1. ## Complex Trignometric Functions

The question
Express tan(1 - i) in the form a + ib, for a and b real

My attempt

I used the fact that:
sin(z) = sin(x)cosh(y) + icos(x)sinh(y)
cos(z) = cos(x)cosh(y) - isin(x)cosh(y)

I substituded these into tan(z) = \frac{sin(z)}{cos(z)}, and after quite a bit of algebra, got the following:

$\frac{sin(1)cos(1) - isinh(1)cosh(1)}{cos^2(1)cosh^2(1)+sin^2(1)sinh^2( 1)}$

I tried simplifying further, but I got a mess involving coth(1), cot(1) and other assorted trig functions. The solution in my text is:

$\frac{tan1^21 - itanh(1)sec^2(1)}{1 + tanh^2(1)tan^2(1)}$

Could someone guide me on how they got to that solution? Thank you.

2. Better:

$\tan(1-i)=\dfrac{\sin (1-i)}{\cos (1-i)}=\ldots=\dfrac{1}{i}\dfrac{ e^{1+i}-e^{-1-i}}{ e^{1+i}+e^{-1-i}}=\ldots$

3. OK, I see how you got that, but how to I simplify it? I tried expanding the exponentials using De Moivre's theorem, but it's looking like I'm getting a big mess again...

4. From the formula for $\tan(\theta-\phi)$, $\tan(1-i) = \frac{\tan1-\tan i}{1+\tan1\tan i}.$ Then $\tan i = \frac{\sin i}{\cos i} = \frac{i\sinh1}{\cosh1} = i\tanh1.$ Then rationalise the fraction.

5. Originally Posted by Opalg
From the formula for $\tan(\theta-\phi)$, $\tan(1-i) = \frac{\tan1-\tan i}{1+\tan1\tan i}.$ Then $\tan i = \frac{\sin i}{\cos i} = \frac{i\sinh1}{\cosh1} = i\tanh1.$ Then rationalise the fraction.
Thank you. What is that formula called, by the way? I tried to find it earlier, but didn't know what to look for. :/

6. Originally Posted by Glitch
What is that formula called, by the way? I tried to find it earlier, but didn't know what to look for. :/
I don't know that it has a particular name, it's just the addition formula for tan. Incidentally, any such formula that hold for real numbers also holds for complex numbers, by the principle of permanence of functional equations.

7. Thanks again!