Question

Solve the equation. (List your answers counterclockwise about the origin starting at the positive real axis.)
z8 − i = 0

Answer 1

In order to solve this kind of equations, it's easier to use de Moivre's formula for complex roots, which says that the equation
\(z^n = r \exp\left(\textsf{i}\theta\right)\)
has solutions given by
\(z = \sqrt[n]{r} \exp\left(\textsf{i} \dfrac{\theta + 2\pi k}{n}\right),\quad \textrm{with } k \in \{0,1,2,\dots,n-1\}.\)

In this case, we have the equation
\(z^8 = \textsf{i} = \exp\left(\textsf{i}\frac{\pi}{2}\right),\)
so we have \(n=8\), \(r=1\) and \(\theta = \frac{\pi}{2}\). Can you now finish the calculation? \(\ddot{\smile}\)

Answer 2

I have no idea what this is and I definitely have worked with radial forms before. If this is Calculus 3 I can't help you here

Answer 3

@Vocaloid (math Q sorry)

Answer 4

\(\color{#0cbb34}{\text{Originally Posted by}}\) @kittybasil
I have no idea what this is and I definitely have worked with radial forms before. If this is Calculus 3 I can't help you here
\(\color{#0cbb34}{\text{End of Quote}}\)
I'm not sure about what course the OP is taking, but this is a question regarding complex numbers, so I don't think it is Calculus III. I would guess it might be a something taught in high school maths or maybe a rather simple review question you can find in complex analysis. \(\ddot{\smile}\)