# $\tan( x + i y ) = a + ib$ then $\tan (x – iy) = a – ib$?

How to prove, if $\tan( x + i y ) = a + ib$ then $\tan (x – iy) = a – ib$ ?

I am not familiar with trignometric identities. So any help will be appreciated.

#### Solutions Collecting From Web of "$\tan( x + i y ) = a + ib$ then $\tan (x – iy) = a – ib$?"

Write $$\tan z={e^{iz}-e^{-iz}\over e^{iz}+e^{-iz}}=1-{2\over 1+e^{-2iz}}$$

From there it follows immediately from

$$\overline{1\over w}={1\over\bar{w}}$$

You’re trying to show that by taking complex conjugates inside the tangent you get the same result as if you put it outside.

So look at the formula, what is $\tan(\bar{z})$? well, it’s:

$$\overline{tan(z)}=\overline{1-{2\over 1+e^{-2i(x+iy)}}}=1-{2\over 1+e^{2y+2ix}}=1-{2\over 1+e^{2i(x-iy)}}=\tan(\bar{z})$$