Conjugate of Complex Number

Operations on Complex Numbers

We can perform several algebraic operations on complex numbers, such as addition, multiplication, division, etc. However, we can find some interesting parameters for any given complex numbers using certain operations. Let’s learn how to find the conjugate and modulus of complex numbers here with help of formulas, examples, and a video lesson.

Conjugate of a Complex Number

Conjugate of a complex number z = x + iy  is x – iy and which is denoted as

For example, the conjugate of the complex number z = 3 – 4i is 3 + 4i.

Consider the complex number z = a + ib. For this, we can define the following formulas.

• Addition:
  \(\begin{array}{l}z + \overline{z}=a + ib + (a – ib)=2a\end{array} \)
  which is a complex number having imaginary part as zero.
  Here,
  \(\begin{array}{l}Re(z + \overline{z})=2a, Im(z + \overline{z}) = 0\end{array} \)

• Subtraction:
  \(\begin{array}{l}z – \overline{z}=a +ib – (a – ib)=2bi\end{array} \)
  
  Here,
  \(\begin{array}{l}Re(z – \overline{z})=0, Im(z -\overline{z})=2b\end{array} \)

• Geometrically, reflection of the complex number z = x + iy in x-axis is the coordinates of
  \(\begin{array}{l}\overline{z}.\end{array} \)
  

Video Lesson

Modulus of Complex Number

Let z = x + iy be a complex number, modulus of a complex number z is denoted as |z| which is equal to √(x² + y²).

Geometrically, the modulus of a complex number z = x + iy is the distance between the corresponding point of z which is (x, y) and the origin (0, 0) in the argand plane.

In the above figure, OP is equal to the distance between the point (x, y) and origin (0, 0) in the argand plane.

Therefore, |z| = OP = √(x² + y²).

Example: 

Find the value of b if the modulus of the complex number, z = 3 + ib is equal to 5.

Solution:

Given,

Z = 3 + ib

|z| = √(3² + b²)

√(9 + b²) = 5 (from the given)

9 + b² = 25

b² = 25 – 9

b = √16

Thus, b = ±4.

From the above example, we can conclude that complex numbers z₁ = x + iy = 3 + 4i and z₂ = x – iy = 3 – 4i will have same modulus which is equal to √(x² + y²). It is because the points corresponding to the above four complex numbers(x, y), (x, -y), (-x, y) and (-x, -y) respectively are at a distance √(x² + y²) away from origin.

Properties

• The modulus of the complex number and its conjugate will be equal.
• Multiplicative inverse of the non-zero complex number z = a + ib is
  \(\begin{array}{l}z^{-1}=\frac{1}{a+ib}=\frac{a-ib}{a^2+b^2}\end{array} \)
• As we know,
  \(\begin{array}{l}\overline{z} = a – ib\end{array} \)
  and
  \(\begin{array}{l}|z|^2=a^2+b^2\end{array} \)
  .
• \(\begin{array}{l}z^{-1}=\frac{z}{|z|^2}\end{array} \)
• \(\begin{array}{l}z\overline{z}=|z|^2\end{array} \)

Example: 

Find the multiplicative inverse of z = 6 + 8i.

Solution:

Given,

z = 6 + 8i

So,

|z| = √(6² + 8²) = √(36 + 64) = √100 = 10

We know that

Thus, z-1 = (6 – 8i)/(10)² = 2(3 – 4i)/100 = (3 – 4i)/50 = (3/50) – (2/25)i

Identities of complex numbers

For any two complex numbers z₁ and z₂,

|z₁ z₂| = |z₁| |z₂|

We can prove this identity as follows.

Let z₁ = a + ib and z₂ = c + id

|z₁| = √(a² + b²)….(1)

|z₂| = √(c² + d²)….(2)

Now,

z₁z₂ = (a + ib)(c + id) = (ac – bd) + (bc + ad)i

|z₁z₂| = √[(ac – bd)² + (bc + ad)²]

|z₁z₂|² = (ac – bd)² + (bc + ad)²

= a²c² + b²d² – 2abcd + b²c² + a²d² + 2abcd

= a²c² + b²d² + b²c² + a²d²

= a²(c² + d²) + b²(c² + d²)

= (a² + b²)(c² + d²)….(3)

From (1) and (2), we have

|z₁|² |z₂|² = (a² + b²)(c² + d²)….(4)

From (3) and (4), we get;

|z₁z₂| = |z₁| |z₂|

Similarly, we can write |z₁/z₂| = |z₁|/ |z₂| provided that z₂ is a non-zero complex number.

Also,

• \(\begin{array}{l}\overline{z_1z_2}=\overline{z_1}\overline{z_2}\end{array} \)
• \(\begin{array}{l}\overline{z_1~ \pm ~z_2}=\overline{z_1}~\pm~\overline{z_2}\end{array} \)
• \(\begin{array}{l}\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}\end{array} \)
  , provided that z₂ is a non-zero complex number.

For any two complex numbers z₁ and z₂,

• \(\begin{array}{l}(z_1~+~z_2)^2=z_1^2~+~z_2^2~+~2z_1 z_2\end{array} \)

We can prove the above identity using the properties of complex numbers.

By using distributive law,

By using the commutative law,

Then (1) will become as,

Similarly,

• \(\begin{array}{l}(z_1~-~z_2)^2=z_1^2~+~z_2^2~-~2z_1 z_2\end{array} \)
• \(\begin{array}{l}(z_1~+~z_2)^3=z_1^3~+~3z_1^2 z_2~+~3z_1 z_2^2~+~z_1^3\end{array} \)
• \(\begin{array}{l}(z_1~-~z_2)^3=z_1^3~-~3z_1^2 z_2~+~3z_1 z_2^2~-~z_1^3\end{array} \)
• \(\begin{array}{l}z_1^2~-~z_2^2=(z_1~+~z_2)(z_1~-~z_2)\end{array} \)

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