Modulus of Complex Number Properties Explain With Proof

Modulus of complex numbers is an important concept in mathematics that helps in finding the magnitude or distance of a complex number from the origin on the complex plane. The properties of modulus of complex numbers make calculations simple and useful in solving algebraic expressions, equations, and geometry-based problems. Understanding modulus, conjugate, product, quotient, and related properties with proofs and solved examples helps students build strong concepts for Class 11 Maths, CBSE board exams, JEE Main, NEET, and other competitive examinations.

What is the Modulus of a Complex Number ?

If z = x + iy ; where x, y are real numbers, then we denote the non-negative square root of x² + y² by | z | and call it the modulus or absolute value of the complex number z. Hence

| Z | = $\sqrt{x^2+y^2}$

It may be remembered that

$z\overline{z}=|z{|}^2$

Proof :

For, if $z=x+iy$ where $x,y\in ℝ$, then

$\overline{z}=x-iy$

$z\overline{z}=(x+iy)(x-iy)$

$z\overline{z}=x^2-i^2{y}^2$

$z\overline{z}=x^2+y^2=(\sqrt{x^2+y^2}{)}^2=|z{|}^2$

Gain deeper understanding by studying Conjugate Properties of Complex Number, Multiplicative or Reciprocal, Additive Inverse of Complex Numbers

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What are the Properties of Modulus of a Complex Number ?

Property.I :

For any complex number $z$, $|z|=0\iff z=0$.

Proof. Let $z=x+iy$

Then

$|z|=0\iff |z{|}^2=0$

$\iff x^2+y^2=0$

$\iff x=0\text{ and }y=0$

$\iff z=x+iy=0+i(0)=0$

Master related concepts such as What are Imaginary Numbers and iota (i), Powers of iota, Solved Examples

Property II :

$|z|=|\overline{z}|=|-z|$

Proof. Let $z=x+iy$. Then $\overline{z}=x-iy$ and $-z=-x-iy$.

$|z|=\sqrt{x^2+y^2}$

$|\overline{z}|=\sqrt{x^2+(-y{)}^2}=\sqrt{x^2+y^2}$

$|-z|=\sqrt{(-x{)}^2+(-y{)}^2}=\sqrt{x^2+y^2}$

Hence,

$|z|=|\overline{z}|=|-z|$

Property III

$|z_1{z}_2|=|z_1||z_2|$

The absolute value of the product of two complex numbers is equal to the product of the absolute values of the numbers.

Proof. Since $|z{|}^2=z\overline{z}$ [Proved above],

$|z_1{z}_2{|}^2=(z_1{z}_2)(\overline{z_1{z}_2})$

$|z_1{z}_2{|}^2=z_1{z}_2\overline{z_1}\overline{z_2}=z_1\overline{z_1}\cdot z_2\overline{z_2}[\because \overline{z_1{z}_2}=\overline{z_1}\overline{z_2}]$

$|z_1{z}_2{|}^2=|z_1{|}^2\cdot |z_2{|}^2$

Hence,

$|z_1{z}_2|=|z_1||z_2|\text{[Taking positive square roots]}$

Understand related topics like Complex Numbers: Addition, Subtraction, Multiplication of Two Complex Numbers With Solved Examples

Property IV

$\left|{\displaystyle \frac{z_1}{z_2}}\right|={\displaystyle \frac{|z_1|}{|z_2|}},z_2\ne 0$

The absolute value of the quotient of two complex numbers (denominator being non-zero) is equal to the quotient of the absolute values of the numbers.

Proof. If $z_2\ne 0$, then

$z_1=\left({\displaystyle \frac{z_1}{z_2}}\right)z_2$

Hence,

$|z_1|=\left|\left({\displaystyle \frac{z_1}{z_2}}\right)z_2\right|=\left|{\displaystyle \frac{z_1}{z_2}}\right||z_2|\text{[By Property III]}$

$\left|{\displaystyle \frac{z_1}{z_2}}\right|={\displaystyle \frac{|z_1|}{|z_2|}}$

Property V

1. $|z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2+2\text{Re}(z_1\overline{z_2})$
2. $|z_1-z_2{|}^2=|z_1{|}^2+|z_2{|}^2-2\text{Re}(z_1\overline{z_2})$
3. $|z_1+z_2{|}^2+|z_1-z_2{|}^2=2(|z_1{|}^2+|z_2{|}^2)$

Proof. Let $z=x+iy$. Then $\overline{z}=x-iy$ so that $z+\overline{z}=2x$ and $z-\overline{z}=2iy$. i.e.,

$x={\displaystyle \frac{z+\overline{z}}{2}}\text{ and }iy={\displaystyle \frac{z-\overline{z}}{2}}$

$\text{Re}(z)={\displaystyle \frac{z+\overline{z}}{2}}\text{ and }i\text{Im}(z)={\displaystyle \frac{z-\overline{z}}{2}}$

$2\text{Re}(z)=z+\overline{z}\text{ and }2i\text{Im}(z)=z-\overline{z}$

(i)

$|z_1+z_2{|}^2=(z_1+z_2)(\overline{z_1+z_2})$

$=(z_1+z_2)(\overline{z_1}+\overline{z_2})=z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+z_2\overline{z_1}$

$=|z_1{|}^2+|z_2{|}^2+z_1\overline{z_2}+(\overline{z_1\overline{z_2}})[\because \overline{z_1\overline{z_2}}=\overline{z_1}\overline{(\overline{z_2})}=\overline{z_1}{z}_2]$

$=|z_1{|}^2+|z_2{|}^2+2\text{Re}(z_1\overline{z_2})\dots (1)\text{ [Using the relation above]}$

(ii)

$|z_1-z_2{|}^2=(z_1-z_2)(\overline{z_1-z_2})=(z_1-z_2)(\overline{z_1}-\overline{z_2})$

$=z_1\overline{z_1}+z_2\overline{z_2}-(z_1\overline{z_2}+\overline{z_1}{z}_2)$

$=|z_1{|}^2+|z_2{|}^2-[z_1\overline{z_2}+(\overline{z_1\overline{z_2}})]$

$=|z_1{|}^2+|z_2{|}^2-2\text{Re}(z_1\overline{z_2})\dots (2)$

(iii) Adding (1) and (2), we get:

$|z_1+z_2{|}^2+|z_1-z_2{|}^2=2(|z_1{|}^2+|z_2{|}^2)$

Property VI (Triangle Inequality)

$|z_1+z_2|\le |z_1|+|z_2|$

The modulus of the sum of two complex numbers cannot be greater than the sum of their moduli.

Proof. Let $z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$, so that

$|z_1|=\sqrt{x_1^2+y_1^2}\text{ and }|z_2|=\sqrt{x_2^2+y_2^2}$

Now,

$z_1+z_2=(x_1+i{y}_1)+(x_2+i{y}_2)=(x_1+x_2)+i(y_1+y_2)$

$|z_1+z_2|=\sqrt{(x_1+x_2{)}^2+(y_1+y_2{)}^2}$

Now, $|z_1+z_2|\le |z_1|+|z_2|$ holds true:

$\text{iff }\sqrt{(x_1+x_2{)}^2+(y_1+y_2{)}^2}\le \sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}$

$\text{iff }(x_1+x_2{)}^2+(y_1+y_2{)}^2\le (x_1^2+y_1^2)+(x_2^2+y_2^2)+2\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}\text{[Squaring]}$

$\text{iff }2x_1{x}_2+2y_1{y}_2\le 2\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}$

$\text{iff }{x}_1{x}_2+y_1{y}_2\le \sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}\text{[Dividing by 2]}$

$\text{iff }(x_1{x}_2+y_1{y}_2{)}^2\le (x_1^2+y_1^2)(x_2^2+y_2^2)\text{[Squaring]}$

$\text{iff }{x}_1^2{x}_2^2+y_1^2{y}_2^2+2x_1{x}_2{y}_1{y}_2\le x_1^2{x}_2^2+x_1^2{y}_2^2+x_2^2{y}_1^2+y_1^2{y}_2^2$

$\text{iff }0\le x_1^2{y}_2^2+x_2^2{y}_1^2-2x_1{x}_2{y}_1{y}_2$

$\text{iff }0\le (x_1{y}_2-x_2{y}_1{)}^2$

which is true. [A perfect square is never negative]

Extension: $|z_1+z_2+\cdots +z_n|\le |z_1|+|z_2|+\cdots +|z_n|$

The modulus of the sum of any number of complex numbers cannot be greater than the sum of their moduli.

Proof.

$|z_1+z_2+\cdots +z_n|=|z_1+(z_2+\cdots +z_n)|$

$\le |z_1|+|z_2+\cdots +z_n|$

$\le |z_1|+|z_2|+|z_3+\cdots +z_n|$

$\le |z_1|+|z_2|+\cdots +|z_n|$

Property VII

$|z_1-z_2|\ge |z_1|-|z_2|$

The modulus of the difference of two complex numbers cannot be less than the difference of their moduli.

Proof. Here,

$z_1-z_2=(x_1+i{y}_1)-(x_2+i{y}_2)=(x_1-x_2)+i(y_1-y_2)$

$|z_1-z_2|=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

Now, $|z_1-z_2|\ge |z_1|-|z_2|$ holds true :

$\text{iff }\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}\ge \sqrt{x_1^2+y_1^2}-\sqrt{x_2^2+y_2^2}$

$\text{iff }(x_1-x_2{)}^2+(y_1-y_2{)}^2\ge (x_1^2+y_1^2)+(x_2^2+y_2^2)-2\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}\text{[Squaring]}$

$\text{iff }-2x_1{x}_2-2y_1{y}_2\ge -2\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}$

$\text{iff }{x}_1{x}_2+y_1{y}_2\le \sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}\text{[Dividing by -2, changing inequality]}$

$\text{iff }(x_1{x}_2+y_1{y}_2{)}^2\le (x_1^2+y_1^2)(x_2^2+y_2^2)\text{[Squaring]}$

$\text{iff }{x}_1^2{x}_2^2+y_1^2{y}_2^2+2x_1{x}_2{y}_1{y}_2\le x_1^2{x}_2^2+x_1^2{y}_2^2+x_2^2{y}_1^2+y_1^2{y}_2^2$

$\text{iff }0\le x_1^2{y}_2^2+x_2^2{y}_1^2-2x_1{x}_2{y}_1{y}_2$

$\text{iff }0\le (x_1{y}_2-x_2{y}_1{)}^2$

which is true. [A perfect square is never negative]

Property VIII

$|z_1-z_2|\ge ||z_1|-|z_2||$

The modulus of the difference of two complex numbers cannot be less than the modulus of the difference of their moduli.

Proof. We know that:

$|z_1-z_2|\ge |z_1|-|z_2|\dots (1)\text{ [By Property VII]}$

Similarly, we can prove that:

$|z_1-z_2|\ge |z_2|-|z_1|\dots (2)$

Combining (1) and (2),

$|z_1-z_2|\ge ||z_1|-|z_2||$

$[\because ||z_1|-|z_2||=|z_1|-|z_2|\text{ or }|z_2|-|z_1|]$

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FAQs on Modulus of Complex Number and Its Properties

What is the modulus of a complex number?

The modulus of a complex number represents its distance from the origin in the Argand plane. It gives the magnitude of the complex number.

How is the modulus of a complex number represented?

The modulus of a complex number is represented by vertical bars around the complex number.

What does the modulus of a complex number indicate geometrically?

Geometrically, the modulus represents the distance of the point corresponding to the complex number from the origin on the complex plane.

Can the modulus of a complex number be negative?

No, the modulus of a complex number is always zero or positive because it represents distance.

What is the modulus of a purely real number?

For a purely real number, the modulus is equal to the absolute value of the real number.

What is the modulus of a purely imaginary number?

For a purely imaginary number, the modulus is equal to the absolute value of the imaginary coefficient.

What is the modulus of zero complex number?

The modulus of the zero complex number is zero.

What is the property of modulus of conjugate complex numbers?

The modulus of a complex number and its conjugate are always equal.

What is the modulus of the product of two complex numbers?

The modulus of the product of two complex numbers is equal to the product of their moduli.

What is the modulus of the quotient of two complex numbers?

The modulus of the quotient of two complex numbers is equal to the quotient of their moduli, provided the denominator is non-zero.

What is the triangle inequality property of modulus?

The modulus of the sum of two complex numbers is always less than or equal to the sum of their individual moduli.

What is the significance of modulus in complex numbers?

The modulus is useful in geometry, vector representation, polar form, trigonometry, and solving advanced mathematical and engineering problems.

Why is modulus important in polar form of complex numbers?

The modulus represents the radial distance of the complex number from the origin in polar representation.

What happens when the modulus of a complex number is squared?

The square of the modulus equals the sum of the squares of the real and imaginary parts.

Where are modulus properties used?

Properties of modulus are widely used in algebra, coordinate geometry, signal processing, electrical engineering, and competitive mathematics.

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