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So I am having trouble visualizing the solutions to $$\cos(z) = 2, z \in \mathbb{C}$$ I know that the solution of this (like mentioned of here Solving $\cos z = 2$ ), but what points does this equation represent on the complex plane? Does it simply represent all the points which are on $x=2$ and $y=0$?

Andrei
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2 Answers2

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Have you thought about considering the polar form of the equation? It might help you think geometrically. $\cos z = \text{Re}(e^{iz})=2$. Then it is important that you consider the complex definition of $\ln$ (not just the principal value) in the absence of a branch cut. Then it should be clear of the periodicity of your solutions.

Joel Biffin
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    If $z \in \mathbb C$, then $\cos z$ is not the real part of $e^{iz}$, rather $$ e^{iz} = e^{i(x+iy)} = e^{-y}(\cos x + i\sin x) $$ $\cos z = \frac{e^{iz}+e^{-iz}}{2}$ is not a real-valued function – Dylan Dec 21 '18 at 18:38
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If $z=x+iy$ then

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

Therefore

\begin{align} \cos(x)\cosh(y) &= 2 \\ \sin(x)\sinh(y) &= 0 \end{align}

Obviously, no real solution exists, so $y\ne 0 \implies \sin x = 0$.

But $\cosh (y) > 0, \forall y \implies \cos x > 0 \implies \cos(x) = 1 \implies x = 2n\pi$, which leaves

$$ \cosh(y) = 2 \implies y = \pm \cosh^{-1}(2) = \pm \ln (2 + \sqrt{3}) $$

So the solution is

$$ x = 2n\pi \pm i\ln(2 + \sqrt{3}) $$

For a visualization, think of the solutions as points on the 2 lines $y = \pm \ln(2+\sqrt{3})$, spaced out evenly by $2n\pi$

Dylan
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  • Why is $\implies \cos(x) = 1$ true? Where does that follow from? Also why there is no real solution? –  Dec 23 '18 at 17:36
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    If $\sin x = 0$ and $\cos x > 0$, what does that imply about the value of $\cos x$? What's the maximum value of $\sin x$ for real $x$? – Dylan Dec 23 '18 at 18:05
  • Ok I understand, when $\sin x = 0$, $\cos x$ can only be 1 or -1. Obvioulsy $|\sin x|$ cannot be greater than 1. –  Dec 23 '18 at 18:09
  • So the justification behind $\sin x =0$ would be that if $\sin x \sinh y= 0$, either $\sin x=0$ or $\sinh y =0$. Assuming $\sinh y= 0$ implies $y = 2\pi n, n \in \mathbb Z$. Therefore $\cosh y = 1$ and $\cos x = 2$ which is impossible, therefore $\sin x =0$. However, since you said that this is obvious is there a "shortcut" to deduce this? –  Dec 23 '18 at 18:42
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    $\sinh y = 0$ only has one solution which is $y=0$ – Dylan Dec 23 '18 at 18:45
  • Why did you add the $\pm$ for the inverse $\cosh$ and not just take the positive side? –  Dec 23 '18 at 18:55
  • It's an even function: $\cosh(-y) = \cosh(y)$. The inverse cosh function only returns the positive solution. It's like saying $x^2 = 2 \implies x = \pm\sqrt 2$ – Dylan Dec 23 '18 at 18:56