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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