- #1

cbarker1

Gold Member

MHB

- 349

- 23

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 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

Last edited: