# Conjugate Complex Number

**conjugate complex number :**The complex number a +ib and a - ib are called conjugates of each other.

From the above you can see that in conjugates there is only difference in the sign of imaginary part.

The conjugate of complex number is denoted by $\bar{z}$.

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

**Examples :**

(i) z = 5 + 8i so $\bar{z}$ = 5 - 8i

(ii) z = 4 -3i so $\bar{z}$ = 4 + 3i

(iii) z = 8i so $\bar{z}$ = -8i

(iv) $\overline{(5+2i)}$ = 5 -2i

## Properties of conjugate complex number

**(I) Conjugate of the conjugate of a complex number Z is the complex number itself.**

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

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

$\overline {\bar{(z)}}$ =$\overline {\overline{(a +ib)}}$

= $\overline {(a -ib)}$

= a + ib = z

∴ $\overline {\bar{(z)}}$ = z

**(II)Conjugate of the sum of two complex numbers z**

_{1},z_{2}is the sum of their conjugates.$\overline{z1 + z2}$ =$\bar{z1}$ +$\bar{z2}$

**Proof :**Let z

_{1}= a+ib and z

_{2}= c + id

z

_{1}+ z

_{2}= (a +ib) +(c+id)

z

_{1}+ z

_{2}= (a+ c) + (b + d)i

∴ $\overline{z1 + z2}$ = (a + c) - (b + d)i ------(1)

z

_{1}= a+ib ⇒ $\bar{z1}$ = a - ib

z

_{2}= c +id ⇒ $\bar{z2}$ = c - id

∴ $\bar{z1}$ + $\bar{z2}$ = (a -ib) +(c - id)

= (a + c) -i(b + d ) -----(2)

Hence from (1) and (2)

$\overline{z1 + z2}$ = $\bar{z1}$+$\bar{z2}$

**(III) conjugate of the product of two complex numbers z**

_{1}and z_{2}is the product of their conjugates i.e. $\overline{z1z2}$ = $\bar{z1}$ . $\bar{z2}$**Proof :**Let z1 = a + ib and z2 = c +id then

z1.z2 = (a + ib).(c +id) = (ac -bd)+ i(ad +bc)

∴ $\overline{z1z2}$ = (ac - bd) - i(ad +bc) -----(1)

$\bar{z1}$ = a - ib and $\bar{z2}$ = c - id

$\bar{z1}$.$\bar{z2}$ = (a -ib).(c - id) =(ac -bd)-i(ad +bc)----(2)

From (1) and (2)

$\overline{z1z2}$ = $\bar{z1}$.$\bar{z2}$

**(IV) Conjugate of the quotient of two complex numbers z1 and z2 (z2 ≠ 0) is the quotient of their conjugates , i.e $\left(\overline{\frac{z1}{z2}}\right)=\frac{\bar{z1}}{\bar{z2}}$**

**Proof :**Let z1 = a + ib and z2 = c +id. Then

$\frac{z1}{z2}$ =$\frac{a+ib}{c +id}$ = $\frac{(a+ib)(c - id)}{(c +id)(c - id)}$ = $\frac{ac+bd}{c^2 + d^2}$ - i $\frac{ad-bc}{c^2 + d^2}$

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