Solving Complex Exponential Equations: Finding Solutions to z^4 = -1

[mathman44]

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.

 

[transphenomen]

Just some hints you might have overlooked.

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

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

 

[mathman44]

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

 

[transphenomen]

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]

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}\,.

 

[mathman44]

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