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Calculate $\tan(1+i)$.

I use the expression $\tan x = -i\dfrac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}$. So it only remains to calculate $e^{i(1+i)}$ (and then $e^{-i(1+i)}$ follows by taking the reciprocal).

So $e^{i(1+i)} = e^ie^{ii} = e^i/e$.

For $e^i$ I use the formula $e^x = 1+x+\dfrac{x^2}{2!}+\ldots$, so $$e^i = 1+i-\dfrac{1}{2!}-\dfrac{i}{3!}+\dfrac{1}{4!} = \left(1-\dfrac{1}{2!}+\dfrac{1}{4!}-\ldots\right)+i\left(1-\dfrac{1}{3!}+\dfrac{1}{5!}-\ldots\right)$$

and there's probably no way to calculate that except say that it's equal to $\cos 1+i\sin 1$.

So $e^i/e = (\cos 1+i\sin 1)/e$, and $e/e^i = e/(\cos 1+i\sin 1)$.

This seems to yield a very ugly expression for the original $\tan(1+i)$. Is there any way to simplify?

Mika H.
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3 Answers3

3

Hint:

$$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}.$$ $$\tan(ix)=i\tanh(x)$$

Ali Caglayan
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2

HINT:

Let $a+ib=\tan(1+i),\implies a-ib=\tan(1-i)$

So using $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ and $\cos(ix)=\cosh x,\sin(ix)=i\sinh x$

$$2a=\tan(1+i)+\tan(1-i)=\frac{\sin2}{\cos(1+i)\cos(1-i)}=\frac{2\sin2}{\cos2i+\cos2}=\frac{2\sin2}{\cosh2+\cos2}$$

$$2ib=\tan(1+i)-\tan(1-i)=\frac{\sin2i}{\cos(1+i)\cos(1-i)}=\frac{2i\sinh2}{\cos2i+\cos2}=\frac{2i\sinh2}{\cosh2+\cos2}$$

lab bhattacharjee
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  • @Alizter, sorry for the typo. Rectified – lab bhattacharjee Aug 27 '13 at 13:28
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    Why is it that $\tan(1+i)=a+ib$ then $\tan(1-i)=a-ib$? – Mika H. Aug 27 '13 at 13:49
  • @MikaH., if $X+iY=f(x+iy)=g(x,y)+ih(x,y)$ what will be $X-iY=?$ – lab bhattacharjee Aug 27 '13 at 13:52
  • We have $f(x-iy) = g(x,-y)+ih(x,-y)$. But that's not necessarily equal to $X-iY$, is it? – Mika H. Aug 27 '13 at 14:01
  • @MikaH., then $X-iY=f(x-iy)=g(x,y)-ih(x,y)$ Btw, what's the end result if you have completed the solution – lab bhattacharjee Aug 27 '13 at 14:36
  • Lab, I haven't completed the problem. I think it's okay to get the idea because the calculation is rather heavy. Thanks anyway! – Mika H. Aug 27 '13 at 14:58
  • @MikaH., Now, I understood your previous comment. You need to replace $i$ with $-i$ – lab bhattacharjee Aug 27 '13 at 15:05
  • But you can't do that, right? $f(x+iy)$ is a function in $x,y$, not in $i$. You can't replace $i$ with $-i$. – Mika H. Aug 27 '13 at 15:17
  • http://books.google.co.in/books?id=zmJkX97FHakC&pg=PA79&lpg=PA79&dq=tan(a%2Bib)&source=bl&ots=Dw98_nl_Dy&sig=Yr-9jx9YUDpiDQh-gABofjEBFhQ&hl=en&sa=X&ei=oOkcUuOMIMbSrQfExoBo&ved=0CDEQ6AEwAQ#v=onepage&q=tan(a%2Bib)&f=false and http://books.google.co.in/books?id=o74n-dcMVIYC&pg=SA5-PA27&lpg=SA5-PA27&dq=tan(a%2Bib)&source=bl&ots=qNrsczGm_A&sig=Gs5Y51fQfL2ye2bRAFM-LaeJ4uc&hl=en&sa=X&ei=oOkcUuOMIMbSrQfExoBo&ved=0CFYQ6AEwBw#v=onepage&q=tan(a%2Bib)&f=false – lab bhattacharjee Aug 27 '13 at 18:05
  • http://books.google.co.in/books?id=qoOOQeRftiYC&pg=PA74&lpg=PA74&dq=tan(a%2Bib)&source=bl&ots=AJ31_zPhBi&sig=y8rl5BG2laqtTs36UqlEU57RV1U&hl=en&sa=X&ei=oOkcUuOMIMbSrQfExoBo&ved=0CFMQ6AEwBg#v=onepage&q=tan(a%2Bib)&f=false and finally just found http://www.askiitians.com/forums/Algebra/22/6987/find-real-part.htm – lab bhattacharjee Aug 27 '13 at 18:05
  • I suppose that your approach can be generalized to work for arbitrary complex number $a+bi$; I have recently posted a more general question if you want to add an answer also there: Real and imaginary part of $\tan(a+bi)$. – Martin Sleziak Nov 04 '18 at 10:09
  • Can we use the Schwarz reflection principle to claim that $\overline{\tan (z)} = \tan (\overline{z})$ because $\tan (z)$ takes real values on the real axis ? – Bijesh K.S May 27 '21 at 08:22
1

use Tan (A+B) Formula

and remember Tan(ix) = i Tanhx

use that in your formula and then simplify as you would normal complex numbers

if you need a detailed answer let me know

cheers

Mr. Math
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