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January 27th, 2017, 06:00 PM   #11
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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
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January 27th, 2017, 06:06 PM   #12
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$$z=\frac{\log(1-i+\sqrt{1-2i})}{i}$$
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January 27th, 2017, 07:05 PM   #13
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$$z=\frac{\text{arsinh}(1-i)}{i}$$

is one solution. I don't know how one would put this in $a+bi$ form.
May I ask how to derived to that?
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January 27th, 2017, 07:06 PM   #14
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$$z=\frac{\log(1-i+\sqrt{1-2i})}{i}$$
How did you reach that?
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January 27th, 2017, 07:20 PM   #15
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$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=-1-i$$

$$e^{iz}-e^{-iz}=-2i+2$$

$$e^{2iz}-1=-2ie^{iz}+2e^{iz}$$

With a suitable rearrangement the quadratic formula can be applied. Can you continue?
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January 27th, 2017, 07:45 PM   #16
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$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=-1-i$$

$$e^{iz}-e^{-iz}=-2i+2$$

$$e^{2iz}-1=-2ie^{iz}+2e^{iz}$$

With a suitable rearrangement the quadratic formula can be applied. Can you continue?
I got it thank you!
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January 28th, 2017, 02:06 AM   #17
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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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