Definition of Modulus of a Complex Number:
Let z = x + iy where x and y are real and i = √-1. Then the non negative square root of (x\(^{2}\)+ y \(^{2}\)) is called the modulus or absolute value of z (or x + iy).
Modulus of a complex number z = x + iy, denoted by mod(z) or |z| or |x + iy|, is defined as |z|[or mod z or |x + iy|] = + \(\sqrt{x^{2} + y^{2}}\) ,where a = Re(z), b = Im(z)
i.e., + \(\sqrt{{Re(z)}^{2} + {Im(z)}^{2}}\)
Sometimes, |z| is called absolute value of z. Clearly, |z| ≥ 0 for all zϵ C.
For example:
(i) If z = 6 + 8i then |z| = \(\sqrt{6^{2} + 8^{2}}\) = √100 = 10.
(ii) If z = -6 + 8i then |z| = \(\sqrt{(-6)^{2} + 8^{2}}\) = √100 = 10.
(iii) If z = 6 - 8i then |z| = \(\sqrt{6^{2} + (-8)^{2}}\) =
√100 = 10.
(iv) If z = √2 - 3i then |z| = \(\sqrt{(√2)^{2} + (-3)^{2}}\) = √11.
(v) If z = -√2 - 3i then |z| = \(\sqrt{(-√2)^{2} + (-3)^{2}}\) = √11.
(vi) If z = -5 + 4i then |z| = \(\sqrt{(-5)^{2} + 4^{2}}\) = √41
(vii) If z = 3 - √7i then |z| = \(\sqrt{3^{2} + (-√7)^{2}}\) =\(\sqrt{9 + 7}\) = √16 = 4.
Note: (i) If z = x + iy and x = y = 0 then |z| = 0.
(ii) For any complex number z we have, |z| = |\(\bar{z}\)| = |-z|.
Properties of modulus of a complex number:
If z, z\(_{1}\) and z\(_{2}\) are complex numbers, then
(i) |-z| = |z|
Proof:
Let z = x + iy, then –z = -x – iy.
Therefore, |-z| = \(\sqrt{(-x)^{2} +(- y)^{2}}\) = \(\sqrt{x^{2} + y^{2}}\) = |z|
(ii) |z| = 0 if and only if z = 0
Proof:
Let z = x + iy, then |z| = \(\sqrt{x^{2} + y^{2}}\).
Now |z| = 0 if and only if \(\sqrt{x^{2} + y^{2}}\) = 0
⇒ if only if x\(^{2}\) + y\(^{2}\) = 0 i.e., a\(^{2}\) = 0and b\(^{2}\) = 0
⇒ if only if x = 0 and y = 0 i.e., z = 0 + i0
⇒ if only if z = 0.
(iii) |z\(_{1}\)z\(_{2}\)| = |z\(_{1}\)||z\(_{2}\)|
Proof:
Let z\(_{1}\) = j + ik and z\(_{2}\) = l + im, then
z\(_{1}\)z\(_{2}\) =(jl - km) + i(jm + kl)
Therefore, |z\(_{1}\)z\(_{2}\)| = \(\sqrt{( jl - km)^{2} + (jm + kl)^{2}}\)
= \(\sqrt{j^{2}l^{2} + k^{2}m^{2} – 2jklm + j^{2}m^{2} + k^{2}l^{2} + 2 jklm}\)
= \(\sqrt{(j^{2} + k^{2})(l^{2} + m^{2}}\)
= \(\sqrt{j^{2} + k^{2}}\) \(\sqrt{l^{2} + m^{2}}\), [Since, j\(^{2}\) + k\(^{2}\) ≥0, l\(^{2}\) + m\(^{2}\) ≥0]
= |z\(_{1}\)||z\(_{2}\)|.
(iv) |\(\frac{z_{1}}{z_{2}}\)| = \(\frac{|z_{1}|}{|z_{2}|}\), provided z\(_{2}\) ≠ 0.
Proof:
According to the problem, z\(_{2}\) ≠ 0 ⇒ |z\(_{2}\)| ≠ 0
Let \(\frac{z_{1}}{z_{2}}\) = z\(_{3}\)
⇒ z\(_{1}\) = z\(_{2}\)z\(_{3}\)
⇒ |z\(_{1}\)| = |z\(_{2}\)z\(_{3}\)|
⇒|z\(_{1}\)| = |z\(_{2}\)||z\(_{3}\)|, [Since we know that |z\(_{1}\)z\(_{2}\)| = |z\(_{1}\)||z\(_{2}\)|]
⇒ \(\frac{|z_{1}}{z_{2}}\) = |z\(_{3}\)|
⇒ \(\frac{|z_{1}|}{|z_{2}|}\) = |\(\frac{z_{1}}{z_{2}}\)|, [Since, z\(_{3}\) = \(\frac{z_{1}}{z_{2}}\)]
11 and 12 Grade Math
From Modulus of a Complex Number 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.
Mar 02, 24 05:31 PM
Mar 02, 24 04:36 PM
Mar 02, 24 03:32 PM
Mar 01, 24 01:42 PM
Feb 29, 24 05:12 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.