Chapter 5.3 - The Modulus and Conjugate of a Complex Number

In the previous section, we saw the  the algebraic operations on complex numbers. We also saw identities related to complex numbers. In this section, we will see modulus and conjugate of a complex number.

Modulus of a complex number

This can be written in 3 steps:
1. Consider the complex number a+bi
• We can take out 'a' and 'b' and then calculate $\sqrt{a^2+b^2}$.
2. Since 'a' and 'b' are real numbers and since they are being squared and added, this $\sqrt{a^2+b^2}$ will be a +ve real number.
• This +ve real number is called the modulus of the complex number a+bi
3. Modulus of any complex number z is denoted as: |z|
• So we can write:
If $z=a+bi$, then $|z|=\sqrt{a^2+b^2}$
• Let us see some examples:
   ♦ If z = 2+i, then |z| = |2+i| = $\sqrt{2^2+1^2}=\sqrt{4+1}=\sqrt{5}$  
   ♦ If z = 3-7i, then |z| = |3-7i| = $\sqrt{3^2+(-7)^2}=\sqrt{9+49}=\sqrt{58}$  

Conjugate of a complex number

This can be written in 4 steps:
1. Consider the complex number a+bi
• We can take out 'a' and 'b' and using them, form a new complex number.
2. This new complex number is such that:
    ♦ Real part is the same a
    ♦ Imaginary part is the same b but with sign reversed
3. So the new complex number will be a-bi
• This new complex number is the conjugate of the given complex number a+bi
4. Conjugate of any complex number is denoted as: $\bar{z}$
• So we can write:
If $z=a+bi$, then $\bar{z}=a-bi$
• Let us see some examples:
   ♦ If z = 2+i, then $\bar{z}=\overline{2+i}=2-i$  
   ♦ If z = 3-7i, then $\bar{z}=\overline{3-7i}=3+7i$  

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Relation between $\mathbf{z,~\frac{1}{z},~\bar{z}~\text{and}~|z|}$

This can be written in 3 steps:
1. Consider the complex number z = a+bi
We know that, it's multiplicative inverse will be: $\frac{1}{z}=\left[\frac{a}{a^2+b^2}~+~\left(\frac{-b}{a^2+b^2} \right)i \right]$
2. The multiplicative inverse consists of two fractions. Since the denominators are the same, we can easily combine them.
• We get: $\frac{1}{z}=\left[\frac{a-bi}{a^2+b^2} \right]$
   ♦ The numerator on the right side is $\bar{z}$
   ♦ The denominator on the right side is $|z|^2$
3. Thus we get:$\frac{1}{z}=\frac{\bar{z}}{|z|^2}$
• Rearranging this, we get: $|z|^2=z\bar{z}$

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In addition to the above relation, we can derive six more results:

1. $\left|z_1 z_2\right| = \left|z_1\right| × \left|z_2\right|$
• Let us see an example:
   ♦ If z₁ = 3+4i and z₂ = 12+5i, then z₁z₂ = 16+63i
   ♦ (We already know how to calculate z₁z₂)
   ♦ |z₁| = $\sqrt{3^2+4^2}=\sqrt{25}$ = 5
   ♦ |z₂| = $\sqrt{12^2+5^2}=\sqrt{169}$ = 13
   ♦ |z₁z₂| = $\sqrt{16^2+63^2}=\sqrt{4225}$ = 65
   ♦ 65 = 5  × 13

2. $\left| \frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$  Provided |z₂| ≠ 0
• Let us see an example:
(i) Let z₁ = 3+4i and z₂ = 12+5i
First we calculate $\frac{1}{z_2}$, which is the multiplicative inverse of z₂:
$\begin{array}{ll} \frac{1}{z_2}&{}={}&\frac{1}{12+5i}& {} &{} \\ \phantom{\frac{1}{z_2}}&{}={}& \frac{12}{12^2+5^2}~+~\frac{-5i}{12^2+5^2} &{} \\ \phantom{\frac{1}{z_2}}&{}={}& \frac{12}{169}+\frac{-5i}{169} &{} \\ \end{array}$
(ii) So $\frac{z_1}{z_2} = z_1 × \frac{1}{z_2}= (3+4i) × \left(\frac{12}{169}+\frac{-5i}{169}\right)$
This works out to $\frac{32}{97}+\frac{17i}{87}$
(iii) Thus $\left| \frac{z_1}{z_2}\right|=\left| \frac{32}{97}+\frac{17i}{87}\right|=\frac{5}{13}$
(iv) Next we calculate individual moduli:
|z₁| = |3+4i| = 5
|z₂| = |12+5i| = 13
(v) Thus we get: $\left| \frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$

3. $\overline{z_1 z_2} = \bar{z_1} × \bar{z_2}$
• Let us see an example:
   ♦ If z₁ = 3+4i and z₂ = 12+5i, then z₁z₂ = 16+63i
   ♦ $\bar{z_1}=\overline{3+4i}=3-4i$
   ♦ $\bar{z_2}=\overline{12+5i}=12-5i$
   ♦ $\bar{z_1} × \bar{z_2}=(3-4i) × (12-5i)=16-63i$ 
   ♦ $\overline{z_1z_2} = \overline{16+63i}=16-63i$

4. $\overline{z_1+z_2}=\bar{z_1}+\bar{z_2}$
• Let us see an example:
   ♦ If z₁ = 3+4i and z₂ = 12+5i, then z₁+z₂ = 15+9i
   ♦ $\bar{z_1}=3-4i$ and $\bar{z_2}=12-5i$
   ♦ $\bar{z_1}+\bar{z_2}=(3-4i)+(12-5i)=15-9i$
   ♦ $\overline{z_1+z_2}=\overline{15+9i}=15-9i$

5. $\overline{z_1-z_2}=\bar{z_1}-\bar{z_2}$
• Let us see an example:
   ♦ If z₁ = 3+4i and z₂ = 12+5i, then z₁-z₂ = -9-i
   ♦ $\bar{z_1}=3-4i$ and $\bar{z_2}=12-5i$
   ♦ $\bar{z_1}-\bar{z_2}=(3-4i)-(12-5i)=-9+i$
   ♦ $\overline{z_1-z_2}=\overline{-9-i}=-9+i$

6. $\overline{\left( \frac{z_1}{z_2}\right)}=\frac{\bar{z_1}}{\bar{z_2}}$  Provided z₂ ≠ 0
• Let us see an example:
(i) Let z₁ = 3+4i and z₂ = 12+5i
First we calculate $\frac{1}{z_2}$, which is the multiplicative inverse of z₂:
$\begin{array}{ll} \frac{1}{z_2}&{}={}&\frac{1}{12+5i}& {} &{} \\ \phantom{\frac{1}{z_2}}&{}={}& \frac{12}{12^2+5^2}~+~\frac{-5i}{12^2+5^2} &{} \\ \phantom{\frac{1}{z_2}}&{}={}& \frac{12}{169}+\frac{-5i}{169} &{} \\ \end{array}$
(ii) So $\frac{z_1}{z_2} = z_1 × \frac{1}{z_2}= (3+4i) × \left(\frac{12}{169}+\frac{-5i}{169}\right)$
This works out to $\frac{32}{97}+\frac{17i}{87}$
(iii) Thus $\overline{\left( \frac{z_1}{z_2}\right)}=\overline{\frac{32}{97}+\frac{17i}{87}}=\frac{32}{97}-\frac{17i}{87}$
(iv) Next we write the individual conjugates:
$\bar{z_1}=3-4i$ and $\bar{z_2}=12-5i$
(v) Now we get: $\frac{\bar{z_1}}{\bar{z_2}}=\frac{3-4i}{12-5i}=\frac{32}{97}-\frac{17i}{87}$
This is the same result obtained in (iii)

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Let us see some solved examples:

Solved example 5.4
Find the multiplicative inverse of 2-3i
Solution:
1. Multiplicative inverse of any complex number z is denoted as: $\frac{1}{z}$
• It can be obtained using the equation: $\frac{1}{z}=\frac{\bar{z}}{|z|^2}$
2. So our first aim is to write $\bar{z}$
We have: $\bar{z} = \overline{2-3i} = 2+3i$
3. Next we calculate $|z|^2$
We have: $|z|^2=|2-3i|^2=(2^2 + (-3)^2)=(4+9)=13$
4. Thus we get: $\frac{1}{z}=\frac{\bar{z}}{|z|^2}=\frac{2+3i}{13}=\frac{2}{13}+\frac{3}{13}i$

Alternate method:
1. We want $\frac{1}{z}$, which is the reciprocal of z
• That means, we want $\frac{1}{2-3i}$
2. We need to write $\frac{1}{2-3i}$ in the form a+bi
• So we need to remove i from the denominator. For that, we can multiply both numerator and denominator by (2+3i)
3. Thus we get:
$\begin{array}{ll} \frac{1}{2-3i}&{}={}&\frac{1(2+3i)}{(2-3i)(2+3i)}& {} &{} \\ \phantom{\frac{1}{2-3i}}&{}={}& \frac{(2+3i)}{2^2-(3i)^2}&{} \\ \phantom{\frac{1}{2-3i}}&{}&\color {green}{(a+b)(a-b)=a^2-b^2} &{} \\ \phantom{\frac{1}{2-3i}}&{}={}& \frac{(2+3i)}{4+9}&{} \\ \phantom{\frac{1}{2-3i}}&{}={}& \frac{(2+3i)}{13}&{} \\ \phantom{\frac{1}{2-3i}}&{}={}& \frac{2}{13}+\frac{3}{13}i&{} \\ \end{array}$

Solved example 5.5
Express the following in the form a+bi
(i) $\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}$  (ii) $i^{-35}$
Solution(i):
$\begin{array}{ll} \frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}&{}={}&\frac{(5+\sqrt{2}\,i)(1+\sqrt{2}\,i)}{(1-\sqrt{2}\,i)(1+\sqrt{2}\,i)}& {} &{} \\ \phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& \frac{(5+\sqrt{2}\,i)(1+\sqrt{2}\,i)}{1^2-(\sqrt{2}\,i)^2}&{} \\ \phantom{\frac{1}{2-3i}}&{}&\color {green}{(a+b)(a-b)=a^2-b^2} &{} \\ \phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& \frac{5+5\sqrt{2}\,i+\sqrt{2}\,i+(\sqrt{2}\,i)^2}{1-(2 × -1)}&{} \\ \phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& \frac{5+5\sqrt{2}\,i+\sqrt{2}\,i+(2 × -1)}{1-(2 × -1)}&{} \\ \phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& \frac{5+5\sqrt{2}\,i+\sqrt{2}\,i-2}{1+2}&{} \\ \phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& \frac{3+6\sqrt{2}\,i}{3}&{} \\ \phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& 1+2\sqrt{2}\,i&{} \\ \end{array}$

Solution(ii):
1. Dividing 35 by 4, we get: $35\div 4 = 8 \frac{3}{4}$
    ♦ So the remainder is 3
2. Thus $i^{-35}=\frac{1}{i^{35}}=\frac{1}{-i}$
3. Now we remove -i from the denominator:
$\frac{1}{-i}=\frac{1}{-i} × \frac{i}{i}=\frac{i}{-1 × i^2}=\frac{i}{-1 × -1}=\frac{i}{1}=i$

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The link below gives some solved examples related to the topics that we have discussed so far in this chapter.

Exercise 5.1

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• In the next section, we will see Argand plane and Polar representation.

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