- #1

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z= exp(-z)

could someone possibly point me in the right direction to start this problem?

this area of math is still new to me so please go easy

thanks

- Thread starter gmans
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- #1

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z= exp(-z)

could someone possibly point me in the right direction to start this problem?

this area of math is still new to me so please go easy

thanks

- #2

berkeman

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- #3

arildno

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Probably, he means:

[tex]z=e^{-z}[/tex]

[tex]z=e^{-z}[/tex]

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quasar987

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That's a fun problem. I say start by defining z=a+ib where a and b are real numbers and also note z=|z|exp(i[itex]\Phi[/itex]) the polar form of z, so that you're looking for the solutions to

[tex]|z|e^{i\Phi}=e^{-a-ib}[/tex]

And use the fact that two complex numbers are equal iff their modulus are the same and their polar angle are the same up to a difference of [itex]2n\pi[/itex], [itex]n\in\mathbb{Z}[/itex].

[tex]|z|e^{i\Phi}=e^{-a-ib}[/tex]

And use the fact that two complex numbers are equal iff their modulus are the same and their polar angle are the same up to a difference of [itex]2n\pi[/itex], [itex]n\in\mathbb{Z}[/itex].

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