Miniwheat Miniwheat
  • 04-05-2017
  • Mathematics
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Solve for z over the imaginary numbers. Please help!

Solve for z over the imaginary numbers Please help class=

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LammettHash
LammettHash LammettHash
  • 04-05-2017
[tex]2e^z+5=0[/tex]
[tex]2e^z=-5[/tex]
[tex]e^z=-\dfrac52[/tex]
[tex]e^{z+2ki\pi}=-\dfrac52[/tex]

where [tex]k\in\mathbb Z[/tex]. Taking the logarithm of both sides gives

[tex]\log e^{z+2ki\pi}=\log\left(-\dfrac52\right)[/tex]
[tex]\log e^{z+2ki\pi}=\log\dfrac52+\log(-1)[/tex]
[tex]\log e^{z+2ki\pi}=\log\dfrac52+\log(e^{i\pi})[/tex]
[tex]z+2ki\pi=\log\dfrac52+i\pi[/tex]
[tex]z=\log\dfrac52+(2k+1)i\pi[/tex]
Answer Link

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