# Conjugate Complex Numbers

Definition of conjugate complex numbers: In any two complex numbers, if only the sign of the imaginary part differ then, they are known as complex conjugate of each other.

Conjugate of a complex number z = a + ib, denoted by $$\bar{z}$$, is defined as

$$\bar{z}$$ = a - ib i.e., $$\overline{a + ib}$$ = a - ib.

For example,

(i) Conjugate of z$$_{1}$$ = 5 + 4i is $$\bar{z_{1}}$$ = 5 - 4i

(ii) Conjugate of z$$_{2}$$ = - 8 - i is $$\bar{z_{2}}$$ = - 8 + i

(iii) conjugate of z$$_{3}$$ = 9i is $$\bar{z_{3}}$$ = - 9i.

(iv) $$\overline{6 + 7i}$$ = 6 - 7i, $$\overline{6 - 7i}$$ = 6 + 7i

(v) $$\overline{-6 - 13i}$$ = -6 + 13i, $$\overline{-6 + 13i}$$ = -6 - 13i

Properties of conjugate of a complex number:

If z, z$$_{1}$$ and z$$_{2}$$ are complex number, then

(i) $$\bar{(\bar{z})}$$ = z

Or, If $$\bar{z}$$ be the conjugate of z then $$\bar{\bar{z}}$$ = z.

Proof:

Let z = a + ib where x and y are real and i = √-1. Then by definition, (conjugate of z) = $$\bar{z}$$ = a - ib.

Therefore, (conjugate of $$\bar{z}$$) = $$\bar{\bar{z}}$$ = a + ib = z. Proved.

(ii) $$\bar{z_{1} + z_{2}}$$ = $$\bar{z_{1}}$$ + $$\bar{z_{2}}$$

Proof:

If z$$_{1}$$ = a + ib and z$$_{2}$$ = c + id then $$\bar{z_{1}}$$ = a - ib and $$\bar{z_{2}}$$ = c - id

Now, z$$_{1}$$ + z$$_{2}$$ = a + ib + c + id = a + c + i(b + d)

Therefore, $$\overline{z_{1} + z_{2}}$$ = a + c - i(b + d) = a - ib + c - id = $$\bar{z_{1}}$$ + $$\bar{z_{2}}$$

(iii) $$\overline{z_{1} - z_{2}}$$ = $$\bar{z_{1}}$$ - $$\bar{z_{2}}$$

Proof:

If z$$_{1}$$ = a + ib and z$$_{2}$$ = c + id then $$\bar{z_{1}}$$ = a - ib and $$\bar{z_{2}}$$ = c - id

Now, z$$_{1}$$ - z$$_{2}$$ = a + ib - c - id = a - c + i(b - d)

Therefore, $$\overline{z_{1} - z_{2}}$$ = a - c - i(b - d)= a - ib - c + id = (a - ib) - (c - id) = $$\bar{z_{1}}$$ - $$\bar{z_{2}}$$

(iv) $$\overline{z_{1}z_{2}}$$ = $$\bar{z_{1}}$$$$\bar{z_{2}}$$

Proof:

If z$$_{1}$$ = a + ib and z$$_{2}$$ = c + id then

$$\overline{z_{1}z_{2}}$$ = $$\overline{(a + ib)(c + id)}$$ = $$\overline{(ac - bd) + i(ad + bc)}$$ = (ac - bd) - i(ad + bc)

Also, $$\bar{z_{1}}$$$$\bar{z_{2}}$$ = (a – ib)(c – id) = (ac – bd) – i(ad + bc)

Therefore, $$\overline{z_{1}z_{2}}$$ = $$\bar{z_{1}}$$$$\bar{z_{2}}$$ proved.

(v) $$\overline{(\frac{z_{1}}{z_{2}}}) = \frac{\bar{z_{1}}}{\bar{z_{2}}}$$, provided z$$_{2}$$ ≠ 0

Proof:

According to the problem

z$$_{2}$$ ≠ 0 ⇒ $$\bar{z_{2}}$$ ≠ 0

Let, $$(\frac{z_{1}}{z_{2}})$$ = z$$_{3}$$

z$$_{1}$$ = z$$_{2}$$ z$$_{3}$$

⇒ $$\bar{z_{1}}$$ = $$\bar{z_{2} z_{3}}$$

⇒ $$\frac{\bar{z_{1}}}{\bar{z_{2}}}$$ = $$\bar{z_{3}}$$

⇒ $$\overline{(\frac{z_{1}}{z_{2}}}) = \frac{\bar{z_{1}}}{\bar{z_{2}}}$$, [Since z$$_{3}$$ = $$(\frac{z_{1}}{z_{2}})$$] Proved.

(vi) |$$\bar{z}$$| = |z|

Proof:

Let z = a + ib then $$\bar{z}$$ = a - ib

Therefore, |$$\bar{z}$$| = $$\sqrt{a^{2} + (-b)^{2}}$$ = $$\sqrt{a^{2} + b^{2}}$$ = |z| Proved.

(vii) z$$\bar{z}$$ = |z|$$^{2}$$

Proof:

Let z = a + ib, then $$\bar{z}$$ = a - ib

Therefore, z$$\bar{z}$$ = (a + ib)(a - ib)

= a$$^{2}$$ – (ib)$$^{2}$$

= a$$^{2}$$ – i$$^{2}$$b$$^{2}$$

= a$$^{2}$$ + b$$^{2}$$, since i$$^{2}$$ = -1

= $$(\sqrt{a^{2} + b^{2}})^{2}$$

= |z|$$^{2}$$. Proved.

(viii) z$$^{-1}$$ = $$\frac{\bar{z}}{|z|^{2}}$$, provided z ≠ 0

Proof:

Let z = a + ib ≠ 0, then |z| ≠ 0.

Therefore, z$$\bar{z}$$ = (a + ib)(a – ib) = a$$^{2}$$ + b$$^{2}$$ = |z|$$^{2}$$

⇒ $$\frac{z\bar{z}}{|z|^{2}}$$ = 1

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

Therefore, z$$^{-1}$$ = $$\frac{\bar{z}}{|z|^{2}}$$, provided z ≠ 0. Proved.

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