# Operations on Complex Number

## Trigonometry # Operations on Complex Number

## Conjugate of Complex numbers ### 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,

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

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