Separate the following trigonometric function into Real and Imaginary Parts
$$\tan^{-1}e^{i\theta} $$
or
$$\tan^{-1}(\cos\theta+i\sin\theta)$$
I Have made till here
Assuming $x+iy$ is the final outcome after separation
$$\therefore \tan^{-1}e^{i\theta}=x+iy$$
$$\therefore e^{i\theta}=\tan(x+iy)$$
$$\therefore e^{i\theta}= \frac{\sin2x}{cos2x+cos2hy}+i \frac{\sin2hy}{cos2x+cos2hy} $$
$$\therefore \cos\theta+i\sin\theta= \frac{\sin2x}{cos2x+cos2hy}+i \frac{\sin2hy}{cos2x+cos2hy} $$
But How to proceed from here to find the value of $ x $ and $ y $ so as to get $x+iy$
Or is there any other way to get the answer other than the steps that I have used.
Best Answer
Using Rasklonikov's very useful suggestion, we have
$$\tan^{-1}e^{i\theta}=\frac{1}{2i}\ln\left(\frac{1+ie^{i\theta}}{1-ie^{i\theta}}\right)=\frac{i}{2}\ln\left(\frac{1-ie^{i\theta}}{1+ie^{i\theta}}\right)=\frac{i}{2}\left[\ln(1-ie^{i\theta})-\ln(1+ie^{i\theta})\right]$$
Examining the natural log terms, the log of a complex number $a+ib$ is given by $$\ln(a+ib)=\ln|a+ib|+i\arg(a+ib)=\ln\sqrt{a^2+b^2}+i\tan^{-1}\left(\frac{b}{a}\right)$$
Thus $$\ln(1-ie^{i\theta})=\ln([1+\sin\theta]-i\cos\theta)=\ln\sqrt{2(1+\sin\theta)}-i\tan^{-1}\left(\frac{\cos\theta}{1+\sin\theta}\right)$$
$$\ln(1+ie^{i\theta})=\ln([1-\sin\theta]+i\cos\theta)=\ln\sqrt{2(1-\sin\theta)}+i\tan^{-1}\left(\frac{\cos\theta}{1-\sin\theta}\right)$$
Substituting the above two expressions into the arctan equation, we have
$$\tan^{-1}e^{i\theta}=\frac{1}{2}\left[\tan^{-1}\left(\frac{\cos\theta}{1+\sin\theta}\right)-\tan^{-1}\left(\frac{\cos\theta}{1-\sin\theta}\right)\right]+\frac{i}{2}\ln\left(2\sqrt{(1-\sin^2\theta)})\right)\\=\frac{1}{2}\left[\tan^{-1}\left(\frac{\cos\theta}{1+\sin\theta}\right)-\tan^{-1}\left(\frac{\cos\theta}{1-\sin\theta}\right)\right]+\frac{i}{2}\ln\left(2\cos\theta\right)$$