# Complex exponential problem

## 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.

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Just some hints you might have overlooked.

|z*z| = |(-z)*(-z)|

i^2=(z^2)^2=(-(z^2))^2

Just some hints you might have overlooked.

|z*z| = |(-z)*(-z)|

i^2=(z^2)^2=(-(z^2))^2
Ok so I can replace

$$|z*z|[cos(2\alpha)+isin(2\alpha)]$$

with $$[cos(2\alpha)+isin(2\alpha)]$$

since $$|z*z|=|z^2|=|i|=1$$

Ok so I can replace

$$|z*z|[cos(2\alpha)+isin(2\alpha)]$$

with $$[cos(2\alpha)+isin(2\alpha)]$$

since $$|z*z|=|z^2|=|i|=1$$
Yup, but it is those negatives that allow you to get more z's that solve the problem z^4=i

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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 $$z^2 = \pm i\,.$$

$$\text{If }\, z=\left|z\right|e^{\alpha\cdot i}\text{ and }w=\left|w\right|e^{\beta\cdot i}, \text{ then } z\cdot w=\left|z\right|\left|w\right| e^{(\alpha+\beta)i}\,.$$

$$\text{Also, }\ -1=\left|-1\right|e^{\pi i}=1e^{\pi i}=e^{\pi i}\,.$$

$$\text{and, }\ i=\left|i\right|e^{\pi i/2}=e^{\pi i/2}\,.$$

$$e^{2\pi n\,i}=\left(e^{2\pi\,i}\right)^n=\left(1\right)^n=1\,,\ \text{ where n is an integer.}$$

$$-1=e^{\pi i}\cdot e^{2\pi n\,i}=e^{(2\pi n+\pi)\,i}$$

$$\left|w\right|e^{i\beta}=\left|v\right|e^{i\phi}\ \ \implies\ \ \left\{\left|w\right|=\left|\ v\right| \ \text{ and }\ \beta =\phi\ \right\}$$.

Solve: $$\left(\left|z\right|e^{i\alpha}\right)^4=e^{(2\pi n+\pi)\,i}\,.$$

Last edited:
Thanks very much Sammy, I've figured it out :)