mathman44 said:
Homework Statement
Let z=|z|e^{\alpha*i}
Using the fact that z*w=|z||w|e^{i(\alpha+\beta)}, find all solutions to
z^4 = -1
The Attempt at a Solution
Not quite sure how to proceed, except for the obvious step
i=z^2=|z*z|e^{i(2\alpha)}= |z*z|[\cos(2\alpha)+\isin(2\alpha)]
Kinda stuck here :s any hints? Thanks.
For one thing, if z^4 = -1\,, then z^2 = \pm i\,.
Seems that the hint might be more helpful had it said:
\text{If }\, z=\left|z\right|e^{\alpha\cdot i}\text{ and }w=\left|w\right|e^{\beta\cdot i}, \text{ then } z\cdot w=\left|z\right|\left|w\right| e^{(\alpha+\beta)i}\,.
\text{Also, }\ -1=\left|-1\right|e^{\pi i}=1e^{\pi i}=e^{\pi i}\,.
\text{and, }\ i=\left|i\right|e^{\pi i/2}=e^{\pi i/2}\,.
Added in edit: Additional helpful facts.
e^{2\pi n\,i}=\left(e^{2\pi\,i}\right)^n=\left(1\right)^n=1\,,\ \text{ where n is an integer.}
-1=e^{\pi i}\cdot e^{2\pi n\,i}=e^{(2\pi n+\pi)\,i}
\left|w\right|e^{i\beta}=\left|v\right|e^{i\phi}\ \ \implies\ \ \left\{\left|w\right|=\left|\ v\right| \ \text{ and }\ \beta =\phi\ \right\}.
Solve: \left(\left|z\right|e^{i\alpha}\right)^4=e^{(2\pi n+\pi)\,i}\,.
