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I tried solving this using the definition of $cos(z)=\frac{e^{iz}+e^{-iz}}{2}$ and equating it to $\frac{3}{4}+\frac{i}{4}$ and converting it to a complex quadratic equation through a substitution $t=e^{iz}$ and finding roots via the complex quadratic formula but it didn't seem to work. I would prefer solutions via elementary methods.

Here is my attempt:

By definition we have $\frac{e^{iz}+e^{-iz}}{2}=\frac{3}{4}+\frac{i}{4} \implies e^{iz}+e^{-iz}=\frac{3}{2}+\frac{i}{2}$. Let $t=e^{iz}$ so then we have $t+\frac{1}{t}=\frac{3}{2}+\frac{i}{2}$ and if we multiply both sides by $t$ we have $t^2+1=(\frac{3}{2}+\frac{i}{2})t$ and hence $t^2+(\frac{3}{2}+\frac{i}{2})t+1=0$ By the quadratic formula for complex numbers we have, $a=1, b=\frac{3}{2}+\frac{i}{2}, c=1 \implies z=\frac{-(\frac{3}{2}+\frac{i}{2}) \pm \sqrt{(\frac{3}{2}+\frac{i}{2})^2-4(1)(1)}}{2(1)}$. Simplyifing we have $z=\frac{-\frac{3}{2}-(\frac{1}{2})i \pm \sqrt{-2+(\frac{3}{2})i}}{2}$ We wish to express $-2+(\frac{3}{3})i$ in polar form so we have $|-2+(\frac{3}{2})i|=\frac{5}{2}$. Now equating the real and imaginary parts we have $\frac{5}{2}\cos(\theta)=-2 \implies \cos(\theta)=-\frac{4}{5}$ and $\frac{5}{2}\sin(\theta)=\frac{3}{2} \implies \sin(\theta)=\frac{3}{5}$. From this we have $\tan(\theta)=-\frac{3}{4} \implies \theta=\arctan(-\frac{3}{4}) \approx -.6435$ rad. So we have $w=-2+(\frac{3}{2})i=\frac{5}{4}(\cos(\theta)+i\sin(\theta))=\frac{5}{4}e^{i\theta}$. By Proposition 1.3.12 we have $\sqrt{w}=\sqrt{\frac{5}{4}}e^{\frac{i\theta}{2}}=\frac{\sqrt{5}}{2}e^{\frac{i\theta}{2}}$. Similarily for $-\frac{3}{2}-(\frac{1}{2})=\frac{\sqrt{10}}{2}e^{i\varphi}$ Where $\varphi=\arctan(\frac{1}{3})$. So finally we have $z=\frac{-(\frac{3}{2}+\frac{i}{2}) \pm \sqrt{(\frac{3}{2}+\frac{i}{2})^2-4(1)(1)}}{2(1)}=\frac{\sqrt{10}e^{i\varphi} \pm \sqrt{5}e^{\frac{i\theta}{2}}}{4}$ as solutions to $\cos(z)=\frac{3}{4}+\frac{i}{4}$.

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    $\begingroup$ Your method is the standard technique. $\endgroup$ Jan 26, 2014 at 21:27
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    $\begingroup$ How can it be more elementary? $\endgroup$
    – J.R.
    Jan 26, 2014 at 21:30
  • $\begingroup$ I wasn't aware that was the "standard" technique. One thing I did differently was instead of using the form $a+bi$ for complex numbers, I opted to use $re^{i\theta}$ as the representation so I could avoid taking $\sqrt{a+bi}$ in the discriminant since $\sqrt{re^{i\theta}}=\sqrt{r}e^{\frac{i\theta}{2}}$ seemed easier to work with. $\endgroup$
    – 1028
    Jan 26, 2014 at 21:37
  • $\begingroup$ The only other reasonable technique I know of would be to compute $\arccos$. However, IMO, the formula for $\arccos$ is pretty annoying. $\endgroup$
    – user14972
    Jan 26, 2014 at 21:40
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    $\begingroup$ If you show the details of the computation that "didn't seem to work", it would be much easier for someone to explain what you did wrong. $\endgroup$ Jan 26, 2014 at 21:42

3 Answers 3

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I agree with your approach for the most part, but I think you've messed up on calculating $b$ in the quadratic formula: $$e^{iz} + e^{-iz} -\frac{3}{2} - \frac{i}{2} = 0$$

Using $t = e^{iz}$: $$t + \frac{1}{t} -\frac{3}{2} - \frac{i}{2} = 0$$ Multiplying by $t$: $$t^2\color{red}{-}\left(\frac{3}{2} + \frac{i}{2}\right)t + 1 = 0$$ This is where your solution starts to go wrong. Those minus signs are just lying in wait for the innocent mathematician! :P

Thus: $$\begin{align} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\ &= \frac{\left(\frac{3}{2} + \frac{i}{2}\right) \pm \sqrt{\left(\frac{3}{2} + \frac{i}{2}\right)^2 - 4}}{2}\\ &= \frac{\left(\frac{3}{2} + \frac{i}{2}\right) \pm \sqrt{\left(\frac{9}{4} - \frac{1}{4}+\frac{3}{2}i\right) - 4}}{2}\\ &= \frac{\left(\frac{3}{2} + \frac{i}{2}\right) \pm \frac{1}{2}\sqrt{-8+6i}}{2}\\ &= \frac{3+i \pm (3i+1)}{4}\\ &= \cdots \end{align}$$

Going from line three to line four: $$\begin{align} \frac{\left(\frac{3}{2} + \frac{i}{2}\right) \pm \sqrt{\left(\frac{9}{4} - \frac{1}{4}+\frac{3}{2}i\right) - 4}}{2} &= \frac{\left(\frac{3}{2} + \frac{i}{2}\right) \pm \sqrt{\left(\frac{8}{4}+\frac{3}{2}i\right) - 4}}{2}\\ &= \frac{\left(\frac{3}{2} + \frac{i}{2}\right) \pm \sqrt{\left(2+\frac{3}{2}i\right) - 4}}{2}\\ &= \frac{\left(\frac{3}{2} + \frac{i}{2}\right) \pm \sqrt{-2+\frac{3}{2}i}}{2}\\ &= \frac{\left(\frac{3}{2} + \frac{i}{2}\right) \pm \sqrt{-\frac{8}{4}+\frac{6}{4}i}}{2}\\ &= \frac{\left(\frac{3}{2} + \frac{i}{2}\right) \pm \frac{1}{2}\sqrt{-8+6i}}{2}\\ \end{align}$$

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  • $\begingroup$ Why can you replace $t$ with $z$ on the LHS of that quadratic formula, when $t = e^{iz}$? $\endgroup$
    – hello all
    Jul 27, 2015 at 9:33
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Two solutions of the system above: (Pi /4 , - ln(2)/2) and (- Pi/4 , ln(2)/2), among others.

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One possible "elementary method" might start:

$$\cos(x+iy) = \cos(x)\cos(iy)-\sin(x)\sin(iy),$$

which simplifies to

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

Now let

$$\cos(x)\cosh(y)-i\sin(x)\sinh(y) \equiv \frac{3}{4}+\frac{i}{4},$$

which gives $$\cos(x)\cosh(y)=3/4$$ and $$\sin(x)\sinh(y)=-1/4.$$

You may be able to solve for $x$ and $y$.

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  • $\begingroup$ I tried this approach but I got bogged down since I went off on a tangent using this. Thank you! $\endgroup$
    – 1028
    Jan 26, 2014 at 22:14
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    $\begingroup$ I highly doubt that a human would be able to solve those nonlinear equations in a reasonable amount of time: see W|A's solutions $\endgroup$
    – apnorton
    Jan 26, 2014 at 22:15

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