Find Modulus of Complex Numbers Examples and Solutions

Modulus of a complex number gives the distance of the complex number from the origin in the argand plane, whereas the conjugate of a complex number gives the reflection of the complex number about the real axis in the argand plane. In this section, we will discuss the modulus and conjugate of a complex number along with a few solved examples.

Conjugate of a Complex Number

Conjugate of a complex number z = x + iy is denoted by $\bar z$ = x – iy. It is the reflection of the complex number about the real axis on Argand's plane or the image of the complex number about the real axis on Argand's plane. If we replace the 'i' with '- i', we get conjugate of the complex number.

Geometrical representation of the complex number is shown in the figure given below:

Properties of the Conjugate of a Complex Number

Below are some properties of the conjugate of complex numbers along with their proof

(1) $\overline{z_1 \pm z_2} = \bar z_1 \pm \bar z_2$

Proof: Let z₁ = a + ib and z₂ = c + id

Then,

$\overline{z_1 \pm z_2} = \overline{{a+ib} \pm {c+id}}$

= $\overline{{a \pm c + ib \pm id}}$

= a ± c – i(b ± d)

= a – ib ± c ± id

= a – ib ± (c – id)

= $\bar z_1 \pm \bar z_2$

(2)

Points to Remember:

• z + $\bar z$ = 2 Re (z)
• z – $\bar z$ = 2i Im (z)
• If z lies in 1st quadrant then $\bar z$ will lie in 4th quadrant and $- \bar z$ will lie in the 2nd quadrant.
• If x + iy = f(a + ib) then x – iy = f(a – ib)
• Further, g(x + iy) = f(a + ib) ⇒g(x – iy) = f(a – ib).

Modulus of Complex Number

Modulus of the complex number is the distance of the point on the argand plane representing the complex number z from the origin.

Let P is the point that denotes the complex number z = x + iy.

Then OP = |z| = √(x² + y² ).

Note:

1. |z| > 0.

2. All the complex number with same modulus lie on the circle with centre origin and radius r = |z|.

Properties of Modulus of Complex Number

Below are few important properties of modulus of complex number and their proofs.

(i) |z₁ z₂| = |z₁||z₂|

Proof: let z₁= a + ib and z₂ = c + id

Then, |z₁ z₂| = |(a + ib)(c + id)|

⇒ |ac + iad + ibc + i²bd|

⇒ |ac + iad + ibc – bd|

⇒ |ac – bd + i(ad + bc)|

⇒ (ac – bd)²+ (ad+bc)²

⇒ (ac)²+ (bd)² – 2abcd + (ad)² + (bc)² + 2abcd

⇒ a² c² + b² d² + a² d²+ b² c²

⇒ a² c² + b² c² + b² d² + a² d²

⇒ (a² + b²)c² + (b² + a²)d²

⇒ (a² + b²) (c² + d²)

⇒ |z₁||z₂|.

(ii) |z₁/ z₂ | = (|z₁|) / (|z₂|).

Proof: |z₁/z₂ | = |z₁ . 1/z₂ |

Using multiplicative property of modulus we have

⇒ |z₁| |1/z₂|

⇒ |z₁| 1/(|z₂|)

⇒ (|z₁ |) / (|z₂|).

Some other important results:

Triangle inequalities:

|z₁ + z₂| ≤ |z₁| + |z₂|

|z₁ + z₂| ≥ |z₁| – |z₂|

|z₁ – z₂| ≥ |z₁| – |z₂|.

Also read

Complex numbers

Complex numbers solved examples

Examples on Modulus and Conjugate of a Complex Number

Example 1: Find the conjugate of the complex number z = (1 + 2i)/(1 – 2i).

Solution: z = (1 + 2i)/(1 – 2i)

Rationalising given complex number we have

⇒ z = ((1 + 2i)/(1 – 2i) )× (1 + 2i)/(1 + 2i)

⇒ z = (1 + 2i)²/(1² – (2i)²)

⇒ z = (1 + 4i² + 4i)/(1 + 4)

⇒ z = (1 – 4 + 4i)/(1 + 4)

⇒ z = (-3 + 4i)/5

⇒ $\bar z$ = (- 3 – 4i)/5.

Example 2: Find the modulus of the complex number z = (3 – 2i)/2i

Solution: z = (3 – 2i)/2i

⇒ z = (3 )/2i – 2i/2i

⇒ z = 3/2i – 1

⇒ z = 3i/(2i² ) – 1

⇒ z = (-3i/2) – 1

Example 3: If z + |z| = 1 + 4i, then find the value of |z|.

Solution: Let z = x + iy

⇒ z + |z| = 1 + 4i

⇒ x + iy + |x + iy| = 1 + 4i

⇒ x + iy + √(x² + y² ) = 1 + 4i

⇒ y = 4 and x + √(x² + y² ) = 1

⇒ x + √(x² + 4² ) = 1

⇒ √(x² + 4² ) = 1 – x

Squaring both side we have

⇒ x² + 4² = 1 + x² – 2x

⇒ 2x = -15

or x = -15/2

Example 4: If $|Z-\frac{4}{Z}|=2$ , then the maximum value of |Z| is equal to

Solution:

$|z-\frac{4}{z}|\geq |z|-|\frac{4}{z}|\\ 2\geq |z|-\frac{4}{|z|}\\ 2|z|\geq |z|^{2}-4\\ |z|^{2}-2|z|-4\leq 0\\ |z|\leq \sqrt{5}+1\\$

Example 5:Let $(\alpha, \beta)$ be real and z be a complex number. If $z^{2}+\alpha z+\beta =0$ has distinct roots on the line Re z = 1, then find the necessary condition.

Solution:

Given Re z = 1

Let z₁ = x + iy and z₂ = x – iy are the roots.

since Re z = 1, x = 1

Let $z^{2}+\alpha z+\beta =0$ has roots (1 + iy) and (1 – iy)

So,

z₁ z₂ =β

(1 + iy)(1 – iy) =β

1 + y² =β

β > 1

$\beta \epsilon (1,\infty)\\$

Example 6:If $\omega (\neq 1)$ is a cube root of unity, and $(1+\omega )^{7}=A+B\omega$ . Then find (A, B).

Solution:

$(1+\omega )^{7}=A+B\omega \\ (-\omega^{2} )^{7}=A+B\omega \\ (-\omega^{2} )=A+B\omega \\ 1+\omega =A+B\omega \\ A=1\\ B=1\\ (A,B)=(1,1)\\$

Example 7: A complex number z is said to be unimodular if |z| = 1. Suppose z₁ and z₂ are complex numbers such that $\frac{|z_1-2z_2|}{|2-z_1\overline{z_{2}}|}$ is unimodular and z₂ is not unimodular. Then the point z₁ lies on a _____.

Solution:

$\frac{|z_1-2z_2|}{|2-z_1\overline{z_{2}}|}=1\\ \Rightarrow |z_1-2z_2|^{2}=|2-z_1{\overline{z_2}}|^{2}\\ \Rightarrow |z_1-2z_2|(\overline{z_1}-2{\overline{z_2}})= (2-z_{1}\overline{z_{2}})(2-\overline{z_{1}}z_{2})\\ |z_1|^{2}+4|z_2|^{2}-4-|z_1|^{2}|z_2|^{2}=0\\ 4[(|z_2|^{2}-1)-|z_1|^{2}(|z_2|^2-1)]=0\\ \Rightarrow |z_1|^{2}-4=0\\ \Rightarrow |z_1|=2\\$

is a circle of radius 2 and centre at origin.

Find Modulus of Complex Numbers Examples and Solutions

Source: https://byjus.com/jee/modulus-and-conjugate-of-a-complex-number/

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