Question

How do you simplify $\left( 4-3i \right)\left( 2+5i \right)$ and write in $a+bi$ form?

Answer 1

Hint: The given expression consists of the product of two complex numbers, which are \[4-3i\] and \[2+5i\]. So in order to simplify the given expression, we have to solve this product. For this, we have to use the distributive law of the algebraic multiplication, which is given by $\left( a+b \right)\left( c+d \right)=a\left( c+d \right)+b\left( c+d \right)$. Then on again applying the distributive law on the two parenthesis, the expression will be further simplified. Since $i$ is a complex number which is equal to the square root of $-1$, we can substitute ${{i}^{2}}=-1$ to finally obtain the simplified expression in the form of $a+bi$.

Complete step by step solution:
Let us consider the given expression as
$\Rightarrow z=\left( 4-3i \right)\left( 2+5i \right)$
Using the distributive law of the algebraic multiplication, given as $\left( a+b \right)\left( c+d \right)=a\left( c+d \right)+b\left( c+d \right)$, we can simplify the above expression as
$\Rightarrow z=4\left( 2+5i \right)-3i\left( 2+5i \right)$
Now, again using the distributive law for the two parenthesis, the above expression will be further simplified as
$\begin{align}
  & \Rightarrow z=4\left( 2 \right)+4\left( 5i \right)-3i\left( 2 \right)-3i\left( 5i \right) \\
 & \Rightarrow z=8+20i-6i-15{{i}^{2}} \\
 & \Rightarrow z=8+14i-15{{i}^{2}} \\
\end{align}$
Now, we know that $i$ is equal to the square root of $-1$. Therefore we can substitute ${{i}^{2}}=-1$ in the above expression to get
$\begin{align}
  & \Rightarrow z=8+14i-15\left( -1 \right) \\
 & \Rightarrow z=8+14i+15 \\
 & \Rightarrow z=23+14i \\
\end{align}$
Hence, the given expression is simplified and written in the form of $a+bi$.

Note:
The simplification of a complex expression basically means to write its real and imaginary parts separately. As given in the above question, we have to write it in the form of $a+bi$. This means that the highest power of $i$ must be equal to one, while all of the higher powers of $i$ must be solved using the identity ${{i}^{2}}=-1$. So after multiplication, do not conclude your final expression as $8+14i-15{{i}^{2}}$.