Question

How do you simplify$\left( 4-5i \right)\left( 3+2i \right)$?

Answer 1

Hint: In the given expression in the above question, we are having the product of two complex numbers. For simplifying the given expression, we have to solve for this product. Since the two complex numbers can be treated as two binomial, we can use the distributive property of the algebraic multiplication given as $\left( a+b \right)\left( c+d \right)=a\left( c+d \right)+b\left( c+d \right)$. Then using the identity ${{i}^{2}}=-1$ to simplify the resulting equation, we will finally obtain the simplified expression in the form of $x+iy$.

Complete step by step solution:
The given expression has a product of two complex numbers. Let us write the given expression as
$\Rightarrow z=\left( 4-5i \right)\left( 3+2i \right)$
Now, the above multiplication is similar to the multiplication of two binomials. So we can use the distributive property of the algebraic multiplication, given by $\left( a+b \right)\left( c+d \right)=a\left( c+d \right)+b\left( c+d \right)$, for simplifying the above expression as
\[\begin{align}
  & \Rightarrow z=4\left( 3+2i \right)-5i\left( 3+2i \right) \\
 & \Rightarrow z=4\left( 3 \right)+4\left( 12i \right)-5i\left( 3 \right)-5i\left( 2i \right) \\
 & \Rightarrow z=12+48i-15i-10{{i}^{2}} \\
 & \Rightarrow z=12+33i-10{{i}^{2}} \\
\end{align}\]
Now, we know that ${{i}^{2}}=-1$. Putting this above, we get
$\begin{align}
  & \Rightarrow z=12+33i-10\left( -1 \right) \\
 & \Rightarrow z=12+33i+10 \\
 & \Rightarrow z=22+33i \\
\end{align}$
Finally on taking $11$ common, we get
$\Rightarrow z=11\left( 2+3i \right)$
Hence, the given expression is simplified as $11\left( 2+3i \right)$.

Note:
We may remember the identity for the multiplication of two complex numbers, given by $\left( a+ib \right)\left( c+id \right)=\left( ac-bd \right)+i\left( ad+bc \right)$ for solving these types of questions. If we cannot remember the identity, then we can proceed to solve these types of problems as in the above solutions. But we must not forget to put ${{i}^{2}}=-1$ while solving. It is a common mistake to take ${{i}^{2}}=1$ due to which the final expression is incorrectly obtained. Also, do not forget to take the common factor $11$ common from the obtained expression $22+33i$.