Question

Express (2+3i) (2-3i) in the form of $a=ib$, $a,b\in R$ and $i=\sqrt{-1}$. State the values of a and b.

Answer 1

Hint: We will multiply the given expression normally as we do. We will use $i=\sqrt{-1}$ to get ${{i}^{2}}$ and hence we will substitute this to get our answer and then represent it in the form of $a=ib$.

Complete Step-by-step answer:
Before proceeding with the question we should understand the concept related to complex numbers.
Complex numbers are the numbers which are expressed in the form of a+ib where i is an imaginary number called iota and has the value of $\sqrt{-1}$. For example, 2+3i is a complex number, where 2 is a real number and 3i is an imaginary number. Therefore, the combination of both the real number and imaginary number is a complex number. The main application of these numbers is to represent periodic motions such as water waves, alternating current, light waves, etc., which relies on sine or cosine waves etc. There are certain formulas which are used to solve the problems based on complex numbers. Also, the mathematical operations such as addition, subtraction and multiplication are performed on these numbers.
$\Rightarrow (2+3i)(2-3i).......(1)$
So we have to represent this in $a=ib$ form so multiplying equation (1) we get,
$\Rightarrow 2(2-3i)+3i(2-3i).......(2)$
Now opening the bracket and multiplying the terms in equation (2) we get,
$\Rightarrow 4-6i+6i-9{{i}^{2}}.......(3)$
So now cancelling the similar terms in equation (3) and substituting ${{i}^{2}}=-1$ as $i=\sqrt{-1}$ in equation (3) we get,

& \Rightarrow 4-9{{i}^{2}} \\\ & \Rightarrow 4-9(-1)=4+9=13 \\\ \end{aligned}$$ Hence $$13=0i$$. So here a is 13 and b is 0. Note: We don’t have to get confused seeing it in the expression. We also have to remember that $${{i}^{2}}=-1$$. We in a hurry can make a mistake by substituting $${{i}^{2}}$$ as 1 and so we need to be careful while doing this step.