January 27th, 2017, 06:00 PM  #11  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,605 Thanks: 817  Quote:
$\Large z=2 \pi k+\dfrac{1}{2} i \log \left(\left(1+\Im\left(\sqrt{12 i}\right)\right)^2+\left(1+\Re\left(\sqrt{12 i}\right)\right)^2\right)\tan ^{1}\left(\frac{1+\Im\left(\sqrt{12 i}\right)}{1+\Re\left(\sqrt{12 i}\right)}\right)$ $\Large z=2 \pi k\dfrac{1}{2} i \log \left(\left(1+\Im\left(\sqrt{12 i}\right)\right)^2+\left(1+\Re\left(\sqrt{12 i}\right)\right)^2\right)+\tan ^{1}\left(\frac{1+\Im\left(\sqrt{12 i}\right)}{1+\Re\left(\sqrt{12 i}\right)}\right)+\pi$ $\Large k \in \mathbb{Z}$ $\Re(),~\Im()~$ are the real and imaginary parts of their arguments respectively  
January 27th, 2017, 06:06 PM  #12 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,642 Thanks: 960 Math Focus: Elementary mathematics and beyond 
$$z=\frac{\log(1i+\sqrt{12i})}{i}$$

January 27th, 2017, 07:05 PM  #13 
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  
January 27th, 2017, 07:06 PM  #14 
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  
January 27th, 2017, 07:20 PM  #15 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,642 Thanks: 960 Math Focus: Elementary mathematics and beyond 
$$\sin(z)=\frac{e^{iz}e^{iz}}{2i}=1i$$ $$e^{iz}e^{iz}=2i+2$$ $$e^{2iz}1=2ie^{iz}+2e^{iz}$$ With a suitable rearrangement the quadratic formula can be applied. Can you continue? 
January 27th, 2017, 07:45 PM  #16 
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  
January 28th, 2017, 02:06 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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