Solving $\sin(z) = 2 + 3i$ using complex numbers

[Dustinsfl]

What is the best way to handle $\sin(z) = 2 + 3i$?

Option (1)
$\sin(z) = \sin(x + yi) = \sin(x)\cos(yi) + \sin(yi)\cos(x) = \sin(x)\cosh(y) + i\sinh(y)\cos(x)$
$\sin(x)\cosh(y) = 2$
$\sinh(y)\cos(x) = 3$

Option (2)
$\displaystyle\sin(z) = \frac{e^{zi}-e^{-zi}}{2i} = 2 + 3i$

Which option is better or is there another option?

Also, I am stuck on (1) and (2) anyways.

 

[Fernando Revilla]

Option (2) $\displaystyle\sin(z) = \frac{e^{zi}-e^{-zi}}{2i} = 2 + 3i$

This is the best option. Write $w=e^{iz}$ and you'll obtain $w^2+(6-4i)w-1=0$ . If $w_1,w_2$ are the solutions of the quadratic equation then, $e^{iz}=w_1,e^{iz}=w_2$ etc.

 

[Dustinsfl]

$$
\displaystyle
w = \frac{4i + 6 \pm\sqrt{(6 - 4i)^2 + 4}}{2} = 2i + 3 \pm\sqrt{6 - 12i}
$$

What can be done about the radical now?

 

[Prove It]

$$
\displaystyle
w = \frac{4i + 6 \pm\sqrt{(6 - 4i)^2 + 4}}{2} = 2i + 3 \pm\sqrt{6 - 12i}
$$

What can be done about the radical now?

Convert it to polars, then take it to the power of 1/2.

 

[Dustinsfl]

Convert it to polars, then take it to the power of 1/2.

So $r = 6\sqrt{5}$. What is $\theta = \tan^{-1}\left(-2\right) =$ ??

 

[Prove It]

So $r = 6\sqrt{5}$. What is $\theta = \tan^{-1}\left(-2\right) =$ ??

It's in the fourth quadrant, so it would be $ - \arctan{(2)}$.

 

[Dustinsfl]

It's in the fourth quadrant, so it would be $ - \arctan{(2)}$.

This is horrible with all the decimals.

$$
\displaystyle
w = 2i + 3 \pm\sqrt{6}\sqrt[4]{5}\exp\left(-\frac{i\tan^{-1}(2)}{2}\right)
$$

Should I convert $2i + 3$ to polar?

 

[Prove It]

This is horrible with all the decimals.

$$
\displaystyle
w = 2i + 3 \pm\sqrt{6}\sqrt[4]{5}\exp\left(-\frac{i\tan^{-1}(2)}{2}\right)
$$

Should I convert $2i + 3$ to polar?

No, now you should convert everything to Cartesians. And keep everything exact.

 

[Dustinsfl]

No, now you should convert everything to Cartesians. And keep everything exact.

Is the half angle formula going to need to be used here?

 

[Prove It]

Is the half angle formula going to need to be used here?

No. You have $\displaystyle \sqrt{6}\sqrt[4]{5}e^{-\frac{i\arctan{2}}{2}} $, which is a complex number with $ \displaystyle r = \sqrt{6}\sqrt[4]{5}$ and $ \displaystyle \theta = -\frac{\arctan{2}}{2} $. So what would this number be in Cartesians?

 

[Dustinsfl]

No. You have $\displaystyle \sqrt{6}\sqrt[4]{5}e^{-\frac{i\arctan{2}}{2}} $, which is a complex number with $ \displaystyle r = \sqrt{6}\sqrt[4]{5}$ and $ \displaystyle \theta = -\frac{\arctan{2}}{2} $. So what would this number be in Cartesians?

So you are saying to leave it in this form:

$\displaystyle\sqrt{6}\sqrt[4]{5}\cos\left(\frac{\arctan(2)}{2}\right) - i\sqrt{6}\sqrt[4]{5}\sin\left(\frac{\arctan(2)}{2}\right)$

 

[Prove It]

So you are saying to leave it in this form:

$\displaystyle\sqrt{6}\sqrt[4]{5}\cos\left(\frac{\arctan(2)}{2}\right) - i\sqrt{6}\sqrt[4]{5}\sin\left(\frac{\arctan(2)}{2}\right)$

Yes. Then perform the addition and subtraction.

On second thought, you probably could simplify these using half angle identities and the Pythagorean Identity, it depends on what's expected of you...

\[ \displaystyle \begin{align*} 1 + \tan^2{\theta} &\equiv \frac{1}{\cos^2{\theta}} \\ \cos^2{\theta} &\equiv \frac{1}{1 + \tan^2{\theta}} \\ \cos{\theta} &\equiv \pm \frac{1}{\sqrt{1 + \tan^2{\theta}}} \end{align*} \]

and

\[ \displaystyle \begin{align*} 1 + \frac{1}{\tan^2{\theta}} &\equiv \frac{1}{\sin^2{\theta}} \\ \frac{1 + \tan^2{\theta}}{\tan^2{\theta}} &\equiv \frac{1}{\sin^2{\theta}} \\ \sin^2{\theta} &\equiv \frac{\tan^2{\theta}}{1 + \tan^2{\theta}} \\ \sin{\theta} &\equiv \pm \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}} \end{align*} \]

 

[Dustinsfl]

Yes. Then perform the addition and subtraction.

On second thought, you probably could simplify these using half angle identities and the Pythagorean Identity, it depends on what's expected of you...

\[ \displaystyle \begin{align*} 1 + \tan^2{\theta} &\equiv \frac{1}{\cos^2{\theta}} \\ \cos^2{\theta} &\equiv \frac{1}{1 + \tan^2{\theta}} \\ \cos{\theta} &\equiv \pm \frac{1}{\sqrt{1 + \tan^2{\theta}}} \end{align*} \]

and

\[ \displaystyle \begin{align*} 1 + \frac{1}{\tan^2{\theta}} &\equiv \frac{1}{\sin^2{\theta}} \\ \frac{1 + \tan^2{\theta}}{\tan^2{\theta}} &\equiv \frac{1}{\sin^2{\theta}} \\ \sin^2{\theta} &\equiv \frac{\tan^2{\theta}}{1 + \tan^2{\theta}} \\ \sin{\theta} &\equiv \pm \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}} \end{align*} \]

I know have this mess:

$$
\begin{array}{lll}
\displaystyle e^{iz} & = & \displaystyle 3 + \sqrt{6}\sqrt[4]{5}\cos\left(\frac{\tan^{-1}(2)}{2}\right) + i\left(2 - \sqrt{6}\sqrt[4]{5}\sin\left(\frac{\tan^{-1}(2)}{2}\right)\right)\\
\displaystyle e^{iz} & = & \displaystyle 3 - \sqrt{6}\sqrt[4]{5}\cos\left(\frac{\tan^{-1}(2)}{2}\right) + i\left(2 + \sqrt{6}\sqrt[4]{5}\sin\left(\frac{\tan^{-1}(2)}{2}\right)\right)
\end{array}
$$

How do I solve for z?

 

[Prove It]

I know have this mess:

$$
\begin{array}{lll}
\displaystyle e^{iz} & = & \displaystyle 3 + \sqrt{6}\sqrt[4]{5}\cos\left(\frac{\tan^{-1}(2)}{2}\right) + i\left(2 - \sqrt{6}\sqrt[4]{5}\sin\left(\frac{\tan^{-1}(2)}{2}\right)\right)\\
\displaystyle e^{iz} & = & \displaystyle 3 - \sqrt{6}\sqrt[4]{5}\cos\left(\frac{\tan^{-1}(2)}{2}\right) + i\left(2 + \sqrt{6}\sqrt[4]{5}\sin\left(\frac{\tan^{-1}(2)}{2}\right)\right)
\end{array}
$$

How do I solve for z?

You could use a Logarithm. Otherwise, convert this complex number to its exponential form (actually that will probably be easier).

 

[Fernando Revilla]

Convert it to polars, then take it to the power of 1/2.

Not necessary for square roots, $\sqrt{6-12i}=x+iy\Leftrightarrow (x+iy)^2=6-12i\Leftrightarrow (x^2-y^2=6)\;\wedge\;(2xy=-12)\Leftrightarrow\ldots$

 

[Dustinsfl]

Not necessary for square roots, $\sqrt{6-12i}=x+iy\Leftrightarrow (x+iy)^2=6-12i\Leftrightarrow (x^2-y^2=6)\;\wedge\;(2xy=-12)\Leftrightarrow\ldots$

So I ended up getting

$x = \pm\sqrt{3\pm 3\sqrt{5}}$ and $y = \pm\sqrt{-3\pm 3\sqrt{5}}$

Something is wrong I believe.

I tested the values and for $x^2-y^2 = 6$

Only $x=\sqrt{3+3\sqrt{5}}$ and $y=\sqrt{-3+3\sqrt{5}}$ worked but when multiplied together they weren't negative 6.

 

[Prove It]

So I ended up getting

$x = \pm\sqrt{3\pm 3\sqrt{5}}$ and $y = \pm\sqrt{-3\pm 3\sqrt{5}}$

Something is wrong I believe.

I tested the values and for $x^2-y^2 = 6$

Only $x=\sqrt{3+3\sqrt{5}}$ and $y=\sqrt{-3+3\sqrt{5}}$ worked but when multiplied together they weren't negative 6.

Actually you should have found that there are four possible solutions, since $ \displaystyle x = \pm \sqrt{3 + 3\sqrt{5}} $ and $\displaystyle y = \pm \sqrt{-3 + 3\sqrt{5}} $ are all possible...

Anyway, if we multiply their absolute values...

\[ \displaystyle \begin{align*} \left(\sqrt{3 + 3\sqrt{5}}\right)\left(\sqrt{-3 + 3\sqrt{5}}\right) &= \sqrt{\left(3 + 3\sqrt{5}\right)\left(-3 + 3\sqrt{5}\right)} \\ &= \sqrt{-9 + 9\sqrt{5} - 9\sqrt{5} + 45} \\ &= \sqrt{36} \\ &= 6 \end{align*} \]

So the possible solutions are those which have differing signs (in order to get a negative when multiplied).

Therefore, the solutions are $\displaystyle (x, y) = \left(\sqrt{3 + 3\sqrt{5}}, -\sqrt{-3 + 3\sqrt{5}}\right)$ or $ \displaystyle (x, y) = \left(-\sqrt{3 + 3\sqrt{5}}, \sqrt{-3 + 3\sqrt{5}}\right)$.

 

[Dustinsfl]

$e^{iz} = w = 3 + \sqrt{3 + 3\sqrt{5}} + i\left(2 - \sqrt{3 + 3\sqrt{5}}\right)$, $3 - \sqrt{3 + 3\sqrt{5}} + i\left(2 + \sqrt{3 + 3\sqrt{5}}\right)$, $3 - \sqrt{3 + 3\sqrt{5}} + i\left(2 + \sqrt{-3 + 3\sqrt{5}}\right)$, and $3 + \sqrt{3 + 3\sqrt{5}} + i\left(2 - \sqrt{-3 + 3\sqrt{5}}\right)$

$z = \tan^{-1}\left(\dfrac{2-\sqrt{3+3\sqrt{5}}}{3+\sqrt{3+3\sqrt{5}}}\right) - i\ln\left(\sqrt{19+6\sqrt{5}+2\sqrt{3(1+\sqrt{5}}}\right)$

$z = \tan^{-1}\left(\dfrac{2+\sqrt{3+3\sqrt{5}}}{3-\sqrt{3+3\sqrt{5}}}\right) - i\ln\left(\sqrt{19+6\sqrt{5}-2\sqrt{3(1+\sqrt{5}}}\right)$

$z = \tan^{-1}\left(\dfrac{2+\sqrt{-3+3\sqrt{5}}}{3-\sqrt{3+3\sqrt{5}}}\right) - i\ln\left(\sqrt{\left(2+\sqrt{-3+3\sqrt{5}}\right)^2+\left(-3+\sqrt{-3+3\sqrt{5}}\right)^2}\right)$

$z = \tan^{-1}\left(\dfrac{2-\sqrt{-3+3\sqrt{5}}}{3+\sqrt{3+3\sqrt{5}}}\right) - i\ln\left(\sqrt{\left(-2+\sqrt{-3+3\sqrt{5}}\right)^2+\left(3+\sqrt{-3+3\sqrt{5}}\right)^2}\right)$

Is this really correct?