Question

Express the complex number $\dfrac{3+2\text{i}}{-2+\text{i}}$ in the standard form of (a + ib).

Answer 1

Hint: We have to rationalize the given equation in the question and then we compare the following to the general term of a complex number which is a + ib. After rationalizing, we multiply the terms in the numerator and obtain our answer.

Complete Step-by-Step solution:
Complex numbers are numbers which are represented on the imaginary plane. They are represented in the following number: a + ib, where a denotes the real part of the complex number and b denotes the imaginary part.
Some of the basic identities we need to remember before we proceed into the question are
1.) i2 = 1
2.) i3 = -i
3.) i4 = 1
With these in mind, let us proceed with the question:
We rationalize the term inside the bracket which means multiplying the number with its conjugate. For example, if we have to rationalize a + ib, we multiply the term with a – ib.
$\left( \dfrac{3+2\text{i}}{-2+\text{i}} \right)\left( \dfrac{-2-\text{i}}{-2-\text{i}} \right)$
= $\dfrac{-\left( 3+2\text{i} \right)\left( 2+\text{i} \right)}{4-{{\left( \text{i} \right)}^{2}}}$,
= $\dfrac{-\left( 6+7\text{i}-2 \right)}{5}$,
= $\dfrac{-4-7\text{i}}{5}$.
So, $\dfrac{3+2\text{i}}{-2+\text{i}}$ in the form of a + ib solves down to $\dfrac{-4-7\text{i}}{5}$.

Note: When we multiply with the conjugate it gives us a simplified solution. So, remember that we have to take the conjugate carefully as it leads to elimination of the imaginary part in the denominator which makes the question more approachable.