Modulus of a Complex Number

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 

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