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 $\bar{z}=\overline{3-7i}=3+7i$
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: $|z|^2=z\bar{z}$
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)
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:
$\frac{1}{-i}=\frac{1}{-i} × \frac{i}{i}=\frac{i}{-1 × i^2}=\frac{i}{-1 × -1}=\frac{i}{1}=i$
The link below gives some solved examples related to the topics that we have discussed so far in this chapter.
• In the next section, we will see Argand plane and Polar representation.
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