$ is equal to $5$.

$|z|$ = $\sqrt{3^2~+~b^2}$

$\sqrt{3^2~+~b^2}$ = $5$
$9~+~b^2$ = $25$

$b^2$ = $25~-~9$ = $16$

$b$ =$±4$

Therefore, $z$ can be $3~+~4i$ or $3~-~4i$.

From the above example, we can conclude that complex numbers $z_1$ = $x~+~iy$, $z_2$ = $x~-~iy$,

$z_3$ = $-x~+~iy$, $z_4$ = $-x~-~iy$ will have same modulus which is equal to $\sqrt{x^2~+~y^2}$. It is because the points corresponding to the above four complex numbers $(x,y)$, $(x,-y)$, $(-x,y)$ and $(-x,-y)$ respectively are at a distance $\sqrt{x^2~+~y^2}$ away from origin.

## Conjugate of a complex number

Conjugate of a complex number $z$ = $x~+~iy$  is  $x~-~iy$and which is denoted as $\overline{z}$.

For example, conjugate of the complex number $z$ = $3~-~4i$ is $3~+~4i$.

• Consider the complex number $z$ = $a~+~ib$,
• $z ~+~ \overline{z}$ = $a ~+ ~ib~+ ~(a~ – ~ib)$ = $2a$ which is a complex number having imaginary part as zero.
• $Re(z ~+~ \overline{z})$ = $2a$, $Im(z ~+ ~\overline{z})$ = $0$
• $z ~-~ \overline{z}$ = $a~ +~ ib~ – ~(a~ -~ ib)$ = $2bi$
• $Re(z~ -~ \overline{z})$ = $0$, $Im(z~ -~ \overline{z})$ = $2b$
• Geometrically, reflection of the complex number $z$ = $x~+~iy$ in $X$ axis is the coordinates of $\overline{z}.$ • Modulus of the complex number and its conjugate will be equal.
• Multiplicative inverse of the non-zero complex number $z$ = $a~+~ib$ is

$z^{-1}$ = $\frac{1}{a~+~ib}$ = $\frac{a~-~ib}{a^2~+~b^2}$

Since,$a~-~ib$ = $\overline{z}$ and $a^2~+~b^2$ = $|z|^2$

$z^{-1}$ = $\frac{z}{|z|^2}$

$z\overline{z}$ = $|z|^2$

Example: Find the multiplicative inverse of $z$ = $6~+~8i$

$z^{-1}$ = $\frac{\overline{z}}{|z|^2}$

$\overline{z} = 6 – 8i$ and |z| = $\sqrt{6^2 + 8^2}$

= $\sqrt{100}$ = 10

$z^{-1}$ = $\frac{6~-~8i}{100}$ = $\frac{3-4i}{50}$ = $\frac{3}{50} – \frac{2}{25} i$

For any two complex number $z_1$ and $z_2$,

• $|z_1 z_2|$ = $|z_1 ||z_2 |$

Let $z_1$ = $a~+~ib$ and $z_2$ = $c~+~id$,

$|z_1|$ = $\sqrt{a^2~+~b^2}$     —(1)

$|z_2|$ = $\sqrt{c^2~+~d^2}$    —(2)

$z_1 z_2$ = $(ac~-~bd)~+(bc~+~ad)i$

$|z_1 z_2|$ = $\sqrt{(ac~-~bd)^2~+~(bc~+~ad)}$

$|z_1 z_2|^2$ =$(ac~-~bd)^2~+~(bc~+~ad)^2$

$=$ $a^2 c^2~+~b^2 d^2~-~2abcd~+~b^2 c^2~+~a^2 d^2~+~2abcd$

$=$ $a^2 c^2~+~b^2 d^2~+~b^2 c^2~+~a^2 d^2$

$=$$a^2(c^2~+~d^2)~+~b^2(c^2~+~d^2)$

$=$$(a^2~+~b^2)(c^2~+~d^2 )$  —(3)

From (1) and (2),

$|z_1|^2 |z_2 |^2$ = $(a^2~+~b^2)(c^2~+~d^2)$                 —(4)

Since, (3) = (4);$|z_1 z_2|$ = $|z_1||z_2|$

• Similarly $z_1$ and $z_2$, $|\frac{z_1}{z_2}|$= $\frac{|z_1 |}{|z_2 |}$ , provided that $z_2$ is a non zero complex number.
• $\overline{z_1z_2}$ = $\overline{z_1}\overline{z_2}$
• $\overline{z_1~ \pm ~z_2}$ = $\overline{z_1}~\pm~\overline{z_2}$
• $\overline{\left(\frac{z_1}{z_2}\right)}$ = $\frac{\overline{z_1}}{\overline{z_2}}$,provided that $z_2$ is a non-zero complex number.

#### Identities of complex numbers

For any two complex numbers $z_1$ and $z_2$,

• $(z_1~+~z_2)^2$ = $z_1^2~+~z_2^2~+~2z_1 z_2$

We can prove the above identity using the properties of complex numbers.

$(z_1~+~z_2)^2$ = $(z_1~+~z_2)(z_1~+~z_2 )$

By using distributive law,

$(z_1~+~z_2)(z_1~+~z_2)$ = $z_1 (z_1~+~z_2)~+~z_2(z_1~+~z_2)$

= $z_1^2~+z_1 z_2~+z_2 z_1~+~z_2^2$—(1)

By using the commutative law, $z_1 z_2$ = $z_2 z_1$

Then (1) will become as,

• $(z_1~+~z_2)^2$ = $z_1^2~+~z_2^2~+~2z_1 z_2$
• $(z_1~-~z_2)^2$ = $z_1^2~+~z_2^2~-~2z_1 z_2$
• $(z_1~+~z_2)^3$ = $z_1^3~+~3z_1^2 z_2~+~3z_1 z_2^2~+~z_1^3$
• $(z_1~-~z_2)^3$ = $z_1^3~-~3z_1^2 z_2~+~3z_1 z_2^2~-~z_1^3$
• $z_1^2~-~z_2^2$ = $(z_1~+~z_2)(z_1~-~z_2)$<