MHB Solving a complex value equation

[cbarker1]

Dear Everyone,

I have a question about how to solve for x near the end of the problem:
\[ 1+2\sinh^{2}(z)=0 \]

Here is the solution and work:
\[ 1+2\sinh^2(z)=0 \\ \sinh^2(z)=\frac{-1}{2}\\ \sqrt{\sinh^2(z)}=\pm \sqrt{\frac{-1}{2}}\\ \sinh(z)=\pm i\frac{1}{\sqrt{2}}\\ \]

Then we can split the positive complex number and the negative complex number.
\[ \sinh(z)= \frac{i}{\sqrt{2}}\ \text{or} \sinh(z)= \frac{-i}{\sqrt{2}} \]

Let's focus on the positive complex number. (The method will be identical to the negative complex):
\[ \sinh(z)=\frac{i}{\sqrt{2}} \]

We know that \( \sinh{z}=\sinh{x} \cos{y}+i \cosh{x} \sin{y} \). So by plugin the identity for $\sinh{z}$, we yield this equation:
\[ \cos{y} \sinh{x} +i \cosh{x} \sin{y}=\frac{i}{\sqrt{2}} \]

We can use this fact that $a+bi=c+di$ is equal if and only if $a=c$ and $b=d$. So the equation can be split off to the real part and the imaginary part on both sides of the equation; we yield:
\[ \cos {y} \sinh{x} =0\ \text{and} \ \sin{y} \cosh{x}=\frac{1}{\sqrt{2}} \]

We need to solve the real part first because we can gain some conditions on $x$ and $y$ for the imaginary part of the equation:
\[ \cos{y} \sinh{x}=0\\ \cos{y}=0\ \text{or} \sinh{x}=0 \\ y=\frac{\pi}{2}+n\pi\, \ \forall n\in \mathbb{Z}\,\ \text{or}\ x=0 \]

For the imaginary part:
For $x=0$, $\cosh(0)\sin{y}=\frac{1}{\sqrt{2}} \implies \sin{y}=\frac{1}{\sqrt{2}}$ Note that $\cosh(0)=1$.
$y=\frac{\pi}{4}+n\pi$, for all $n\in\mathbb{Z}$.

For $y=\frac{\pi}{2}+n\pi$,
Case 1: $n=2m$ , where $m\in\mathbb{Z}$
We know that $\sin({\frac{\pi}{2}+2m\pi})=1$, so we can simplify the equation:
\[ \cosh{x}=\frac{1}{\sqrt{2}} \]
We can use the definition of $\cosh{x}$ in the form of the exponential functions nd solve the quadratic equations in terms of $e^x$:
\[ \frac{e^{x}+e^{-x}}{2}=\frac{\sqrt{2}}{2}\\ e^x+e^{-x}=\sqrt{2} \\ e^x+\frac{1}{e^x}=\sqrt{2}\\ e^{2x}-\sqrt{2}e^x+1=0\\ e^x=\frac{\sqrt{2}\pm\sqrt{2-4}}{2} \]
\[ e^x=\frac{\sqrt{2}\pm i\sqrt{2}}{2} \]

We want to solve the $x$. I am confused by the next step.




For $y=\frac{\pi}{2}+n\pi$,
Case 2: $n=2m+1$ , where $m\in\mathbb{Z}$
We know that $\sin({\frac{\pi}{2}+(2m+1)\pi})=-1$, so we can simplify the equation:
\[ \cosh{x}=\frac{-1}{\sqrt{2}} \]
We can use the definition of $\cosh{x}$ in the form of the exponential functions nd solve the quadratic equations in terms of $e^x$:
\[ \frac{e^{x}+e^{-x}}{2}=\frac{-\sqrt{2}}{2}\\ e^x+e^{-x}=-\sqrt{2} \\ e^x+\frac{1}{e^x}=-\sqrt{2}\\ e^{2x}+\sqrt{2}e^x+1=0\\ e^x=\frac{-\sqrt{2}\pm\sqrt{2-4}}{2} \]
\[ e^x=\frac{-\sqrt{2}\pm i\sqrt{2}}{2} \]

We want to solve the $x$. I am confused by the next step as well.

Do we need to worry about the negative complex numbers as well?Thanks for your help,
Cbarker1

 

[I like Serena]

\[ e^x=\frac{-\sqrt{2}\pm i\sqrt{2}}{2} \]

We want to solve the $x$. I am confused by the next step as well.

Do we need to worry about the negative complex numbers as well?

The $x$ here is supposed to be a real number. Since your solution will be complex, it means there is no solution for $x$ in this case.
Note that $\cosh(x)\ge 1$ for all real $x$, so from $\cosh x=\frac 1{\sqrt 2}$, we could already see that it would not have a solution.

Either way, you're making it a bit more complicated than it needs to be.
Consider that \( \sinh(iw)=\frac 12(e^{iw}-e^{-iw}) = i\frac 1{2i}(e^{iw}-e^{-iw}) = i\sin(w) \).

So substitute $z=iw$ to find:
\[ \sinh z=\pm\frac i{\sqrt 2} \\ \sinh(iw)=\pm i\frac 1{\sqrt 2} \\ i\sin(w)=\pm i\frac 1{\sqrt 2} \\ \sin(w)=\pm\frac 1{\sqrt 2} \\ w=\frac\pi 4+k\frac\pi 2 \\ z=i\Big(\frac\pi 4+k\frac\pi 2\Big) \]

 

[cbarker1]

OK. I learned about my method in my university's complex analysis course as an undergrad. Your method is much simpler.

 

[cbarker1]

sin(w)=±12–√w=π4+kπ2z=i(π4+kπ2)

Did you used the $\arcsin(u)$ , where $u$ is some function, on both side of the equation to find $w$. How did you found out that $w$ needs to include this part $\frac{k\pi}{2}$?

 

[I like Serena]

Did you used the $\arcsin(u)$ , where $u$ is some function, on both side of the equation to find $w$. How did you found out that $w$ needs to include this part $\frac{k\pi}{2}$?

It's from the unit circle in combination with the fact that $\sin\frac\pi 4=\frac 1{\sqrt 2}$ is a 'standard' formula.
\begin{tikzpicture}[scale=4]
\draw[help lines] ({1/sqrt(2)},{1/sqrt(2)}) -- ({-1/sqrt(2)},{1/sqrt(2)}) -- ({-1/sqrt(2)},{-1/sqrt(2)}) -- ({1/sqrt(2)},{-1/sqrt(2)}) -- cycle;
\draw[help lines] ({-1/sqrt(2)},{-1/sqrt(2)}) -- ({1/sqrt(2)},{1/sqrt(2)}) ({-1/sqrt(2)},{1/sqrt(2)}) -- ({1/sqrt(2)},{-1/sqrt(2)}) -- cycle;
\path ({1/sqrt(2)},0) node[fill=black!5,below] {$\frac 1{\sqrt 2}$};
\path (0,{1/sqrt(2)}) node[fill=black!5,left] {$+\frac 1{\sqrt 2}$};
\path (0,{-1/sqrt(2)}) node[fill=black!5,left] {$-\frac 1{\sqrt 2}$};
\draw[-latex] (0,0) node[above right,xshift=.8cm] {$\frac\pi 4$} (0.2,0) arc (0:45:0.2);
\draw[-latex] (-1.2,0) -- (1.2,0) node[above] {$x$-axis};
\draw[-latex] (0,-1.2) -- (0,1.2) node[ right ] {$y$-axis};
\draw[blue, ultra thick] (0,0) circle (1);
\end{tikzpicture}

In the unit circle we can see that there are 4 angles that correspond to $\sin \theta=\pm\frac 1{\sqrt 2}$, and they have an angle of $\frac\pi 2$ between them.