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[SOLVED] all z such that sin = 2 + 3i

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\displaystyle
w = \frac{4i + 6 \pm\sqrt{(6 - 4i)^2 + 4}}{2} = 2i + 3 \pm\sqrt{6 - 12i}
$$

What can be done about the radical now?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
$$
\displaystyle
w = \frac{4i + 6 \pm\sqrt{(6 - 4i)^2 + 4}}{2} = 2i + 3 \pm\sqrt{6 - 12i}
$$

What can be done about the radical now?
Convert it to polars, then take it to the power of 1/2.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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)}$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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?
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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

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

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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

Well-known member
MHB Math Helper
Jan 29, 2012
661
Convert it to polars, then take it to the power of 1/2.
Not necessary for square roots, $\sqrt{6-12i}=x+iy\Leftrightarrow (x+iy)^2=6-12i\Leftrightarrow (x^2-y^2=6)\;\wedge\;(2xy=-12)\Leftrightarrow\ldots$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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)$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$e^{iz} = w = 3 + \sqrt{3 + 3\sqrt{5}} + i\left(2 - \sqrt{3 + 3\sqrt{5}}\right)$, $3 - \sqrt{3 + 3\sqrt{5}} + i\left(2 + \sqrt{3 + 3\sqrt{5}}\right)$, $3 - \sqrt{3 + 3\sqrt{5}} + i\left(2 + \sqrt{-3 + 3\sqrt{5}}\right)$, 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?