How to solve $2e^3\cos (6z)-e^6=1$, being $z\in\mathbb{C}$.

Question

I need to find a solution to this equation, being $z\in\mathbb{C}$, $$2e^3\cos (6z)-e^6=1.$$ I don't know what method can I follow to solve it.

Answer 1

By Euler's Formula, we can say $$\cos x = \frac{e^{ix}+e^{-ix}}{2}$$

So the equation becomes:

$$e^3(e^{6iz}+e^{-6iz}) = 1 + e^6 $$

From here divide both sides by $e^3$ and compare powers, or if you dont fancy that, put $e^{6iz}$ as $x$

$$e^3t^2 -(1+e^6)t +e^3 = 0$$

Solve by quadratic formula:

$$t=\frac{(1+e^6) \pm \sqrt{e^{12} +2e^6 +1 -4e^6}}{2e^3}$$

$$t=e^{\pm 3}$$

$$e^{6iz} = e^{\pm 3}$$

$$z = \frac{2n\pi \pm 3i}{6}$$

Here $n \in N$

EDIT: The OP asked for these details through the comments.

$$e^{ix} = \cos x + i\sin x $$

So for the equation:

$$z^n = w$$

Here $z=e^{i\theta}$ and $w = e^{i\alpha}$

$$(\cos \theta + i \sin \theta)^n = (\cos \alpha + i \sin \alpha)$$

By De Moivre's Theorem :

$$\cos n\theta + i\sin n\theta = cos \alpha +i \sin \alpha $$

So by the periodicity of sine and cosine functions,

$$n\theta = \alpha +2k\pi $$

For some integer $k$

$$\therefore \theta = \frac{\alpha +2k\pi}{n}$$

So in the problem here,

$$e^{6iz} = e^{\pm 3}$$

$$(e^{iz})^6 = e^{\pm i\left( \frac{3}{i} \right)}$$

Here $\frac{3}{i}$ can be written as $(-3i)$ by multiplying num. and den. by $i$

So this is like the form of $[z^n = w]$ I mentioned above.

From there its easy to see where the answer came from.

NOTE: Refer here for furthur understanding

Answer 2

HINT: for every $w\in\Bbb C$ $$ \cos(w)=\frac{e^{iw}+e^{-iw}}{2} $$

Answer 3

Note

$$\cos (6z)=\frac{ 1+e^6}{2e^3}=\cosh(3)=\cos(i 3)$$

which yields

$$6z =\pm i 3+2\pi n\implies z= \frac13\pi n \pm \frac i2$$

Answer 4

Just replace $\cos 6z$ by $(e^{6iz}+e^{-6iz})/2$ Then if you call $t=e^{6iz}$ you get the equation $$ t+\frac 1{t}=e^3+e^{-3} $$ With obvious solutions $t=e^3$ and $t=e^{-3}$. That is $$ e^{6iz}=e^{\pm 3} $$ Which implies $$ z=\frac {k\pi}{3}\pm \frac i{2} $$

Answer 5

Hint: Express $\cos(6z)$ as:

$$\cos(6z) = \frac{e^{6iz} + e^{-6iz}}{2}$$