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Complex exponential problem

  • Thread starter mathman44
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  • #1
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Homework Statement




Let [tex]z=|z|e^{\alpha*i}[/tex]

Using the fact that [tex]z*w=|z||w|e^{i(\alpha+\beta)}[/tex], find all solutions to
[tex]z^4 = -1[/tex]

The Attempt at a Solution



Not quite sure how to proceed, except for the obvious step

[tex]i=z^2=|z*z|e^{i(2\alpha)}= |z*z|[cos(2\alpha)+isin(2\alpha)][/tex]

Kinda stuck here :s any hints? Thanks.
 

Answers and Replies

  • #2
Just some hints you might have overlooked.

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

i^2=(z^2)^2=(-(z^2))^2
 
  • #3
206
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Just some hints you might have overlooked.

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

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

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

with [tex][cos(2\alpha)+isin(2\alpha)][/tex]

since [tex]|z*z|=|z^2|=|i|=1[/tex]
 
  • #4
Ok so I can replace

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

with [tex][cos(2\alpha)+isin(2\alpha)][/tex]

since [tex]|z*z|=|z^2|=|i|=1[/tex]
Yup, but it is those negatives that allow you to get more z's that solve the problem z^4=i
 
  • #5
SammyS
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Homework Statement




Let [tex]z=|z|e^{\alpha*i}[/tex]

Using the fact that [tex]z*w=|z||w|e^{i(\alpha+\beta)}[/tex], find all solutions to
[tex]z^4 = -1[/tex]

The Attempt at a Solution



Not quite sure how to proceed, except for the obvious step

[tex]i=z^2=|z*z|e^{i(2\alpha)}= |z*z|[\cos(2\alpha)+\isin(2\alpha)][/tex]

Kinda stuck here :s any hints? Thanks.
For one thing, if [tex]z^4 = -1\,,[/tex] then [tex]z^2 = \pm i\,.[/tex]

Seems that the hint might be more helpful had it said:

[tex]\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}\,.[/tex]

[tex]\text{Also, }\ -1=\left|-1\right|e^{\pi i}=1e^{\pi i}=e^{\pi i}\,.[/tex]

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

Added in edit: Additional helpful facts.

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

[tex]-1=e^{\pi i}\cdot e^{2\pi n\,i}=e^{(2\pi n+\pi)\,i}[/tex]

[tex]\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\}[/tex].

Solve: [tex]\left(\left|z\right|e^{i\alpha}\right)^4=e^{(2\pi n+\pi)\,i}\,.[/tex]
 
Last edited:
  • #6
206
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Thanks very much Sammy, I've figured it out :)
 

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