## Precalculus (6th Edition) Blitzer

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 $0$.
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 $0$.