tanh(2+3i)

**dwsmith** · December 19th 2010, 04:56 PM

$\displaystyle tanh(2+3\mathbf{i})=\frac{\mathbf{i}sinh(2+3\mathb f{i})}{cosh(2+3\mathbf{i})}$

$\displaystyle \frac{cos(3)sinh(2)+\mathbf{i}sin(3)cosh(2)}{cos(3 )cosh(2)+\mathbf{i}sin(3)sinh(2)}$

How can I simplify this further without multiplying by the conjugate?

**Drexel28** · December 19th 2010, 06:14 PM

Originally Posted by dwsmith

$\displaystyle tanh(2+3\mathbf{i})=\frac{\mathbf{i}sinh(2+3\mathb f{i})}{cosh(2+3\mathbf{i})}$

$\displaystyle \frac{cos(3)sinh(2)+\mathbf{i}sin(3)cosh(2)}{cos(3 )cosh(2)+\mathbf{i}sin(3)sinh(2)}$

How can I simplify this further without multiplying by the conjugate?

Hint: Use the sum formula for tangent and recall that $\text{tanh}(x)=-i\tan(ix)$ .

**Prove It** · December 19th 2010, 07:36 PM

Also, $\displaystyle \tanh{(2+3i)} = \frac{\sinh{(2+3i)}}{\cosh{(2+3i)}}$ , not $\displaystyle \frac{i\sinh{(2+3i)}}{\cosh{(2+3i)}}$ .

Anyway...

$\displaystyle \tanh{(2+3i)} = \frac{e^{2(2+3i)}-1}{e^{2(2+3i)}+1}$

$\displaystyle = 1 - \frac{2}{e^{2(2+3i)}-1}$

$\displaystyle = 1 - \frac{2}{e^{4 + 6i}-1}$

$\displaystyle = 1 - \frac{2}{e^4\,e^{6i} - 1}$

$\displaystyle = 1 - \frac{2}{e^4(\cos{6} + i\sin{6}) - 1}$

$\displaystyle = 1 - \frac{2}{e^4\cos{6}-1 + i\,e^4\sin{6}}$

$\displaystyle = 1 - \left(\frac{2}{e^4\cos{6} - 1 + i\,e^4\sin{6}}\right)\left(\frac{e^4\cos{6} - 1 - i\,e^4\sin{6}}{e^4\cos{6} - 1 - i\,e^4\sin{6}}\right)$

$\displaystyle = 1 - \frac{2(e^4\cos{6} - 1 - i\,e^4\sin{6})}{(e^4\cos{6} - 1)^2 + (e^4\sin{6})^2}$

$\displaystyle = 1 - \frac{2e^4\cos{6} - 2 - 2i\,e^4\sin{6}}{e^8\cos^2{6} - 2e^4\cos{6} + 1 + e^8\sin^2{6}}$

$\displaystyle = 1 - \frac{2e^4\cos{6} - 2 - 2i\,e^4\sin{6}}{e^8 - 2e^4\cos{6} + 1}$

$\displaystyle = \frac{e^8 - 2e^4\cos{6} + 1 - (2e^4\cos{6} - 2 - 2i\,e^4\sin{6})}{e^8 - 2e^4\cos{6} + 1}$

$\displaystyle = \frac{e^8 - 4e^4\cos{6} + 3 + 2i\,e^4\sin{6}}{e^8 - 2e^4\cos{6} + 1}$

$\displaystyle = \frac{e^8 - 4e^4\cos{6} + 3}{e^8 - 2e^4\cos{6} + 1} + \left(\frac{2e^4\sin{6}}{e^8-2e^4\cos{6} + 1}\right)i$

**dwsmith** · December 19th 2010, 08:02 PM

$\displaystyle tan(\mathbf{i}z)=\frac{\mathbf{i}sinh(z)}{cosh(z)}$

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