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$$_{3}$$ = $$\frac{z_{1}}{z_{2}}$$]

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