# Complex Trignometric Functions

• Apr 11th 2011, 11:47 PM
Glitch
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.
• Apr 12th 2011, 12:13 AM
FernandoRevilla
Better:

$\tan(1-i)=\dfrac{\sin (1-i)}{\cos (1-i)}=\ldots=\dfrac{1}{i}\dfrac{ e^{1+i}-e^{-1-i}}{ e^{1+i}+e^{-1-i}}=\ldots$
• Apr 12th 2011, 01:01 AM
Glitch
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...
• Apr 12th 2011, 03:27 AM
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.
• Apr 12th 2011, 03:51 AM
Glitch
Quote:

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. :/
• Apr 12th 2011, 04:02 AM
Opalg
Quote:

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.
• Apr 12th 2011, 04:16 AM
Glitch
Thanks again!