Question

Solve for z over the imaginary numbers. Please help!

Answer 1

`2e^z+5=0`

`2e^z=-5`

`e^z=-\frac52`

`e^(z+2ki\pi)=-\frac52`

where
`k\in\mathbb Z`. Taking the logarithm of both sides gives


`\log e^(z+2ki\pi)=\log\left(-\frac52\right)`

`\log e^(z+2ki\pi)=\log\frac52+\log(-1)`

`\log e^(z+2ki\pi)=\log\frac52+\log(e^(i\pi))`

`z+2ki\pi=\log\frac52+i\pi`

`z=\log\frac52+(2k+1)i\pi`