Question

If \[p=9+3i\] and \[q=2-i\], then \[pq\] is equal to
1. \[15+15i\]
2. \[21-3i\]
3. \[18+3i\]
4. \[1-i\]

Answer 1

Hint: Various properties of complex numbers are used to solve this problem. The properties used in this problem are:
\[\begin{align}
  & \Rightarrow \sqrt{-1}=i \\
 & \Rightarrow {{i}^{2}}=-1 \\
\end{align}\]

Complete Step-by-Step solution:
A complex number $z$ has two parts. A real part and an imaginary part. If ${{z}_{1}}=a+ib$ and ${{z}_{2}}=c+id$ are two complex numbers then multiplication of ${{z}_{1}}$ and ${{z}_{2}}$ is denoted by ${{z}_{1}}{{z}_{2}}$.
$\begin{align}
  & \Rightarrow {{z}_{1}}{{z}_{2}}=(a+ib)(c+id) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=(ac+iad+ibc+{{i}^{2}}bd) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=(ac-bd+iad+ibc) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=ac-bd+i(ad+bc) \\
\end{align}$
This means that the real part of ${{z}_{1}}{{z}_{2}}$ is $ac-bd$ and the imaginary part of ${{z}_{1}}{{z}_{2}}$ is $ad+cb$. We will use this property to solve this problem.

The symbol $i$(iota) is used to represent the square root of $\sqrt{-1}$. This also implies, ${{i}^{2}}=-1$, which means $i$ is the solution of the quadratic equation ${{x}^{2}}+1=0$. In this way the square root of any negative number can be expressed using $i$.
$\begin{align}
  & \Rightarrow \sqrt{-1}=i.......(i) \\
 & \Rightarrow {{i}^{2}}=-1........(ii) \\
\end{align}$
Here we have,
\[\Rightarrow p=9+3i......(iii)\]
\[\Rightarrow q=2-i......(iv)\]
In this question we have to find \[pq\].
We know , if ${{z}_{1}}=a+ib$ and ${{z}_{2}}=c+id$ are two complex numbers then multiplication of ${{z}_{1}}$ and ${{z}_{2}}$ is denoted by ${{z}_{1}}{{z}_{2}}$.
$\begin{align}
  & \Rightarrow {{z}_{1}}{{z}_{2}}=(a+ib)(c+id) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=(ac+iad+ibc+{{i}^{2}}bd) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=(ac-bd+iad+ibc) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=ac-bd+i(ad+bc) \\
\end{align}$
Here ${{z}_{1}}$ is $p$ and ${{z}_{2}}$ is $q$.
So comparing equations (i) and (ii) with ${{z}_{1}}$and ${{z}_{2}}$ we can write \[pq\] as,
$\begin{align}
  & \Rightarrow pq=(9+3i)(2-i) \\
 & \Rightarrow pq=(9\times 2-9i+3i\times 2-3{{i}^{2}}).......(v) \\
\end{align}$
Here, in equation (v) we have ${{i}^{2}}$. Substituting this with equation (ii) in equation (v) and simplifying, we get,
$\begin{align}
  & \Rightarrow pq=(18-9i+6i-3\times -1) \\
 & \Rightarrow pq=(18-9i+6i+3) \\
 & \Rightarrow pq=(21-3i).........(vi) \\
\end{align}$
 Hence, the value of \[pq\] is $21-3i$.
$\therefore $the correct answer is option 2.

Note: In this problem we can also find the answer using the formula, $pq=(ac-bd)+i(ad+bc)$.
Here, \[p=9+3i\], this implies \[a=9,b=3\]. Similarly, \[q=2-i\] implies \[c=2,d=-1\].
Directly substituting this we get,
$\begin{align}
  & \Rightarrow pq=(ac-bd)+i(ad+bc) \\
 & \Rightarrow pq=(9\times 2-(3\times -1))+i(9\times -1+3\times 2) \\
 & \Rightarrow pq=(18+3)+i(-9+6) \\
 & \Rightarrow pq=21-3i. \\
\end{align}$