In the previous section, we saw complex numbers and some of their algebraic operations like addition, subtraction and multiplication. We also saw additive inverse and multiplicative inverse. In this section, we will see some more algebraic operations.
D. Division of two complex numbers.
This can be written in 2 steps:
1. If z1 and z2 are two complex numbers,
Then the quotient ${z_1}\div{z_2}$ is defined as: ${z_1}\div{z_2}=z_1 × \frac{1}{z_2}$
(Recall that, for 'division of real numbers' also, we use the reciprocal.
For example: $5 \div 2=5 × \frac{1}{2}$)
2. We have seen that, $\frac{1}{z_2}$ is the multiplicative inverse of z2
• So we can write:
♦ If we want to divide z1 by z2,
♦ We must multiply z1 by the multiplicative inverse of z2
• Let us see an example:
If z1 = 6+3i and z2 = 2-i, find $\frac{z_1}{z_2}$
Solution:
• We have:
Multiplicative inverse of a+bi = $\frac{a}{a^2+b^2}~+~\left(\frac{-b}{a^2+b^2} \right)i$
• So the multiplicative inverse of 2-i = $\frac{2}{2^2+(-1)^2}~+~\left(\frac{-(-1)}{2^2+(-1)^2} \right)i$
= $\frac{2}{4+1}~+~\left(\frac{1}{4+1} \right)i$
= $\frac{2}{5}~+~\left(\frac{1}{5} \right)i$
• Thus we get:
$\begin{array}{ll}
\frac{6+3i}{2-i}&{}={}& (6+3i) × \left[\frac{2}{5}~+~\left(\frac{1}{5} \right)i \right]& {} \\
\phantom{\frac{6+3i}{2-i}}&{}={}& 6×\frac{2}{5}~-~3×\frac{1}{5}~+~\left[6×\frac{1}{5}~+~3×\frac{2}{5}\right] i &{} & {} \\
\phantom{\frac{6+3i}{2-i}}&{}={}& \frac{12}{5}~-~\frac{3}{5}~+~\left[\frac{6}{5}~+~\frac{6}{5}\right] i &{}& {} \\
\phantom{\frac{6+3i}{2-i}}&{}={}& \frac{9}{5}~+~\left[\frac{12}{5}\right] i &{}& {} \\
\end{array}$
E. Power of i
◼ First we will consider positive powers. It can be written in 7 steps:
1. We know that: $i=\sqrt{-1}$
2. So $i^2=\left(\sqrt{-1}\right)^2 = -1$
3. Then $i^3 = (i^2) i = (-1)i = -i$
4. $i^4 = \left(i^2 \right)^2 = (-1)^2 = 1$
5. Then $i^5 = (i^4)i = (1)i = i$
6. Then $i^6 = (i^5)i = (i)i = i^2 = -1$
7. We see that the result can be -1, -i, 1 and i
• Based on the above results, we can write:
If k is any integer, the general form can be written as:
♦ $i^{4k}=1$
♦ $i^{4k+1}=i$
♦ $i^{4k+2}=-1$
♦ $i^{4k+3}=-i$
• That means, we have to divide the power by 4
♦ If the remainder is 0, the result is 1
♦ If the remainder is 1, the result is i
♦ If the remainder is 2, the result is -1
♦ If the remainder is 3, the result is -i
• Let us see an example:
Find the value of $i^{15}$
Solution:
(i) Dividing 15 by 4, we get: $\frac{15}{4}=3\frac{3}{4}$
♦ So the remainder is 3
♦ Thus we get: $i^{15}=-i$
(ii) Check:
$i^{15}=\left(i^{14}\right)i=\left(i^{2}\right)^7i=\left(-1\right)^7
i=\left[(-1)^6 × -1 \right]i=\left[\left(-1^2\right)^3 × -1 \right]i$
${}=\left[\left(1\right)^3 × -1 \right]i=\left[1 × -1 \right]i=\left[-1 \right]i=-i$
◼ Next we will consider negative powers.
Negative powers can be converted into positive powers by taking the reciprocals. Once the powers become positive, we can use the above method.
• Let us see an example:
Find the value of $i^{-18}$
Solution:
(i) $i^{-18}= \frac{1}{i^{18}}$
(ii) Dividing 18 by 4, we get: $\frac{18}{4}=4\frac{2}{4}$
♦ So the remainder is 2
♦ Thus we get: $i^{-18}= \frac{1}{i^{18}}=\frac{1}{-1}=-1$
(iii) Check:
$i^{-18}= \left[ i^{-1}\right]^{18}=(-i)^{18}=(-1)^{18}(i)^{18}=(1)(i)^{18}$
${}=i^{18}=\left(i^{2}\right)^9=\left(-1\right)^9=-1$
F. The square roots of negative real numbers
• In the first section of this chapter, we derived the result:
If '$a$' is a positive real number, $\sqrt{-a}=\sqrt{a} × \sqrt{-1}=\sqrt{a}\,i$
• Using this result, we can write the square root of any real number.
• However, there is a possible contradiction when we take the square root of the product of two -ve real numbers. This can be explained in 2 steps:
1. We know that, $\sqrt{a} × \sqrt{b}=\sqrt{ab}$
• This result is applicable when:
♦ Both a and b are greater than zero.
♦ a is greater than 0 but b is less than zero.
✰ $\sqrt{a} × \sqrt{-b}=\sqrt{a} × \sqrt{b}\,i = \sqrt{ab}\,i$
♦ b is greater than 0 but a is less than zero.
✰ $\sqrt{-a} × \sqrt{b}=\sqrt{a}\,i × \sqrt{b} = \sqrt{ab}\,i$
2. What happens if both a and b are less than zero?
Answer: If both are less than zero, it is possible to write as:
$\sqrt{-a} × \sqrt{-b}=\sqrt{a}\,i × \sqrt{b}\,i = \sqrt{ab} × i^2 =\sqrt{ab} × -1 =-\sqrt{ab}$
• But the square root of the product of two -ve real numbers is never negative. So there is a contradiction.
• Based on this, mathematicians have decided that:
The formula $\sqrt{a} × \sqrt{b}=\sqrt{ab}$ is not applicable when both a and b are less than zero.
G. Identities related to complex numbers.
• Identities help us to perform algebraic operations quickly. Some important identities related to complex numbers are given below:
1. $(z_1+z_2)^2=z_1^2 +2z_1 z_2 + z_2^2 $
• Let us write the proof:
$\begin{array}{ll}
(z_1+z_2)^2&{}={}& (z_1+z_2) × (z_1+z_2)& {} &{} \\
\phantom{(z_1+z_2)^2}&{}={}& (z_1+z_2)z_1~+~(z_1+z_2)z_2 &\color
{green}{\text{Distributive law}} &{} \\
\phantom{(z_1+z_2)^2}&{}={}& z_1^2+z_2z_1+z_1z_2+z_2^2 &\color
{green}{\text{Distributive law}} &{} \\
\phantom{(z_1+z_2)^2}&{}={}& z_1^2+z_1z_2+z_1z_2+z_2^2 &\color
{green}{\text{Commutative law}} &{} \\
\phantom{(z_1+z_2)^2}&{}={}& z_1^2+2z_1z_2+z_2^2 &{} &{} \\
\end{array}$
• Two points can be written about this identity:
(i) We know how to multiply two complex numbers z1 and z2
• If z1 = (a+bi) and z2 = (c+di), then:
z1z2 = (ac-bd)+(ad+bc)i
• So the second term in the above identity can be easily calculated.
(ii) z12 and/or z22 can be calculated in a similar way:
$\begin{array}{ll}
(a+bi)(c+di)&{}={}& (ac-bd)~+~(ad+bc)i & {} \\
(a+bi)(a+bi)&{}={}& a×a~-~b×b~+~(a×b~+~b×a)i &\color
{green}{\text{Put c = a and d = b}} & {} \\
\Rightarrow (a+bi)^2 &{}={}& a^2~-~b^2~+~(ab+ba)i&{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& a^2~-~b^2~+~(ab+ab)i&{} & {} \\
\phantom{(a+bi)(c+di)}&{}={}& \left(a^2-b^2\right)+(2ab)i&{} & {} \\
\end{array}$
2. $(z_1-z_2)^2=z_1^2 -2z_1 z_2 + z_2^2 $
3. $(z_1+z_2)^3=z_1^3 +3z_1^2 z_2+3z_1 z_2^2 + z_2^3 $
4. $(z_1-z_2)^3=z_1^3 -3z_1^2 z_2+3z_1 z_2^2 - z_2^3 $
5. $z_1^2-z_2^2=(z_1+z_2)(z_1-z_2)$
• Identities (2) to (5) above can be proved using methods similar to that we used for (1).
Now we will see some solved examples.
Solved example 5.1
Express the following in the form a+bi
(i) $(-5i)\left(\frac{1}{8}i\right)$ (ii) $(-i)(2i)\left(\frac{1}{8}i\right)^3$
Solution (i):
$\begin{array}{ll}
(-5i)\left(\frac{1}{8}i\right)&{}={}& -1 × 5 × i × \frac{1}{8} × i& {} &{} \\
\phantom{(-5i)\left(\frac{1}{8}i\right)}&{}={}& -1 × 5 × \frac{1}{8} × i^2 &{} \\
\phantom{(-5i)\left(\frac{1}{8}i\right)}&{}={}& -1 × 5 × \frac{1}{8} × -1 &{} \\
\phantom{(-5i)\left(\frac{1}{8}i\right)}&{}={}& \frac{5}{8} &{} \\
\phantom{(-5i)\left(\frac{1}{8}i\right)}&{}={}& \frac{5}{8}+0i &{} \\
\end{array}$
Solution (ii):
$\begin{array}{ll}
(-i)(2i)\left(\frac{1}{8}i\right)^3&{}={}& -1 × i × 2 × i × (-1)^3 × \frac{1}{8^3} × i^3& {} &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& -1 × 2 × (-1)^3 × \frac{1}{8^3} × i^5 &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& -1 × 2 × -1 × \frac{1}{8^3} × i &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}&\color
{green}{5\div4=1 \frac{1}{4}\text{ Remainder is 1. So }i^5=i} &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& 2 × \frac{1}{(2^3)^3} × i &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& 2 × \frac{1}{2^9} × i &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& \frac{1}{2^8} × i &{} \\
\phantom{(-i)(2i)\left(\frac{1}{8}i\right)^3}&{}={}& \frac{1}{256}i &{} \\
\end{array}$
Solved example 5.2
Express (5-3i)3 in the form a+bi
Solution:
We can use the identity: $(z_1-z_2)^3=z_1^3 -3z_1^2 z_2+3z_1 z_2^2 - z_2^3 $
$\begin{array}{ll}
(5-3i)^3&{}={}&5^3~-~3 × 5^2 × (3i)~+~3 × 5 × (3i)^2~-~ (3i)^3 & {} &{} \\
\phantom{(5-3i)^3}&{}={}&125~-~75 × (3i)~+~15 × 3^2 × i^2~-~ 3^3 × i^3 &{} \\
\phantom{(5-3i)^3}&{}={}&125~-~225i~+~135 × i^2~-~ 27 × i^3 &{} \\
\phantom{(5-3i)^3}&{}={}&125~-~225i~+~135 × -1~-~ 27 × (-i) &{} \\
\phantom{(5-3i)^3}&{}={}&125~-~225i~+~-135~-~ -27i &{} \\
\phantom{(5-3i)^3}&{}={}&125~-135~-~225i~+27i &{} \\
\phantom{(5-3i)^3}&{}={}&-10-198i &{} \\
\end{array}$
Solved example 5.3
Express $(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)$ in the form a+bi
Solution:
$\begin{array}{ll}
(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)&{}={}&(-\sqrt{3}+\sqrt{2}\,i)(2\sqrt{3}-i)& {} &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}&\color
{green}{\because~\sqrt{-2}=\sqrt{2}\,i} &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}={}&[(-\sqrt{3}
× 2\sqrt{3})-(\sqrt{2} × -1)]~+~[(-\sqrt{3} × -1)+(\sqrt{2} ×
2\sqrt{3})]i &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}&\color
{green}{\because~(a+bi)(c+di)=(ac-bd)~+~(ad+bc)i} &{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}={}&[(-2(\sqrt{3})^2)--(\sqrt{2})]~+~[(\sqrt{3})+(2\sqrt{2}×\sqrt{3})]i
&{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}={}&[(-2 × 3)+\sqrt{2}]~+~[(\sqrt{3})i+(2\sqrt{2}×\sqrt{3})i]
&{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}={}&[-6+\sqrt{2}]~+~[\sqrt{3}\,i+(2\sqrt{2}×\sqrt{3}) × i]
&{} \\
\phantom{(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)}&{}={}&(-6+\sqrt{2})+\sqrt{3}(1+2\sqrt{2})i
&{} \\
\end{array}$
• So we have completed a discussion on the algebra of complex numbers.
• In the next section, we will see Modulus and conjugate of a complex
number.
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