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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