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January 26th, 2017, 11:37 PM   #1
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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
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January 27th, 2017, 04:56 AM   #2
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$\displaystyle \begin{align*} \sin{(z)} &= -1 - \mathrm{i} \\ \sin{ \left( x + \mathrm{i}\,y \right) } &= -1 - \mathrm{i} \\ \sin{ \left( x \right) } \cos{ \left( \mathrm{i}\,y \right) } + \cos{ \left( x \right) } \sin{ \left( \mathrm{i}\,y \right) } &= -1 - \mathrm{i} \end{align*}$

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.
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January 27th, 2017, 10:43 AM   #3
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Originally Posted by Prove It View Post
$\displaystyle \begin{align*} \sin{(z)} &= -1 - \mathrm{i} \\ \sin{ \left( x + \mathrm{i}\,y \right) } &= -1 - \mathrm{i} \\ \sin{ \left( x \right) } \cos{ \left( \mathrm{i}\,y \right) } + \cos{ \left( x \right) } \sin{ \left( \mathrm{i}\,y \right) } &= -1 - \mathrm{i} \end{align*}$

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.

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..
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January 27th, 2017, 10:45 AM   #4
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Quote:
Originally Posted by Lalaluye View Post
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..
If I set cos(x) = -1, sin(x) = 0, if I set sin(x) = -1, cos(x) = 0.
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January 27th, 2017, 03:22 PM   #5
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Quote:
Originally Posted by Lalaluye View Post

cosh(y)sin(x) = -1
isinh(y)cos(x) = -i
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 03:33 PM.
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January 27th, 2017, 03:42 PM   #6
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How did you get sinh(y) = -1/cos(y). What's the equivalent trig for cosh(y)?

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January 27th, 2017, 03:50 PM   #7
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Quote:
Originally Posted by Lalaluye View Post
How did you get sinh(y) = -1/cos(y). What's the equivalent trig for cosh(y)?

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he divided the first equation by $\cos(y)$

Based on the answer Mathematica returns this won't be a simple algebra problem to solve.
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January 27th, 2017, 04:17 PM   #8
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Quote:
Originally Posted by Lalaluye View Post
How did you get sinh(y) = -1/cos(y). What's the equivalent trig for cosh(y)?
I edited that a couple of times so I might have posted the wrong thing at some point.

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?


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Based on the answer Mathematica returns this won't be a simple algebra problem to solve.
Thanks I thought it was me!

Last edited by Maschke; January 27th, 2017 at 04:34 PM.
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January 27th, 2017, 05:49 PM   #9
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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.
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January 27th, 2017, 05:51 PM   #10
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Originally Posted by greg1313 View Post
$$z=\frac{\text{arsinh}(1-i)}{i}$$

is one solution. I don't know how one would put this in $a+bi$ form.
well yeah.. you can solve it as $\arcsin(-1-i)$ too

but what does that expand to?
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