Find all solutions to Log(z^3-9)=(pi)*i ,z = a +bi

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SUMMARY

The equation Log(z^3-9)=(pi)*i can be solved by first rewriting it as z^3-9 = e^(pi*i), which simplifies to z^3 = 8. The roots of this equation are found to be -1 - sqrt(3)i, -1 + sqrt(3)i, and 2. This confirms that the solutions to the original logarithmic equation are indeed correct.

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Homework Statement



Sorry to repost this but no one replied to my other thread.

find all solutions to Log(z^3-9)=(pi)*i ,z = a +bi

The Attempt at a Solution



z^3-9 = e^(pi*i)

e^(pi*i)=-1

z^3 = -1 + 9

then the problem becomes finding the roots of

z^3 = 8

which are -1 - sqrt(3)i, -1 + sqrt(3)i, and 2

did I get this right? Thanks.
 
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Yes, that is correct.
 

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