January 26th, 2017, 10:37 PM  #1 
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  Find all numbers in z such that sin(z) = 1  i
Hi, I tried this question for several hours but could not get the answer. Detailed steps would be appreciated. Thank you 
January 27th, 2017, 03:56 AM  #2 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 
$\displaystyle \begin{align*} \sin{(z)} &= 1  \mathrm{i} \\ \sin{ \left( x + \mathrm{i}\,y \right) } &= 1  \mathrm{i} \\ \sin{ \left( x \right) } \cos{ \left( \mathrm{i}\,y \right) } + \cos{ \left( x \right) } \sin{ \left( \mathrm{i}\,y \right) } &= 1  \mathrm{i} \end{align*}$ Now some fiddling with Euler's Formula $\displaystyle \begin{align*} \mathrm{e}^{\mathrm{i}\,\theta} &= \cos{ \left( \theta \right) } + \mathrm{i}\,\sin{ \left( \theta \right) } \\ \\ \mathrm{e}^{\mathrm{i}\,\theta} &= \cos{ \left( \theta \right) } + \mathrm{i}\,\sin{ \left( \theta \right) } \\ &= \cos{ \left( \theta \right) }  \mathrm{i}\,\sin{\left( \theta \right) } \\ \\ \mathrm{e}^{\mathrm{i}\,\theta} + \mathrm{e}^{\mathrm{i}\,\theta} &= \cos{ \left( \theta \right) } + \mathrm{i}\,\sin{ \left( \theta \right) } + \cos{ \left( \theta \right) }  \mathrm{i}\,\sin{\left( \theta \right) } \\ &= 2\cos{ \left( \theta \right) } \\ \cos{ \left( \theta \right) } &= \frac{\mathrm{e}^{\mathrm{i}\,\theta} + \mathrm{e}^{\mathrm{i}\,\theta }}{2} \\ \\ \cos{ \left( \mathrm{i}\,\theta \right) } &= \frac{\mathrm{e}^{\mathrm{i}^2\,\theta} + \mathrm{e}^{\mathrm{i}^2\,\theta}}{2} \\ &= \frac{\mathrm{e}^{\theta} + \mathrm{e}^{\theta}}{2} \\ &= \cosh{ \left( \theta \right) } \\ \\ \mathrm{e}^{ \mathrm{i}\,\theta }  \mathrm{e}^{\mathrm{i}\,\theta} &= \cos{ \left( \theta \right) } + \mathrm{i}\,\sin{ \left( \theta \right) }  \left[ \cos{ \left( \theta \right) }  \mathrm{i}\,\sin{ \left( \theta \right) } \right] \\ &= 2\,\mathrm{i}\,\sin{ \left( \theta \right) } \\ \sin{ \left( \theta \right) } &= \frac{\mathrm{e}^{\mathrm{i}\,\theta}  \mathrm{e}^{\mathrm{i}\,\theta}}{2\,\mathrm{i}} \\ \\ \sin{ \left( \mathrm{i}\,\theta \right) } &= \frac{\mathrm{e}^{\mathrm{i}^2\,\theta}  \mathrm{e}^{\mathrm{i}^2\,\theta}}{2\,\mathrm{i}} \\ &= \frac{\mathrm{e}^{\theta}  \mathrm{e}^{\theta}}{2\,\mathrm{i}} \\ &= \frac{\left[ \mathrm{e}^{\theta}  \mathrm{e}^{\theta} \right]}{2\,\mathrm{i}} \\ &= \frac{\mathrm{i}^2\,\left[ \mathrm{e}^{\theta}  \mathrm{e}^{\theta} \right] }{2\,\mathrm{i}} \\ &= \mathrm{i}\,\frac{\mathrm{e}^{\theta}  \mathrm{e}^{\theta}}{2} \\ &= \mathrm{i}\,\sinh{ \left( \theta \right) } \end{align*}$ so $\displaystyle \begin{align*} \sin{ \left( x \right) } \cos{ \left( \mathrm{i}\,y \right) } + \cos{ \left( x \right) } \sin{ \left( \mathrm{i}\,y \right) } &= 1  \mathrm{i} \\ \sin{ \left( x \right) } \cosh{ \left( y \right) } + \mathrm{i}\,\cos{ \left( x \right) } \sinh{ \left( y \right) } &= 1  \mathrm{i} \\ \\ \sin{ \left( x \right) } \cosh{ \left( y \right) } = 1 \textrm{ AND } \cos{ \left( x \right) } \sinh{ \left( y \right) } &= 1 \end{align*}$ I will now leave it to you to solve this system for x and y. 
January 27th, 2017, 09:43 AM  #3  
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  Quote:
Hi there thank you for taking your time to prove, I got that far by googling the equivalent equation of sinx but could not proceed as I did not know how to set cosh(y)sin(x) = 1 isinh(y)cos(x) = i I'm only used to dealing with one equation being 0 so that I can find the 0 points..  
January 27th, 2017, 09:45 AM  #4  
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  Quote:
 
January 27th, 2017, 02:22 PM  #5 
Senior Member Joined: Aug 2012 Posts: 1,438 Thanks: 353  Note that the second line is better written as $\sinh(y) \cos(x) = 1$. Thats the imaginary part of $\sin(x + iy)$, as the top line is the real part. Remember "imaginary part" is a real number. That doesn't make it any more clear what to do next ... Wait, now $\sinh(y) = \frac{1}{\cos(x)} =  \sec(x)$. At least that gets rid of the hyperbolic functions, about which many of us have no intuition at all. Maybe this approach is helpful. Last edited by Maschke; January 27th, 2017 at 02:33 PM. 
January 27th, 2017, 02:42 PM  #6 
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0 
How did you get sinh(y) = 1/cos(y). What's the equivalent trig for cosh(y)? Thanks 
January 27th, 2017, 02:50 PM  #7 
Senior Member Joined: Sep 2015 From: CA Posts: 1,304 Thanks: 669  
January 27th, 2017, 03:17 PM  #8  
Senior Member Joined: Aug 2012 Posts: 1,438 Thanks: 353  Quote:
From $\sinh(y) \cos(x) = −1$ I divided through by $\cos(x)$. Note that one of the $y$'s changed to $x$ by the time I got it right. But I don't think this is going to help much. As a meta clue, where did you get this problem? Thanks I thought it was me! Last edited by Maschke; January 27th, 2017 at 03:34 PM.  
January 27th, 2017, 04:49 PM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,551 Thanks: 926 Math Focus: Elementary mathematics and beyond 
$$z=\frac{\text{arsinh}(1i)}{i}$$ is one solution. I don't know how one would put this in $a+bi$ form. 
January 27th, 2017, 04:51 PM  #10 
Senior Member Joined: Sep 2015 From: CA Posts: 1,304 Thanks: 669  

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