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 ˉz, is defined as

ˉz = a - ib i.e., ¯a+ib = a - ib.

For example,

(i) Conjugate of z1 = 5 + 4i is ¯z1 = 5 - 4i

(ii) Conjugate of z2 = - 8 - i is ¯z2 = - 8 + i

(iii) conjugate of z3 = 9i is ¯z3 = - 9i.

(iv) ¯6+7i = 6 - 7i, ¯67i = 6 + 7i

(v) ¯613i = -6 + 13i, ¯6+13i = -6 - 13i

Properties of conjugate of a complex number:

If z, z1 and z2 are complex number, then

(i) ¯(ˉz) = z

Or, If ˉz be the conjugate of z then ˉˉz = z.

Proof:

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

Therefore, (conjugate of ˉz) = ˉˉz = a + ib = z. Proved.

 

(ii) ¯z1+z2 = ¯z1 + ¯z2

Proof:

If z1 = a + ib and z2 = c + id then ¯z1 = a - ib and ¯z2 = c - id

Now, z1 + z2 = a + ib + c + id = a + c + i(b + d)

Therefore, ¯z1+z2 = a + c - i(b + d) = a - ib + c - id = ¯z1 + ¯z2


(iii) ¯z1z2 = ¯z1 - ¯z2

Proof:

If z1 = a + ib and z2 = c + id then ¯z1 = a - ib and ¯z2 = c - id

Now, z1 - z2 = a + ib - c - id = a - c + i(b - d)

Therefore, ¯z1z2 = a - c - i(b - d)= a - ib - c + id = (a - ib) - (c - id) = ¯z1 - ¯z2


(iv) ¯z1z2 = ¯z1¯z2

Proof:

If z1 = a + ib and z2 = c + id then

¯z1z2 = ¯(a+ib)(c+id) = ¯(acbd)+i(ad+bc) = (ac - bd) - i(ad + bc)

Also, ¯z1¯z2 = (a – ib)(c – id) = (ac – bd) – i(ad + bc)

Therefore, ¯z1z2 = ¯z1¯z2 proved. 


(v) ¯(z1z2)=¯z1¯z2, provided z2 ≠ 0   

Proof:

According to the problem

z2 ≠ 0 ⇒ ¯z2 ≠ 0

Let, (z1z2) = z3

z1 = z2 z3

¯z1 = ¯z2z3

¯z1¯z2 = ¯z3

¯(z1z2)=¯z1¯z2, [Since z3 = (z1z2)] Proved.


(vi) |ˉz| = |z|

Proof:

Let z = a + ib then ˉz = a - ib

Therefore, |ˉz| = a2+(b)2 = a2+b2 = |z| Proved.


(vii) zˉz = |z|2

Proof:

Let z = a + ib, then ˉz = a - ib

Therefore, zˉz = (a + ib)(a - ib)

= a2 – (ib)2

= a2 – i2b2

= a2 + b2, since i2 = -1

= (a2+b2)2

= |z|2. Proved.

 

(viii) z1 = ˉz|z|2, provided z ≠ 0

Proof:

Let z = a + ib ≠ 0, then |z| ≠ 0.

Therefore, zˉz = (a + ib)(a – ib) = a2 + b2 = |z|2

⇒ zˉz|z|2 = 1

⇒ ˉz|z|2 = 1z = z1

Therefore, z1 = ˉz|z|2, provided z ≠ 0. Proved.







11 and 12 Grade Math 

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