Question

Find z that is complex number : cos z = 5

Answer 1

Recall that

`\cos z=\frac{e^(iz)+e^(-iz)}2`

Then

`\frac{e^(iz)+e^(-iz)}2=5\implies e^(iz)+e^(-iz)=10\implies e^(2iz)-10e^(iz)+1=0`

Let
`\zeta=e^(iz)`, then

`\zeta^2-10\zeta+1=0\implies\zeta=5\pm2\sqrt6`

Solving back for
`z` gives

`e^(iz)=5+2\sqrt 6\implies z=2n\pi-i\log(5+2\sqrt6)`

or

`e^(iz)=5-2\sqrt6\implies z=2n\pi+i\log(5-2\sqrt6)`

where
`n` is any integer.