Question

Find the value of real number y such that (3 + 2i)(1 + iy) is:
(a) Real
(b) Imaginary

Answer 1

Hint: We have to find the value of y in the two given case, so first we will multiply the two complex number and then we will separate them in real part and imaginary part and after that for case (a) we will equate imaginary part equal to zero and for case (b) we will equate real part equal to zero to find the value of y.

Complete step-by-step answer:
First let’s multiply them and see what we get,
\[\begin{align}
  & \Rightarrow \left( 3\text{ }+\text{ }2i \right)\left( 1\text{ }+\text{ }iy \right) \\
 & \Rightarrow (3-2y)+\left( 2+3y \right)i \\
\end{align}\]
Imaginary part is the coefficient of i and the real part is which does not contain i.
Now in this question we will use these two definitions.
Now for the case (a) we will equate imaginary part equal to zero, therefore,
$\begin{align}
  & \Rightarrow 2+3y=0 \\
 & \Rightarrow 3y=-2 \\
 & \Rightarrow y=\dfrac{-2}{3} \\
\end{align}$
Hence the value of y in case (a) is $\dfrac{-2}{3}$ .
Now for the case (b) we will equate real part equal to zero, therefore,
$\begin{align}
  & \Rightarrow 3-2y=0 \\
 & \Rightarrow 3=2y \\
 & \Rightarrow y=\dfrac{3}{2} \\
\end{align}$
Hence the value of y in case (b) is $\dfrac{3}{2}$ .
So, the answer to this question of case (a) and (b) is $\dfrac{-2}{3}$ and $\dfrac{3}{2}$ respectively.

Note: There are some concepts that one needs to understand this question that is the value of i is $\sqrt{-1}$ and ${{i}^{2}}=-1$, ${{i}^{3}}=-i$ and ${{i}^{4}}=1$, after that it repeats itself. One should also know how to distinguish between real and imaginary parts.