# Find Value of Complex Trig

• ver_mathstats

#### 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.

ver_mathstats said:
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 .
ver_mathstats said:
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
Mark44 said:
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.

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.

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