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I couldn't find any, I tried to write $\cos(z)$ as $\cos(x)\cos(iy)-\sin(x)\sin(iy)$ which then gave me

$\cos(x)\cosh(y) - i\sin(x)\sinh(y) = -2$

$\sin(x)=0$ so that imaginary part become $0$

now we have to find $\cosh(y) = -2$ which is not true for no $y$.

is it right or i made a mistake in my substitutions?

Boris Valderrama
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no0ob
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3 Answers3

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I should have used $x=(2k+1)\pi$ for the $x$ and then had the

$\cos( (2k+1)\pi ) \cosh(y) = -2$

which made it $\cosh(y) = 2$

Boris Valderrama
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no0ob
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    Right, and here you get two solutions for $y$ with $\pm$ (recall $2\pm\sqrt{3}$ in another approach). – A.Γ. Jan 02 '19 at 22:12
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    Yes; since $\cos x\cosh y<0$, you need to have $\cos x<0$ and therefore $x=\pi+2k\pi$. – egreg Jan 02 '19 at 22:33
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Hint: start with $$ \cos z=\frac{e^{iz}+e^{-iz}}{2} $$ and obtain a quadratic equation for $e^{iz}$.

A.Γ.
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  • i got $\exp(iz) = -2+\sqrt(3)$ and $\exp(iz) = -2-\sqrt(3)$ which gets me that $e^x = -2+\sqrt(3)$ which is negative on the right side and impossible :-? – no0ob Jan 02 '19 at 21:41
  • If $e^{iz}=-2\pm\sqrt{3}$ then $$e^{iz+i\pi}=e^{i\pi}(-2\pm\sqrt{3})=(-1)(-2\pm\sqrt{3})=2\pm\sqrt{3}$$ whith positive RHS. – A.Γ. Jan 02 '19 at 21:43
  • What should $z$ be btw? – no0ob Jan 02 '19 at 21:48
  • @no0ob Do you follow the argumentation? If $$e^{i(z+\pi)}=2\pm\sqrt{3}$$ then how to find $i(z+\pi)$? – A.Γ. Jan 02 '19 at 21:52
  • Ok i think i had a mistake i just assumed $x = 0$ in my original post, i could use $x=\pi$ too – no0ob Jan 02 '19 at 21:52
  • Cause with your path i just have to solve the thing again but this time with a new $x$ like $x' = x+\pi$ so if $x=0$ then $ x'=\pi$ – no0ob Jan 02 '19 at 21:55
  • @no0ob You could use $x=\pi k$ for any integer $k$ (all solutions to $\sin x=0$). – A.Γ. Jan 02 '19 at 22:02
  • as you see i couldn't done that with $0$ cause it made $cos(x)ch(y) = 1*ch(y) = -2$ which is negative RHS – no0ob Jan 02 '19 at 22:08
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    @no0ob Right, sorry, it should be every second zero of $\sin x=0$ to make $\cos x=-1$, i.e. $x=\pi+2\pi k$. – A.Γ. Jan 02 '19 at 22:10
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Be: $$ \cos z=\frac{e^{iz}+e^{-iz}}{2}=-2 $$ Then: \begin{eqnarray} \frac{e^{iz}+e^{-iz}}{2} &=& -2 \\ e^{iz}+e^{-iz} &=& -4 \\ e^{2iz} + 1 &=& -4e^{iz} \\ \left(e^{iz}\right)^2 + 4\left(e^{iz}\right) + 1 &=& 0 \\ e^{iz} &=& \frac{-4 \pm \sqrt{16-4}}{2} \\ e^{iz} &=& -2 \pm \sqrt{3} \\ \end{eqnarray} Then since $e^{iz}=\cos(z)+i\sin(z)$, $e^{iz}=e^{i(z+2k\pi)}$ with $k\in\mathbb{Z}$: \begin{eqnarray} e^{i(z+2k_1\pi)} &=& -2 \pm \sqrt{3} \\ i(z +2k_1\pi) &=& \ln\left(-2 \pm \sqrt{3}\right) \\ z &=& 2k_1\pi - i\ln\left(-2 \pm \sqrt{3}\right) \\ \end{eqnarray} There are two set solution to differents solutions: \begin{eqnarray} z_{+} &=& 2k_1\pi - i\ln\left(-2 + \sqrt{3}\right) \\ z_{-} &=& 2k_1\pi - i\ln\left(-2 - \sqrt{3}\right) \\ \end{eqnarray} For $z_{-}$: \begin{eqnarray} z_{-} &=& 2k_1\pi - i\ln\left(\left(2 + \sqrt{3}\right)e^{-i(1+2k_2)\pi}\right) \\ z_{-} &=& 2k_1\pi - i\left[\ln\left(2 + \sqrt{3}\right)-i(1+2k_2)\pi\right] \\ z_{-} &=& 2k_1\pi - (1+2k_2)\pi - i\ln\left(2 + \sqrt{3}\right) \\ z_{-} &=& (1+2k)\pi - i\ln\left(2 + \sqrt{3}\right) \\ \end{eqnarray} Then, with $k\in\mathbb{Z}$ the solutions are: \begin{eqnarray} z_{+} &=& 2k\pi - i\ln\left(-2 + \sqrt{3}\right) \\ z_{-} &=& (1+2k)\pi - i\ln\left(2 + \sqrt{3}\right) \\ \end{eqnarray}

Boris Valderrama
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