How do I separate tan(x+iy) into real and imaginary parts and solve problems based on it?

In one line

Multiply numerator and denominator by the conjugate \(\cos(x-iy)\) to get \(\tan(x+iy)=\dfrac{\sin2x+i\sinh2y}{\cos2x+\cosh2y}\).

Explained simply

For \(\tan(x+iy)=\dfrac{\sin(x+iy)}{\cos(x+iy)}\) the denominator is complex. Just as you rationalise a surd, you "realise" this denominator by multiplying by the conjugate. Then the product-to-sum formulas \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\) and \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\) give a neat answer.

Alternative approach used in many problems: if \(\tan(x+iy)=a+ib\), then \(\tan(x-iy)=a-ib\), and you can use \(\tan2x=\tan[(x+iy)+(x-iy)]\) and \(\tan2iy=\tan[(x+iy)-(x-iy)]\). This second route is usually faster when the question gives you the value of \(\tan(x+iy)\) and asks you to prove something about \(x\) and \(y\), because the sum and difference of the two angles are purely real (\(2x\)) and purely imaginary (\(2iy\)).

Why bother? The function \(\tan\) appears in transmission-line and filter formulas in electrical engineering, and these formulas are only usable once you know which part is real (resistive) and which is imaginary (reactive).

How it works

1. \(\tan(x+iy)=\dfrac{2\sin(x+iy)\cos(x-iy)}{2\cos(x+iy)\cos(x-iy)}\).
2. Numerator: \(\sin2x+\sin2iy=\sin2x+i\sinh2y\).
3. Denominator: \(\cos2x+\cos2iy=\cos2x+\cosh2y\) (real).
4. Split: real part \(\dfrac{\sin2x}{\cos2x+\cosh2y}\), imaginary part \(\dfrac{\sinh2y}{\cos2x+\cosh2y}\).
5. For the conjugate method: \(\tan2x=\dfrac{2a}{1-a^2-b^2}\), and \(\tan2iy=i\tanh2y=\dfrac{2ib}{1+a^2+b^2}\).

Key formula

\(\tan(x+iy)=\dfrac{\sin2x+i\sinh2y}{\cos2x+\cosh2y}\), \(\ \tanh(x+iy)=\dfrac{\sinh2x+i\sin2y}{\cosh2x+\cos2y}\), with \(x,y\) real.

Quick example

Separate \(\tan\left(\dfrac\pi4+i\log2\right)\). Here \(2x=\dfrac\pi2\), \(2y=2\log2=\log4\). So \(\sin2x=1\), \(\cos2x=0\), \(\cosh2y=\dfrac{4+\frac14}2=\dfrac{17}8\), \(\sinh2y=\dfrac{4-\frac14}2=\dfrac{15}8\). Hence \(\tan\left(\dfrac\pi4+i\log2\right)=\dfrac{1+\frac{15}8i}{\frac{17}8}=\dfrac8{17}+\dfrac{15}{17}i\).

Reverse problem: if \(\tan(x+iy)=a+ib\) with \(a=0,\ b=\tfrac12\): \(\tan2x=\dfrac{0}{1-\frac14}=0\Rightarrow x=\dfrac{n\pi}2\), and \(\tanh2y=\dfrac{2(1/2)}{1+\frac14}=\dfrac45\Rightarrow y=\dfrac12\tanh^{-1}\dfrac45=\dfrac14\log\dfrac{1+4/5}{1-4/5}=\dfrac14\log9=\dfrac12\log3\).

Common mistakes

• Multiplying by \(\cos(x+iy)\) instead of its conjugate \(\cos(x-iy)\).
• Using \(\tan2iy=\tanh2y\) without the factor \(i\).
• Forgetting the general solution \(n\pi/2\) when solving \(\tan2x=\) a value.

How MU asks it

Standard 6-mark question: "If \(\tan(\theta+i\phi)=\cos\alpha+i\sin\alpha\), prove \(\theta=\dfrac{n\pi}2+\dfrac\pi4\) and \(\phi=\dfrac12\log\tan\left(\dfrac\pi4+\dfrac\alpha2\right)\)", or "Separate \(\tanh(x+iy)\) into real and imaginary parts". For full marks show the conjugate step or the \(\tan2\theta\), \(\tan2i\phi\) step, simplify \(\tanh^{-1}\) using its log form, and state the general solution.