Complex Trignometric Functions

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April 11th 2011, 10:47 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.
April 11th 2011, 11:13 PM
FernandoRevilla

Better:

$\tan(1-i)=\dfrac{\sin (1-i)}{\cos (1-i)}=\ldots=\dfrac{1}{i}\dfrac{ e^{1+i}-e^{-1-i}}{ e^{1+i}+e^{-1-i}}=\ldots$
April 12th 2011, 12:01 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...
April 12th 2011, 02:27 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.
April 12th 2011, 02:51 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. :/
April 12th 2011, 03:02 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.
April 12th 2011, 03:16 AM
Glitch

Thanks again!