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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