How do I find real and imaginary parts of inverse circular functions like tan⁻¹(α + iβ)?

In one line

Write the inverse function as an ordinary equation, e.g. \(\tan^{-1}(\alpha+i\beta)=x+iy\) means \(\tan(x+iy)=\alpha+i\beta\), then use the conjugate and separation results to find \(x\) and \(y\).

Explained simply

An inverse function answers "which input gives this output?". For a complex output, the input is also complex, so we want its real part \(x\) and imaginary part \(y\). Instead of working with the inverse directly, flip the equation around so you work with \(\tan\), \(\sin\) or \(\cos\), which you already know how to separate.

Technically, \(w=\sin^{-1}z\) means \(\sin w=z\), \(w=\cos^{-1}z\) means \(\cos w=z\), and \(w=\tan^{-1}z\) means \(\tan w=z\). These have log forms too, for example \(\tan^{-1}z=\dfrac1{2i}\log\dfrac{1+iz}{1-iz}\), but at MU level the "flip and separate" method is enough.

How it works

1. Let the inverse function equal \(x+iy\) and flip: \(\tan(x+iy)=\alpha+i\beta\).
2. Take the conjugate: \(\tan(x-iy)=\alpha-i\beta\).
3. Add the angles: \(\tan2x=\tan[(x+iy)+(x-iy)]=\dfrac{2\alpha}{1-(\alpha^2+\beta^2)}\).
4. Subtract the angles: \(\tan2iy=i\tanh2y=\dfrac{(\alpha+i\beta)-(\alpha-i\beta)}{1+(\alpha^2+\beta^2)}=\dfrac{2i\beta}{1+\alpha^2+\beta^2}\).
5. Solve each for \(x\) and \(y\). For \(\sin^{-1}\) or \(\cos^{-1}\) problems, expand \(\sin(x+iy)\) and equate parts.

Key formula

If \(\tan^{-1}(\alpha+i\beta)=x+iy\): \(\ \tan2x=\dfrac{2\alpha}{1-\alpha^2-\beta^2}\), \(\ \tanh2y=\dfrac{2\beta}{1+\alpha^2+\beta^2}\), with \(\alpha,\beta,x,y\) real.

Quick example

Separate \(\tan^{-1}(e^{i\theta})\) into real and imaginary parts (\(\theta\) real, \(\cos\theta\ne0\)). Here \(\alpha=\cos\theta\), \(\beta=\sin\theta\), so \(\alpha^2+\beta^2=1\).

\(\tan2x=\dfrac{2\cos\theta}{0}\to\infty\) when \(\cos\theta>0\), so \(2x=\dfrac\pi2\) and \(x=\dfrac\pi4\) (principal value; if \(\cos\theta<0\) it tends to \(-\infty\) and \(x=-\dfrac\pi4\); in general \(x=\dfrac{n\pi}{2}+\dfrac\pi4\)).

\(\tanh2y=\dfrac{2\sin\theta}{2}=\sin\theta\Rightarrow y=\dfrac12\tanh^{-1}(\sin\theta)=\dfrac14\log\dfrac{1+\sin\theta}{1-\sin\theta}\).

So \(\tan^{-1}(e^{i\theta})=\dfrac\pi4+\dfrac i4\log\dfrac{1+\sin\theta}{1-\sin\theta}\) (principal value).

Common mistakes

• Trying to split \(\tan^{-1}(\alpha+i\beta)\) as \(\tan^{-1}\alpha+i\tan^{-1}\beta\); inverse functions do not split like that.
• Forgetting the factor \(i\) in \(\tan2iy=i\tanh2y\).
• Not converting \(\tanh^{-1}\) into its log form when the question asks for the answer in log form.

How MU asks it

Usually a 6-mark question: "If \(\tan^{-1}(\alpha+i\beta)=x+iy\), prove \(x=\dfrac12\tan^{-1}\dfrac{2\alpha}{1-\alpha^2-\beta^2}\)" or "Separate \(\sin^{-1}(e^{i\theta})\) into real and imaginary parts". For full marks: write the flipped equation, the conjugate equation, the compound angle step, and the final values. The syllabus says "simple examples", so expect clean values.