# Thread: sin(z) = -i, z = x + yi

1. ## sin(z) = -i, z = x + yi

$\displaystyle sin(z)=\mathbf{i}$

$\displaystyle \displaystyle \frac{e^{\mathbf{i}z}-e^{-\mathbf{i}z}}{2\mathbf{i}}=\mathbf{i}\Rightarrow e^{2\mathbf{i}z}+2e^{\mathbf{i}z}-1=0$

$\displaystyle \displaystyle e^{\mathbf{i}z}=\frac{2\pm\sqrt{8}}{2}=-1\pm\sqrt{2}$

$\displaystyle \displaystyle \mathbf{i}z=ln(-1+\sqrt{2})\Rightarrow z=-\mathbf{i}(ln(-1+\sqrt{2})+\mathbf{i}arg(z))\Rightarrow -\mathbf{i}(ln(-1+\sqrt{2})+\mathbf{i}(0+2\pi k), \ k\in\mathbb{Z}$

$\displaystyle \displaystyle z=2\pi k -\mathbf{i}ln(-1+\sqrt{2})$

$\displaystyle \displaystyle z=(2k+1)\pi-\mathbf{i}ln(1+\sqrt{2})$

Why the discrepancy?

2. Originally Posted by dwsmith
$\displaystyle sin(z)=\mathbf{i}$

$\displaystyle \displaystyle \frac{e^{\mathbf{i}z}-e^{-\mathbf{i}z}}{2\mathbf{i}}=\mathbf{i}\Rightarrow e^{2\mathbf{i}z}+2e^{\mathbf{i}z}-1=0$

$\displaystyle \displaystyle e^{\mathbf{i}z}=\frac{2\pm\sqrt{8}}{2}=1\pm\sqrt{2 }$

$\displaystyle \displaystyle \mathbf{i}z=ln(1+\sqrt{2})\Rightarrow z=-\mathbf{i}(ln(1+\sqrt{2})+\mathbf{i}arg(z))\Righta rrow -\mathbf{i}(ln(1+\sqrt{2})+\mathbf{i}(0+2\pi k), \ k\in\mathbb{Z}$

$\displaystyle \displaystyle z=2\pi k -\mathbf{i}ln(1+\sqrt{2})$

$\displaystyle \displaystyle z=(2k+1)\pi-\mathbf{i}ln(1+\sqrt{2})$

Why the discrepancy?
Dear dwsmith,

In the title of the thread you have written, $\displaystyle \sin(z)=-i$ but the equation used in the calculation is $\displaystyle \sin(z)=i$. Which is the correct one?

3. It is supposed to be i.

4. Originally Posted by dwsmith
It is supposed to be i.
Ok. Note that $\displaystyle \displaystyle e^{\mathbf{i}z}=\frac{2\pm\sqrt{8}}{2}=1\pm\sqrt{2 }$ should be, $\displaystyle \displaystyle e^{\mathbf{i}z}=\frac{-2\pm\sqrt{8}}{2}=-1\pm\sqrt{2}$

5. I am glad you noticed that but it doesn't solve my initial dilemma.

6. Second solution solves the problem except the inside term of the log is negative not positive.

$\displaystyle \displaystyle z=-\mathbf{i}(ln(-1-\sqrt{2}))$

$\displaystyle \displaystyle z=-\mathbf{i}(ln(-1-\sqrt{2})+\mathbf{i}(\pi+2\pi k))=-\mathbf{i}ln(-1-\sqrt{2})+\pi+2\pi k$$\displaystyle =-\mathbf{i}ln(-1-\sqrt{2})+(1+2k)\pi Should I also do this: \displaystyle \displaystyle ln((-1-\sqrt{2})^{-\mathbf{i}}) and solve the expression \displaystyle z^{\alpha}\mbox{?} 7. Originally Posted by dwsmith Second solution solves the problem except the inside term of the log is negative not positive. \displaystyle \displaystyle z=-\mathbf{i}(ln(-1-\sqrt{2})) \displaystyle \displaystyle z=-\mathbf{i}(ln(-1-\sqrt{2})+\mathbf{i}(\pi+2\pi k))=-\mathbf{i}ln(-1-\sqrt{2})+\pi+2\pi k$$\displaystyle =-\mathbf{i}ln(-1-\sqrt{2})+(1+2k)\pi$

Should I also do this:

$\displaystyle \displaystyle ln((-1-\sqrt{2})^{-\mathbf{i}})$

and solve the expression $\displaystyle z^{\alpha}\mbox{?}$
Your have made errors in your calculations. The second solution gives the answer stated in the book. Follow what I have done below and try to find out where you have done wrong.

$\displaystyle e^{\mathbf{i}z}=-1-\sqrt{2}$

$\displaystyle \displaystyle \mathbf{i}z=\ln(-1-\sqrt{2})+2\pi\mathbf{i}k~where~k\in Z$

$\displaystyle \displaystyle z=-\mathbf{i}\left[\ln(1+\sqrt{2})+\ln(-1)\right]+2\pi k$

$\displaystyle \displaystyle z=-\mathbf{i}\left[\ln(1+\sqrt{2})+\mathbf{i}\pi\right]+2\pi k$

$\displaystyle \displaystyle z=(2k+1)\pi-\mathbf{i}ln(1+\sqrt{2})~where~k\in Z$

The first answer gives, $\displaystyle \displaystyle z=(2k+1)\pi-\mathbf{i}ln(1-\sqrt{2})~where~k\in Z$

Hope this helps you to understand.

8. I forgot to take the modulus of the second solution.

9. You need to learn the basic equivalences.
$\displaystyle \sin(z)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)=-i$.
From that we get $\displaystyle \sin(x)\cosh(y)=0~\&~\cos(x)\sinh(y)=-1$.
Two equations in two unknowns.
Because $\displaystyle \cosh(y)\not=0$ we get $\displaystyle x=k\pi$.
Use that to find values of $\displaystyle y$.

10. Originally Posted by Plato
You need to learn the basic equivalences.
$\displaystyle \sin(z)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)=-i$.
From that we get $\displaystyle \sin(x)\cosh(y)=0~\&~\cos(x)\sinh(y)=-1$.
Two equations in two unknowns.
Because $\displaystyle \cosh(y)\not=0$ we get $\displaystyle x=k\pi$.
Use that to find values of $\displaystyle y$.
I need to learn a lot of things, but I do know that equation.

Additionally, anything I learn wont happen in one try either. It will take a little time, like most things in life.

11. Plato you are l little abrasive with your help sometimes.

"You need to learn basic equivalences," when read by the reader, sounds rude.

If I may suggest, saying if you learn or practice the basic identities more, you may see that some methods would be easier to tackle by using this....

12. Originally Posted by dwsmith
Plato you are l little abrasive with your help sometimes.
If you think that that remark is at all abrasive, then you need give up any idea of graduate study. Because you will be eaten alive.

13. I truly doubt that. Almost every professor at my university is very kind in how they treat the students regardless of them being undergraduate or graduate. That is also why visiting schools is good so prospective students can identify which institution meshes with their interests, goals, and personality.

14. Originally Posted by dwsmith
I truly doubt that. Almost every professor at my university is very kind in how they treat the students regardless of them being undergraduate or graduate. That is also why visiting schools is good so prospective students can identify which institution meshes with their interests, goals, and personality.
In a really good graduate school it is not the professors that matter but your fellow students. In a good school it is all about competition, professors have little to do with that.

15. Even with that being the case, students are still capable of being respectful and kind with help and suggestions.

The only school's students that act in such a manner is the business school at my University. I am both a Finance and Math major. The students in the math and science programs don't act as you describe, only the self centered business students do.

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