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.

11 and 12 Grade Math 

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