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January 27th, 2017, 05:00 PM   #11
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Quote:
 Originally Posted by romsek well yeah.. you can solve it as $\arcsin(-1-i)$ too but what does that expand to?
Mathematica shows 2 solution templates that produce countably infinite solutions each

$\Large z=2 \pi k+\dfrac{1}{2} i \log \left(\left(1+\Im\left(\sqrt{1-2 i}\right)\right)^2+\left(-1+\Re\left(\sqrt{1-2 i}\right)\right)^2\right)-\tan ^{-1}\left(\frac{1+\Im\left(\sqrt{1-2 i}\right)}{-1+\Re\left(\sqrt{1-2 i}\right)}\right)$

$\Large z=2 \pi k-\dfrac{1}{2} i \log \left(\left(1+\Im\left(\sqrt{1-2 i}\right)\right)^2+\left(-1+\Re\left(\sqrt{1-2 i}\right)\right)^2\right)+\tan ^{-1}\left(\frac{1+\Im\left(\sqrt{1-2 i}\right)}{-1+\Re\left(\sqrt{1-2 i}\right)}\right)+\pi$

$\Large k \in \mathbb{Z}$

$\Re(),~\Im()~$ are the real and imaginary parts of their arguments respectively

 January 27th, 2017, 05:06 PM #12 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,788 Thanks: 1037 Math Focus: Elementary mathematics and beyond $$z=\frac{\log(1-i+\sqrt{1-2i})}{i}$$ Thanks from romsek
January 27th, 2017, 06:05 PM   #13
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Quote:
 Originally Posted by greg1313 $$z=\frac{\text{arsinh}(1-i)}{i}$$ is one solution. I don't know how one would put this in $a+bi$ form.
May I ask how to derived to that?

January 27th, 2017, 06:06 PM   #14
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Quote:
 Originally Posted by greg1313 $$z=\frac{\log(1-i+\sqrt{1-2i})}{i}$$
How did you reach that?

 January 27th, 2017, 06:20 PM #15 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,788 Thanks: 1037 Math Focus: Elementary mathematics and beyond $$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=-1-i$$ $$e^{iz}-e^{-iz}=-2i+2$$ $$e^{2iz}-1=-2ie^{iz}+2e^{iz}$$ With a suitable rearrangement the quadratic formula can be applied. Can you continue? Thanks from Maschke
January 27th, 2017, 06:45 PM   #16
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Quote:
 Originally Posted by greg1313 $$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=-1-i$$ $$e^{iz}-e^{-iz}=-2i+2$$ $$e^{2iz}-1=-2ie^{iz}+2e^{iz}$$ With a suitable rearrangement the quadratic formula can be applied. Can you continue?
I got it thank you!

 January 28th, 2017, 01:06 AM #17 Member   Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 I would probably divide the first equation by the second to get \displaystyle \begin{align*} \sin{(x)}\cosh{(y)} &= -1 \\ \cos{(x)}\sinh{(y)} &= -1 \\ \\ \frac{\sin{(x)}\cosh{(y)}}{\cos{(x)}\sinh{(y)}} &= \frac{-1}{-1} \\ \frac{\tan{(x)}}{\tanh{(y)}} &= 1 \\ \tan{(x)} &= \tanh{(y)} \\ \tan{(x)} &= \frac{\mathrm{e}^{y} - \mathrm{e}^{-y}}{\mathrm{e}^{y} + \mathrm{e}^{-y}} \\ \tan{(x)}\,\left( \mathrm{e}^y + \mathrm{e}^{-y} \right) &= \mathrm{e}^y - \mathrm{e}^{-y} \\ \tan{(x)}\,\mathrm{e}^y + \tan{(x)}\,\mathrm{e}^{-y} &= \mathrm{e}^y - \mathrm{e}^{-y} \\ \mathrm{e}^y\,\left[ \tan{(x)}\,\mathrm{e}^y + \tan{(x)}\,\mathrm{e}^{-y} \right] &= \mathrm{e}^y\,\left( \mathrm{e}^y - \mathrm{e}^{-y} \right) \\ \tan{(x)}\,\left( \mathrm{e}^y \right) ^2 + \tan{(x)} &= \left( \mathrm{e}^y \right) ^2 - 1 \\ 1 + \tan{(x)} &= \left( \mathrm{e}^y \right) ^2 - \tan{(x)} \,\left( \mathrm{e}^y \right) ^2 \\ 1 + \tan{(x)} &= \left( \mathrm{e}^y \right) ^2 \,\left[ 1 -\tan{(x)} \right] \\ \left( \mathrm{e}^y \right) ^2 &= \frac{1 + \tan{(x)}}{1 - \tan{(x)}} \\ \mathrm{e}^y &= \pm \sqrt{ \frac{1 + \tan{(x)}}{1 - \tan{(x)}} } \\ y &= \ln{ \left( \pm \sqrt{ \frac{1 + \tan{(x)}}{1 - \tan{(x)}} } \right) } \end{align*}

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