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 $x = ( 1 - \sqrt{2})(-i)$