Finding all the roots from a complex equation

Question

I'm struggling a lot with complex numbers recently. How do I find all the roots for equations like:
(1) $\cos z = 3$
(2) $e^{2z} = -e$
(3) $e^z+6e^{-z} = 5$

Thanks

Answer 1

1°) If the input of your question is real, then you need to ask yourself what's the minimum and maximum values that cos can reach ?
If the input of your question is complex, $$\cos{z}=\frac{\exp(iz)+\exp(-iz)}{2}=3$$ then split real and imaginary parts ($z=x+iy$)

imaginary part equation : $$\exp(-y)\sin x=\sin x\exp(y)$$
so $x\equiv 0\bmod(\pi)$ or $y=0$
If $y=0$, go to the case : input is real.

Let's consider $x\equiv 0\bmod(\pi)$ real part equation : $$\cos(x)(\exp(-y)+\exp(y))=6$$ Let $Y=\exp(y)$ so $$\cos(x)(1+Y^2)-6Y=0$$ so $Y_1=\frac{6-\sqrt{23}}{\pm 2}$ and $Y_2=\frac{6+\sqrt{23}}{\pm 2}$
hence $y_1=\ln(\frac{6-\sqrt{23}}{2})$ and $y_2=\ln(\frac{6+\sqrt{23}}{2})$

2°) $$\exp(2z) = -e\\ \exp(2z-1)=-1$$ if you write $z=x+iy$, you get : $$\exp(2x-1+2iy)=-1$$ Then apply the absolute and arg function to get :
$\exp(2x-1)=1$ and $2y\equiv \pi \bmod(2\pi)$

3°)$$e^z+6e^{-z} = 5\\ e^{2z}-5e^z+6=0$$ Let $X=e^z$ so : $$X^2-5X+6=0$$ so $X_1=2$ and $X_2=3$
so $z_1=\ln2$ and $X_2=\ln3$

Answer 2

For 1) set $w=e^{iz}$, then $w+w^{-1}=6$ or $w^2-6w+1=0$ which has the solutions $$ e^{iz}=w=3\pm\sqrt{8}. $$ Now you need to apply the complex logarithm.

2) just is the application of the complex logarithm.

3) set $w=e^z$ to find the quadratic equation $0=w^2-5w+6=(w-2)(w-3)$. Again apply the complex logarithm.

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The main difference between real and complex logarithm that is to consider here is that $e^{i2\pi}=1$, so that there are always infinitely many solutions.