For any real numbers $a$, $b$, $c$ and $d$. Let $z=a+bi$, where $i^2 =-1$.
- Equality of complex number
If $a+ib=c+id$, then $a=c$ and $b=d$.
If $(a+bi)(i^2-3i)=1$, find the values of $a$ and $b$.$\begin{array}{rcl}
(a+bi)(i^2-3i) & = & 1 \\
ai^2-3ai+bi^3-3bi^2 & = & 1 \\
a(-1)-3ai+b(-i)-3b(-1) & = & 1 \\
(-a+3b)+(-3a-b)i & = & 1
\end{array}$By comparing the real parts and the imaginary parts of both sides, we have
$\left\{ \begin{array}{ll}
-a+3b=1 & \ldots \unicode{x2460} \\
-3a-b=0 & \ldots \unicode{x2461}
\end{array}\right.$$\unicode{x2460}+3\times\unicode{x2461}$, we have
$\begin{array}{rcl}
-a-9a & = & 1 \\
-10a & = & 1 \\
a & = & \dfrac{-1}{10}
\end{array}$Sub. $a=\dfrac{-1}{10}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
-3\left(\dfrac{-1}{10}\right)-b & = & 0 \\
b & = & \dfrac{3}{10}
\end{array}$$\therefore a=\dfrac{-1}{10}$ and $b=\dfrac{3}{10}$.
- The zero of complex number
If $a+ib=0$, then $a=b=0$.
- Purely imaginary number
If $z$ is a purely imaginary number, then $a=0$ and $b\neq 0$.
Let $z=\dfrac{a-3i}{3+i}$, where $a$ is a real number. If $z$ is a pure imaginary number, find the value of $a$.$\begin{array}{cl}
& \dfrac{a-3i}{3+i} \\
= & \dfrac{a-3i}{3+i} \times \dfrac{3-i}{3-i} \\
= & \dfrac{3a-ai-9i+3i^2}{9-i^2} \\
= & \dfrac{3a+(-a-9)i+3(-1)}{9-(-1)} \\
= & \dfrac{(3a-3)+(-a-9)i}{10} \\
= & \dfrac{3a-3}{10}+\dfrac{-a-9}{10}i
\end{array}$Since $z$ is a pure imaginary number, then we have
$\begin{array}{rcl}
\dfrac{3a-3}{10} & = & 0 \\
a & = & 1
\end{array}$ - Real number
If $z$ is a real number, then $b=0$.
Let $z=\dfrac{a-3i}{3+i}$, where $a$ is a real number. If $z$ is a real number, find the value of $a$.$\begin{array}{cl}
& \dfrac{a-3i}{3+i} \\
= & \dfrac{a-3i}{3+i} \times \dfrac{3-i}{3-i} \\
= & \dfrac{3a-ai-9i+3i^2}{9-i^2} \\
= & \dfrac{3a+(-a-9)i+3(-1)}{9-(-1)} \\
= & \dfrac{(3a-3)+(-a-9)i}{10} \\
= & \dfrac{3a-3}{10}+\dfrac{-a-9}{10}i
\end{array}$Since $z$ is a real number, then we have
$\begin{array}{rcl}
\dfrac{-a-9}{10} & = & 0 \\
a & = & -9
\end{array}$
