Answer
The complex conjugate of the complex number $2+3i$ is $2-3i$. The multiplication of the complex number $2+3i$ with its complex conjugate $2-3i$ gives a real part equal to $13$ and an imaginary part equal to $0$.
Work Step by Step
Consider the complex number, $2+3i$
If the complex number is in standard form $a+bi$, where $a$ is real part and $b$ is the imaginary part then the complex conjugate of a complex number $a+bi$ is equal to $a-bi$.
So, the complex conjugate of the complex number $2+3i$ is $2-3i$.
The multiplication of the complex number with its complex conjugate is $\left( 2+3i \right)\left( 2-3i \right)$.
FOIL method to multiply the four terms of the product:
\[\left( a+b \right)\left( c+d \right)=\overbrace{ac}^{\text{F}}+\overbrace{ad}^{\text{O}}+\overbrace{bc}^{\text{I}}+\overbrace{bd}^{\text{L}}\]
F is the first terms of each binomial.
O is the outside terms or first term of the first binomial and second term of the second binomial.
I is the inside terms or second term of the first binomial and first term of the second binomial.
L is the last terms of each binomial.
Use the FOIL method.
$\begin{align}
& \left( 2+3i \right)\left( 2-3i \right)=2\cdot 2+2\left( -3i \right)+3i\cdot 2+3i\cdot \left( -3i \right) \\
& =4-6i+6i-9{{i}^{2}}
\end{align}$
The imaginary unit is $i=\sqrt{-1}$, where ${{i}^{2}}=-1$.
Replace the value ${{i}^{2}}=-1$.
$\left( 2+3i \right)\left( 2-3i \right)=4-6i+6i-9\left( -1 \right)$
Make a group of real and imaginary terms.
$\left( 2+3i \right)\left( 2-3i \right)=4+9+6i-6i$
Simplify the real and imaginary terms.
$\begin{align}
& \left( 2+3i \right)\left( 2-3i \right)=\left( 4+9 \right)+\left( 6-6 \right)i \\
& =13+0i \\
& =13
\end{align}$
Therefore, the complex conjugate of the complex number $2+3i$ is $2-3i$. The multiplication of the complex number $2+3i$ with its complex conjugate $2-3i$ gives a real part equal to $13$ and an imaginary part equal to $0$.
