13th February 2011 - 04:21 PM
I have a question as follows,
find the value of tan[(-3pi/4) - 2i], where i is a complex number.
The steps I have taken are as follows,
derived tan[ix] = i tanh[x]
using tan[A-B] = (tanA - tanB)/(1+tanAtanB)
substitute itanh[real] for tan[complex], after rearranging and solving I get an approximate answer of...
tan[(-3pi/4) - 2i] = 0.04 + i (2dp)
Would someone please be kind enough to tell me if I am correct? If wrong, no work throughs or correct answers please. Will go away and have another stab at it.
Many thanks in advance,
14th February 2011 - 07:31 AM
tan(-3pi/4) = tan(pi/4-pi) = tan(pi/4) = 1
tan( - 2i) = -i tanh = (-i)(e^2-e^-2)/(e^2 + e^-2)
tan(A+B) = (tanA + tanB)/(1-(tanA)(tanB))
I checked, and these steps worked for an exact answer in terms of i and powers of e.
Tricks to help simplify:
(a + b)/(a - b) = (a^2 + 2ab + b^2)/(a^2 - b^2)
1/(a^2 + b^2) = 1/((a + bi)(a - bi))
The solution is close to (6 decimal places), but not exactly:
13/355 - 1490 i / 1491
so I guess your answer is fine except you seem to have gotten the sign of the imaginary part wrong.