# [SOLVED]all z such that sin = 2 + 3i

#### dwsmith

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

#### Prove It

##### Well-known member
MHB Math Helper
$$\displaystyle w = \frac{4i + 6 \pm\sqrt{(6 - 4i)^2 + 4}}{2} = 2i + 3 \pm\sqrt{6 - 12i}$$

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

#### dwsmith

##### Well-known member
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
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
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?

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#### Prove It

##### Well-known member
MHB Math Helper
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
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
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
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
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*}

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#### dwsmith

##### Well-known member
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
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
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
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.

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#### Prove It

##### Well-known member
MHB Math Helper
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
$e^{iz} = w = 3 + \sqrt{3 + 3\sqrt{5}} + i\left(2 - \sqrt{3 + 3\sqrt{5}}\right)$, $3 - \sqrt{3 + 3\sqrt{5}} + i\left(2 + \sqrt{3 + 3\sqrt{5}}\right)$, $3 - \sqrt{3 + 3\sqrt{5}} + i\left(2 + \sqrt{-3 + 3\sqrt{5}}\right)$, 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?