Equations in complex plane

[Camillo]

Solve following equations

\( \sin z = 2 \)
\( \sinh z = i \)
\( \cosh z= -1 \)
\( e^z =-1 \) .

[j18eos]

It's a greatiful resolution!

[Raptorista1]

I'll try the third one:
\[ \cosh z = -1\]
\[
\cosh z = \frac{e^z + e^{-z}}{2} = -1 \Rightarrow e^z + e^{-z} + 2 = 0 \Rightarrow e^{2z} + 2e^{z} + 1 = (e^z + 1)^2 = 0
\]
That means solutions are given by
\[
e^z = -1
\]
which has been solved just above.

[Camillo]

[j18eos]

I prefer the fourth equation!

By means of Euler's formulas \(-1=e^z=e^{x+iy}=e^x(\cos y+i\sin y)\Rightarrow\begin{cases}e^x\cos y=-1\\e^x\sin y=0\end{cases}\) therefore \(\sin y=0\Rightarrow y=k\pi\) when \(k\in\mathbb{Z}\), but \(e^x\cos y=-1\) then \(x=0\) and \(y=(2k+1)\pi\) when \(k\in\mathbb{Z}\).

The solutions are \(z=0+i(2k+1)\pi\) when \(k\in\mathbb{Z}\).

[Camillo]

My solution for ex.1
Taking advantage of formula:
$sin z= sin x*\cosh y +i cos x*\sinh y $ we have to solve the system:

$sinx\cosh y=2$
$ cos x\sinh y=0 rarr y=0 $ or $ x=pi/2+kpi$
If $ y=0 $ the first equation $ rarr sinx=2 $ impossible since $x inRR$
if $x= pi/2+kpi $ then first equation $ rarr sin(pi/2+kpi)\cosh y=2 rarr (-1)^k \cosh y=2 $ .
This implies $k $ even and $\cosh y=2 rarr e^y=2+-sqrt(3) rarr y=ln(2+-sqrt(3))$.
Consequently :
$z_1= pi/2 +2npi+iln(2+sqrt(3)) $
$z_2= pi/2+2npi+iln(2-sqrt(3)) $.
$n in ZZ$.

[Camillo]

@ gugo : Yes I was intending hyperbolic sin and cos , have just corrected.

Now I wait for solutions of 3 others exercise .

[gugo82]

An alternative approach to the first equation.

Remembering the fundamental relation:
\[\sin z= \frac{1}{2\imath}\ (e^{\imath z}-e^{-\imath z})\; ,\]
one can write \(\sin z=2\) as \(e^{\imath z} -e^{-\imath z}=4\imath\) and finally, using some algebra, as:
\[ \tag{1} e^{2\imath z} -4\imath e^{\imath z}-1=0\; .\]
Equation (1) can be solved using standard tools: e.g., one can write (1) in the form:
\[(e^{\imath z}-2\imath)^2+3=0\]
therefore its solutions come from the following two exponential equations:
\[e^{\imath z}-(2+\sqrt{3})\imath=0\quad \text{and} \quad e^{\imath z}-(2-\sqrt{3})\imath =0\; ,\]
which respectively give:
\[\begin{split} z&=\frac{1}{\imath}\ \text{Ln} ((2+\sqrt{3})\imath) =\frac{\pi}{2}+2k\pi+\imath\ \ln (2+\sqrt{3}) \quad \text{and} \\
z&= \frac{1}{\imath}\ \text{Ln} ((2-\sqrt{3})\imath) =\frac{\pi}{2}+2h\pi+\imath\ \ln (2-\sqrt{3}) \quad \text{with } k,h\in \mathbb{Z} \; .\end{split}\]

@Camillo: What is the meaning of \(Sh z\) and \(Ch z\) in #2 and #3?
Maybe hyperbolic sine and cosine? If it is the case, may I remind you that LaTeX has the \sinh and \cosh command?

[fireball1]

Let's start with the first equation, which is, incidentally, a good example to show that, unlike the sine of a real number, the sine of a complex number can be larger than 1 in its norm.

On setting \(x:=\text{Re }z\in\mathbb R\) and \(y:=\text{Im }z\in\mathbb R\), it is straightforward to get the following system of equations: \begin{cases}\cos x \left(e^{-y}-e^{y}\right)=0 \\ \sin x = \frac{4}{e^y+e^{-y}}, \end{cases} the second equation having solutions for all \(y\) such that \(\frac{4}{e^y+e^{-y}} \le 1\). The first equation is satisfied if either \(\cos x = 0\) or \(y=0\). If \(y=0\), then the second equation has no solution. If \(\cos x = 0\), i.e. \(x=\frac{\pi}{2}+n\pi,\) \(n\in \mathbb Z\), the second equation becomes, on setting \(\alpha:=e^y\), \[ (-1)^n \alpha^2 - 4\alpha + (-1)^n=0,\]
whose solutions have the form \(\alpha_{\pm} := (-1)^n (2\pm\sqrt 3)\). Hence, we have to choose \(n\) even, for satisfying the requirement that \(\alpha_{\pm}\) be nonnegative, and so we have \(e^{y_{\pm}} = 2\pm\sqrt 3\Rightarrow y_{\pm} = \ln(2\pm\sqrt 3)\). All in all, on setting \(n:=2k,\,k\in\mathbb Z\), we find that the \(z\in\mathbb C\) which satisfy \(\sin z = 2\) have the forms \(\begin{equation}\begin{split}z_+ := \frac{\pi}{2} + 2k\pi + i\,\ln(2+\sqrt 3),\\z_- := \frac{\pi}{2} + 2k\pi + i\,\ln(2-\sqrt 3),\end{split}\end{equation}\), with \(k\in\mathbb Z.\) Hope there are no mistakes...

A faster method could have been to note that, when \(\cos x=0\), we must have \(\sin x = \pm 1\) (in this case, \(\sin x = 1\) since the right-hand side of the second equation is nonnegative). This holds if and only if \(e^y+e^{-y}=4\), which can be rapidly solved with respect to \(y\). Condition \(\sin x = 1\) also implies that we must replace \(n\) by an even number \(2k\).


(How can I centre a system of equations? \begin{equation} doesn't work...)