1. Definitions

For any real numbers $a$, $b$, $c$ and $d$. Let $z=a+bi$, where $i^2=-1$.

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 $b$.

   $\begin{array}{rcl}(a+bi)(i^2-3i)& =& 1\\ a{i}^2-3ai+b{i}^3-3b{i}^2& =& 1\\ a(-1)-3ai+b(-i)-3b(-1)& =& 1\\ (-a+3b)+(-3a-b)i& =& 1\end{array}$

   By comparing the real parts and the imaginary parts of both sides, we have

   $\{\begin{array}{ll}-a+3b=1& \dots \text{①}\\ -3a-b=0& \dots \text{②}\end{array}$

   $\text{①}+3\times \text{②}$, we have

   $\begin{array}{rcl}-a-9a& =& 1\\ -10a& =& 1\\ a& =& {\displaystyle \frac{-1}{10}}\end{array}$

   Sub. $a={\displaystyle \frac{-1}{10}}$ into $\text{②}$, we have

   $\begin{array}{rcl}-3\left({\displaystyle \frac{-1}{10}}\right)-b& =& 0\\ b& =& {\displaystyle \frac{3}{10}}\end{array}$

   $\therefore a={\displaystyle \frac{-1}{10}}$ and $b={\displaystyle \frac{3}{10}}$.

2. The zero of complex number

   If $a+ib=0$, then $a=b=0$.

3. Purely imaginary number

   If $z$ is a purely imaginary number, then $a=0$ and $b\ne 0$.

   Let $z={\displaystyle \frac{a-3i}{3+i}}$, where $a$ is a real number. If $z$ is a pure imaginary number, find the value of $a$.

   $\begin{array}{cl}& {\displaystyle \frac{a-3i}{3+i}}\\ =& {\displaystyle \frac{a-3i}{3+i}}\times {\displaystyle \frac{3-i}{3-i}}\\ =& {\displaystyle \frac{3a-ai-9i+3i^2}{9-i^2}}\\ =& {\displaystyle \frac{3a+(-a-9)i+3(-1)}{9-(-1)}}\\ =& {\displaystyle \frac{(3a-3)+(-a-9)i}{10}}\\ =& {\displaystyle \frac{3a-3}{10}}+{\displaystyle \frac{-a-9}{10}}i\end{array}$

   Since $z$ is a pure imaginary number, then we have

   $\begin{array}{rcl}{\displaystyle \frac{3a-3}{10}}& =& 0\\ a& =& 1\end{array}$

4. Real number

   If $z$ is a real number, then $b=0$.

   Let $z={\displaystyle \frac{a-3i}{3+i}}$, where $a$ is a real number. If $z$ is a real number, find the value of $a$.

   $\begin{array}{cl}& {\displaystyle \frac{a-3i}{3+i}}\\ =& {\displaystyle \frac{a-3i}{3+i}}\times {\displaystyle \frac{3-i}{3-i}}\\ =& {\displaystyle \frac{3a-ai-9i+3i^2}{9-i^2}}\\ =& {\displaystyle \frac{3a+(-a-9)i+3(-1)}{9-(-1)}}\\ =& {\displaystyle \frac{(3a-3)+(-a-9)i}{10}}\\ =& {\displaystyle \frac{3a-3}{10}}+{\displaystyle \frac{-a-9}{10}}i\end{array}$

   Since $z$ is a real number, then we have

   $\begin{array}{rcl}{\displaystyle \frac{-a-9}{10}}& =& 0\\ a& =& -9\end{array}$

Same Topic:

2. Basic Operations of Complex Number Direct Variation Partial Variation Percentage Changed