Solving for z in the Equation tan z = 1 + 2i

[ver_mathstats]

Homework Statement

Find values of tan^-1(1+2i)

Relevant Equations

tan^-1(1+2i)

Find the values of tan^-1(1+2i).

We can use the fact: tan^-1z = (i/2)log((i+z)/(i-z)).

Then with substitutions we have (i/2)log((1+3i)/(-i-1)).

Then I think the next step would be (i/2)(log(1+3i)-log(-1-i)).

Do we then just proceed to solve log(1+3i) and log(-1-i)? I'm just a little confused what the next step should be, any help is appreciated, thank you.

 

[Mark44]

Then with substitutions we have (i/2)log((1+3i)/(-i-1)).

Then I think the next step would be (i/2)(log(1+3i)-log(-1-i)).

Instead, simplify the argument to the log function. That fraction simplifies to -2 - i .

Do we then just proceed to solve log(1+3i) and log(-1-i)?

You're not "solving" these expressions since they're not equations.

 

[ver_mathstats]

Instead, simplify the argument to the log function. That fraction simplifies to -2 - i .

You're not "solving" these expressions since they're not equations.

Sorry, I miswrote that.

 

[anuttarasammyak]

Let z be the value we want, I would write the equation
$$\tan z =1+2i=-i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$$
to get z.