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 (x2+ 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|] = + √x2+y2 ,where a = Re(z), b = Im(z)

i.e., + √Re(z)2+Im(z)2

Sometimes, |z| is called absolute value of z. Clearly, |z| β‰₯ 0 for all zΟ΅ C.

For example:

(i) If z = 6 + 8i then |z| = √62+82 = √100 = 10.

(ii) If z = -6 + 8i then |z| = √(βˆ’6)2+82 = √100 = 10.

(iii) If z = 6 - 8i then |z| = √62+(βˆ’8)2 = √100 = 10.

(iv) If z = √2 - 3i then |z| = √(√2)2+(βˆ’3)2 = √11.

(v) If z = -√2 - 3i then |z| = √(βˆ’βˆš2)2+(βˆ’3)2 = √11.

(vi) If z = -5 + 4i then |z| = √(βˆ’5)2+42 = √41

(vii) If z = 3 - √7i then |z| = √32+(βˆ’βˆš7)2 =√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| = |Λ‰z| = |-z|.

 

Properties of modulus of a complex number:

If z, z1 and z2 are complex numbers, then

(i) |-z| = |z|

Proof:

Let z = x + iy, then –z = -x – iy.

Therefore, |-z| = √(βˆ’x)2+(βˆ’y)2 = √x2+y2 = |z|


(ii) |z| = 0 if and only if z = 0

Proof:

Let z = x + iy, then |z| = √x2+y2.

Now |z| = 0 if and only if √x2+y2 = 0

β‡’ if only if x2 + y2 = 0 i.e., a2 = 0and b2 = 0

β‡’ if only if x = 0 and y = 0 i.e., z = 0 + i0

β‡’ if only if z = 0.

 

(iii) |z1z2| = |z1||z2|

Proof:

Let z1 = j + ik and z2 = l + im, then

z1z2 =(jl - km) + i(jm + kl)

Therefore, |z1z2| = √(jlβˆ’km)2+(jm+kl)2

= √j2l2+k2m2–2jklm+j2m2+k2l2+2jklm

= √(j2+k2)(l2+m2

= √j2+k2 √l2+m2, [Since, j2 + k2 β‰₯0, l2 + m2 β‰₯0]

= |z1||z2|.


(iv) |z1z2| = |z1||z2|, provided z2 β‰  0.

Proof:

According to the problem, z2 β‰  0 β‡’ |z2| β‰  0

Let z1z2 = z3

β‡’ z1 = z2z3

β‡’ |z1| = |z2z3|

β‡’|z1| = |z2||z3|, [Since we know that |z1z2| = |z1||z2|]

β‡’ |z1z2 = |z3|

β‡’ |z1||z2| = |z1z2|, [Since, z3 = z1z2]





11 and 12 Grade Math 

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