Question

Multiply: $\left( 5-3i \right)\left( 7+2i \right)$

Answer 1

Hint: In this question, we will use the concept of multiplication of two complex numbers using distributive law.

Complete step-by-step solution -

In a given question, we have two complex numbers $5-3i$ and $7+2i$ .
We know that i is an imaginary number such that, $i=\sqrt{-1}$ .
Therefore, squaring both sides of this, we get, ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$ .
Now, multiplying $5-3i$ and $7+2i$ , we get,
$\left( 5-3i \right)\left( 7+2i \right)$
Applying distributive law, we get,
$\left( 5-3i \right)\left( 7+2i \right)=5\left( 7+2i \right)-3\left( 7+2i \right)$
Applying distributive law again, we get,
$\begin{aligned} & \left( 5-3i \right)\left( 7+2i \right)=5\times 7+5\times 2i-3i\times 7-3i\times 2i \\\ & =35+10i-21i-6{{i}^{2}} \\\ \end{aligned}$
Using ${{i}^{2}}=-1$ here, we get
$\begin{aligned} & \left( 5-3i \right)\left( 7+2i \right)=35+10i-21i-6\left( -1 \right) \\\ & =35+10i-21i+6 \\\ \end{aligned}$
Taking i common we get
$\begin{aligned} & \left( 5-3i \right)\left( 7+2i \right)=35+6+i\left( -10-21 \right) \\\ & =41+i\left( -31 \right) \\\ & =41-31i \\\ \end{aligned}$
Hence, on multiplying complex numbers $5-3i$ and $7+2i$ , we get a complex number $41-31i$ , where 41 is its real part and -31 is its imaginary part.

Note: For multiplying any two complex numbers, such that the numbers are $a+ib$ and $c+id$ , we can use a formula for the product which is $ac-bd+i\left( ad+bc \right)$ .