College Algebra, Chapter 4, 4.5, Section 4.5, Problem 70

Find all solutions of equation

$\begin{equation} \begin{aligned} &\text{(a)} 2x + 4i = 1\\ \\ &\text{(b)} x^2 - ix = 0\\ \\ &\text{(c)} x^2 + 2ix - 1 = 0\\ \\ &\text{(d)} ix^2 - 2x + i = 0 \end{aligned} \end{equation}$


$\begin{equation} \begin{aligned} \text{a.) } 2x + 4i &=1 \\ \\ 2x &= 1 - 4i && \text{Subtract } 4i\\ \\ x &= \frac{1-4i}{2} && \text{Divide by 2} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{b.) } x^2 - i x &= 0\\ \\ x(x-i) &=0 && \text{Factor out } x\\ \\ x &= 0 \text{ and } x - i = 0 && \text{Zero product property}\\ \\ x &= 0 \text{ and } x = i && \text{Solve for } x \end{aligned} \end{equation}$


c.) To find the solution for $x^2 + 2ix - 1 = 0$, we use quadratic formula

$\begin{equation} \begin{aligned} x &= \frac{-2i \pm \sqrt{(2i)^2 - 4(1)(-1)}}{2(1)}\\ \\ &= \frac{-2i \pm \sqrt{4i^2 + 4}}{2}\\ \\ &= \frac{-2i \pm \sqrt{-4 + 4}}{2}\\ \\ &= \frac{-2i}{2} = -i \end{aligned} \end{equation}$

Thus, the solution is $x = -i$

d.) To find the solution for $ix^2 - 2x + i = 0$, we use quadratic formula

$\begin{equation} \begin{aligned} x &= \frac{-(-2) \pm \sqrt{(-2)^2 - 4(i)(i)}}{2i} \\ \\ &= \frac{2\pm \sqrt{4-4i^2}}{2i}\\ \\ &= \frac{2 \pm \sqrt{4+4}}{2i}\\ \\ &= \frac{2 \pm 2 \sqrt{2}}{2i}\\ \\ &= \frac{1\pm\sqrt{2}}{i} \end{aligned} \end{equation}$

By multiplying the complex conjugate

$\begin{equation} \begin{aligned} x &= \frac{1 \pm \sqrt{2}}{i} \left( \frac{-i}{-i} \right) \\ \\ &= \frac{(1 \pm \sqrt{2})(-i)}{-i^2}\\ \\ &= \frac{(1 \pm \sqrt{2})(-i)}{1} \\ \\ &= (1 \pm \sqrt{2})(-i) \end{aligned} \end{equation}$

Thus, the complex solutions are $x = (1 + \sqrt{2})(-i)$ and $x = ( 1 - \sqrt{2})(-i)$