Question

How do you simplify $\dfrac{4+9i}{2+6i}$?

Answer 1

Hint: In the above question, we are given an expression which is a complex fraction, in which both the numerator and the denominator are fractions, and are respectively equal to $4+9i$ and $2+6i$. We first need to rationalize the denominator, by multiplying the numerator and the denominator by the conjugate of the denominator $2+6i$, which is equal to $2-6i$. For simplifying the denominator, we have to use the algebraic property $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ and the identity ${{i}^{2}}=-1$ and for simplifying the numerator, we have to use the distributive property of multiplication, given by $\left( a+b \right)\left( c+d \right)=a\left( c+d \right)+b\left( c+d \right)$. Final simplified expression will be obtained in the form of $x+iy$

Complete step by step solution:
Let us write the complex number given in the above question as
$\Rightarrow z=\dfrac{4+9i}{2+6i}$
For simplifying the above expression, we have to rationalize the denominator by multiplying the numerator and the denominator by the conjugate of $2+6i$, which is equal to $2-6i$. On multiplying the numerator and denominator of the above expression by $2-6i$, we get
$\Rightarrow z=\dfrac{\left( 4+9i \right)\left( 2-6i \right)}{\left( 2+6i \right)\left( 2-6i \right)}$
Now, using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ in the denominator, the above expression becomes
$\begin{align}
  & \Rightarrow z=\dfrac{\left( 4+9i \right)\left( 2-6i \right)}{{{\left( 2 \right)}^{2}}-{{\left( 6i \right)}^{2}}} \\
 & \Rightarrow z=\dfrac{\left( 4+9i \right)\left( 2-6i \right)}{4-\left( -36 \right)} \\
 & \Rightarrow z=\dfrac{\left( 4+9i \right)\left( 2-6i \right)}{40} \\
\end{align}$
Now, we use the distributive property of multiplication, given by $\left( a+b \right)\left( c+d \right)=a\left( c+d \right)+b\left( c+d \right)$ on the numerator to get
\[\begin{align}
  & \Rightarrow z=\dfrac{4\left( 2-6i \right)+9i\left( 2-6i \right)}{40} \\
 & \Rightarrow z=\dfrac{8-24i+18i-54{{i}^{2}}}{40} \\
 & \Rightarrow z=\dfrac{8-6i-54{{i}^{2}}}{40} \\
\end{align}\]
Now, since \[{{i}^{2}}=-1\], we have
\[\begin{align}
  & \Rightarrow z=\dfrac{8-6i-54\left( -1 \right)}{40} \\
 & \Rightarrow z=\dfrac{8-6i+54}{40} \\
 & \Rightarrow z=\dfrac{62-6i}{40} \\
 & \Rightarrow z=\dfrac{62}{40}-\dfrac{6i}{40} \\
 & \Rightarrow z=\dfrac{31}{20}-\dfrac{3i}{20} \\
\end{align}\]
Hence, the given complex fraction is simplified as \[\dfrac{31}{20}-\dfrac{3i}{20}\].

Note:
While applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ to simplify the denominator $\left( 2+6i \right)\left( 2-6i \right)$, do not write it as ${{2}^{2}}-{{6}^{2}}$ since a factor of $i$ is also there which is actually ${{2}^{2}}-{{\left( 6i \right)}^{2}}$. And since the square of $i$ is equal to $-1$, the denominator became equal to ${{2}^{2}}+{{6}^{2}}$. We may also remember the identity $\left( a+ib \right)\left( a-ib \right)={{a}^{2}}+{{b}^{2}}$ so as to avoid chances of calculation mistakes.