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
From Conjugate Complex Numbers to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Sep 14, 24 04:31 PM
Sep 14, 24 03:39 PM
Sep 14, 24 02:12 PM
Sep 13, 24 02:48 AM
Sep 12, 24 03:07 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.