Problems while solving equation with $e^z$

Question

I'm trying to solve this equation:

$e^z-2ie^{-z}=i-2$

Here's what I've done: $z=x+iy$

$e^x(\cos y + i \sin y) - 2ie^{-x} (\cos y - i \sin y) = i-2$

(1) $ \ \ e^x \cos y - 2e^{-x} \sin y = -2 \ $ and $ \ \ $ (2) $ \ \ e^x \sin y - 2e^ {-x} \cos y = 1$

Now I substitute $e^x = w \neq 0$

(1) $ \ \ w^2 \cos y + 2w - 2 \sin y = 0 \ $ and $ \ \ $ (2) $ \ \ w^2 - w - 2 \cos y = 0$

$\Delta_1 = 4+8 \sin y \cos y = 4 + 4 \sin 2y, \ \ \Delta_2 = 1+4 \sin 2y$

From the first equation we have $w_1 = -1- \sqrt{1+\sin 2y}, \ \ w_2 = -1+ \sqrt{1+ \sin 2y}$.

But after I plug one of the $w_1, w_2$ I get a nasty equation which I don't know how to solve.

Could you help?

Answer 1

$$(e^z)^2-(i-2)e^z-2i=0$$

$$e^z=\frac{i-2\pm\sqrt{(i-2)^2+8i}}2=\frac{i-2\pm(i+2)}2$$

as $(a-b)^2+4ab=(a+b)^2$

$+\displaystyle\implies e^z=i=e^{\left(2n\pi i+i\dfrac\pi2\right)}$

$\displaystyle-\implies e^z=-2=e^{(\ln2+2n\pi i+i\pi)}$ where $n$ is any integer

Answer 2

Hint

If you multiply the both sides of the equation by $e^z$ you get

$$e^{2z}-2i=(i-2)e^z$$ or, equivalently,

$$e^{2z}-(i-2)e^z-2i=0.$$

Now, with the substitution, $w=e^z$ you have

$$w^2-(i-2)w-2i=0,$$ which should be easy to solve.

Answer 3

If we temporarily denote $u := e^z$, the equation becomes $$u - \frac{2i}{u} = i - 2,$$ and multiplying both sides by $u$ gives $$u^2 - 2i = (i - 2)u,$$ which is quadratic in $u$. (Note that $u \neq 0$ so multiplying through by $u$ does not introduce any spurious solutions.) You can use the quadratic formula to solve for $u$, and then take logarithms to solve for $z$.