How do you multiply $(a - bi)(a + bi)$?

Hint: Solve the question by multiplying $(a - bi)(a + bi)$ and then simplify the equation to get the answer. You should also take care of the fact that $i \times i = {i^2} = - 1$.

Complete step by step solution:
Let $y = (a - bi)(a + bi)$
Therefore on solving the equation we get,
 $y = a.a + a.bi - a.bi - bi.bi$
$y = {a^2} - {b^2}{(i)^2}$
As we know that $i \times i = {i^2} = - 1$
Therefore after substituting the value of iota in the above expression we get
$y = {a^2} + {b^2}$
which is the required answer.

Note:
The square root of a negative real number is called an imaginary quantity or imaginary number.
The quantity $\sqrt { - 1} $ is an imaginary number , denoted by ‘i’ and is called iota.
$i = \sqrt { - 1} $
${i^2} = - 1$
${i^3} = i$
${i^4} = 1$
In other words,
${i^n} = {( - 1)^{n/2}}$ , if n is an even integer
${i^n} = {( - 1)^{(n - 1)/2}}$ , if n is an odd number
A number in the form of $z = x + iy$, where $x,y \in R$ is called a complex number.
The numbers x and y are called the real part and imaginary part of the complex number z respectively.
i.e $x = \operatorname{Re} (z)$ and $y = \operatorname{Im} g(z)$
MULTIPLICATION OF COMPLEX NUMBERS
Let ${z_1} = {x_1} + i{y_1}$ and be any two complex numbers , then there multiplication is defined as
${z_1}{z_2} = ({x_1} + i{y_1})({x_2} + i{y_2})$
$ = ({x_1}{x_2} - {y_1}{y_2}) + i({x_1}{y_2} + {x_2}{y_1})$
Properties of multiplication
1. Commutative : ${z_1}{z_2} = {z_2}{z_1}$
2. Associative: $({z_1}{z_2}){z_3} = {z_1}({z_2}{z_3})$
3. Multiplicative Identity: $z.1 = 1.z = z$
Here, 1 is the multiplicative identity of z.
4. Multiplicative Inverse: Every non-zero complex number z there exists a complex number ${z_1}$ such that ${z_1}z = 1 = z{z_1}$.
5. Distribution Law:
(a) ${z_1}({z_2} + {z_3}) = {z_1}{z_2} + {z_1}{z_3}$ (left distribution).
(b) $({z_1} + {z_2}){z_3} = {z_1}{z_3} + {z_2}{z_3}$(right distribution).