 > > Modulus and Conjugate of a Complex Number

# Modulus and Conjugate of a Complex Number

Can we somehow represent the absolute value of a complex number? How can we use imaginary numbers to represent real-life situations? Turns out there is a way to do it. Let us build our skills of complex analysis in the following section and find out more.

### Suggested Videos        Concept of Complex Numbers Modulus and it's Properties Complex Numbers Problem and its Solution ## Conjugate of A Complex Number

Complex numbers are represented in a binomial form as (a + ib). It almost invites you to play with that ‘+’ sign. What happens if we change it to a negative sign? Let z = a + ib be a complex number. We define another complex number $$\bar{z}$$ such that $$\bar{z}$$ = a – ib. We call $$\bar{z}$$ or the complex number obtained by changing the sign of the imaginary part (positive to negative or vice versa), as the conjugate of z. Let us now find the product $$z \bar{z}$$ = (a + ib)×(a – ib)

Hence,  = {a2 -i(ab) + i(ab) + b2 } = (a2 + b2 )   …(1)

If a and b are large numbers, the sum in (1) will be greater. So one can use this equation to measure the value of a complex number. The complex conjugates of complex numbers are used in “ladder operators” to study the excitation of electrons!

Learn the Basics of Complex Numbers here in detail.

## Modulus of A Complex Number

There is a way to get a feel for how big the numbers we are dealing with are. We take the complex conjugate and multiply it by the complex number as done in (1). Hence, we define the product  as the square of the Absolute value or modulus of a complex number. Therefore, we can write  = |z|2

This is done because as we just discussed,  gives a way to measure the Absolute value or magnitude of our complex number. The actual reason for this definition will get clear once you learn about the Argand Plane.

Therefore, |z|2 = (a2 + b2 )      [Using (1)]

Hence, |z| = $$\sqrt[]{(a^2 + b^2)}$$  …(2)

The above equation represents the modulus or the absolute value of our complex number z.

## Important Identities

If |z|2 = 1 = |z|, i.e. z is a complex number of unit modulus, then from (1), we have

z = $$\frac{1}{\bar{z}}$$ and $$\bar{z} = \frac{1}{z}$$

Also If z1 and z2 is any complex number, we have the following identities:

• |z1 z2| = |z1| |z2|
• Modulus (z1 / z2)= Modulus of z1 /Modulus of z2
• $$\overline{z_1z_2} = \bar{z}_1 \bar{z}_2$$
• $$\overline{z_1 \pm z_2} = \bar{z}_1 \pm \bar{z}_2$$
• $$\overline{\frac{z_1}{z_2}} = \frac{\bar{z}_1}{\bar{z}_2}$$; with the accord that z2 ≠ 0

You can download Complex Numbers Cheat Sheet by clicking on the download button below  ## Solved Examples For You

Example 1: If z = 2 – i, find ?

Answer : All we need to do is to change the sign of the imaginary term. Therefore, we denote it as follows:

z = 2 – i or (  )= ( $$\overline{2-i}$$ )

Now $$\bar{}$$ or ‘bar’ or ‘dagger’ becomes an operation that transforms the imaginary part of a complex number in sign. In other words, we can write:  = 2 + i. Note that there is no change in the sign of 2.

Question 2: If z = a+ib, show that |z|2.

Answer : We know that $$\frac{1}{z}$$ = $$\frac{1}{a + ib}$$ = $$\frac{1}{a + ib}$$ × $$\frac{a – ib}{a – ib}$$

Therefore, we have $$\frac{1}{z}$$ =  $$\frac{\bar{z}}{z\bar{z}}$$       [using the definition of conjugate]

Hence, $$\frac{1}{z}$$ = $$\frac{\bar{z}}{(|z|)^2}$$ Or |z|2 =

Question 3: The maximum value of z when z satisfies the condition z+ 2/z = 2 is:

A) 1             B) $$\sqrt[]{3}$$ + 1             C) $$\sqrt[]{3}$$ – 1          D) -1

Answer: B) We have the condition that modulus ( z + 2/z ) = 2. Therefore, can we not treat this as an equation in one variable? We can but the equation would be second order in z and we won’t be able to get a relation for |z| alone. So we do the following:

|z| can be written as = |z + [2/z] – [2/z]| ≤ |z + [2/z]| ≤ |z + [2/z]| + 2/|z|

Therefore, |z| ≤ 2 + 2/|z| Or |z| ≤ (2|z| + 2)/|z|

Hence, |z|2 ≤ 2|z| + 2  Or |z|2 – 2|z| ≤  2

To get the value of |z|, let’s convert the LHS of the above equation into a perfect square. After adding 1 on both the side, we have |z|2 – 2|z|(1) + (1)2 ≤  2 + (1)2

(|z| – 1)≤ 3

Therefore,  – $$\sqrt[]{3}$$ ≤ (|z| – 1) ≤ $$\sqrt[]{3}$$

Hence, the maximum value of |z| will be $$\sqrt[]{3}$$ + 1

Question 4: What is said to be an absolute value of a number?

Answer:The Absolute value of a number tells that the distance of a number at the number line that is starting from zero (‘0’) without considering which direction from the zero (‘0’) the number is lying.

Question 5: What is the absolute value of the number 3, 0, and -156?

Answer: The absolute value of the number 3 is ‘3’. The absolute value of the number 0 is ‘0’ and the absolute value of the number −156 is ‘156’.

Question 6: What is the absolute value of the number negative one (-1) and one third (1/3)?

Answer: The absolute value of the number negative one (-1) and one third (1/3) is ‘1’ and one third (‘1/3’) respectively.This is becausewe just count the spaces from zero to find an absolute value and an absolute value is never negative.

Question 7: How to solve the absolute value?

Answer:The steps to solve the absolute value are as follows:

1st step: firstly, isolate the absolute value expression.

2nd step: Then, Set the amount inside the absolute value notation equal to (+) and (-) the amount on the opposite side of the equation.

3rd step: Solve the unknowns in both the equations.

Learn the Operations on Complex Numbers here.

Share with friends

## Customize your course in 30 seconds

##### Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.  Ashhar Firdausi
IIT Roorkee
Biology  Dr. Nazma Shaik
VTU
Chemistry  Gaurav Tiwari
APJAKTU
Physics
Get Started

## Browse

##### Complex Numbers and Quadratic Equations Subscribe
Notify of

## Question Mark?

Have a doubt at 3 am? Our experts are available 24x7. Connect with a tutor instantly and get your concepts cleared in less than 3 steps. 