How can I solve for the complex number z in sin(z) = i?

[Lavabug]

Homework Statement

Calculate sin(z) = i

Homework Equations

Sin, cos, sinh, cosh exponential formulas?

The Attempt at a Solution

I tried expanding the sin out with exponentials, I think I could take e^iz and substitute that for z, so I'd have:

(z+ z^-1)/2i = i

Maybe I could multiply everything by z and find the roots of that polynomial aaand...?

 

[tiny-tim]

Hi Lavabug!

sin(z) = Im(…) ?

 

[Hurkyl]

Homework Statement

Calculate sin(z) = i

Homework Equations

Sin, cos, sinh, cosh exponential formulas?

The Attempt at a Solution

I tried expanding the sin out with exponentials, I think I could take e^iz and substitute that for z, so I'd have:

(z+ z^-1)/2i = i

Maybe I could multiply everything by z and find the roots of that polynomial aaand...?

What's with the ellipsis? Why'd you stop here?

 

[vela]

I tried expanding the sin out with exponentials, I think I could take e^iz and substitute that for z, so I'd have:

(z+ z^-1)/2i = i

Maybe I could multiply everything by z and find the roots of that polynomial aaand...?

It would be better to use a different variable, say w=e^iz, so that you have

(w+w^-1)/2i = i

Then do as you thought and solve for w. Once you have w, you can solve for z.

 

[Lavabug]

Hello there.

sin(z) is the imaginary part of... e^(iz) ? Here's a swing in the dark: Should I take ln on both sides, giving me:
iz = ln (i)

then writing i inside the log as e^(i*pi/2)?

so z = pi/2 ? No imaginary part?

 

[Quinzio]

I looked on Wiki and found this

 

[hunt_mat]

So you equate:
$$\sin x\cosh y+i\cos x\sinh y=i$$
This tells you the form of x and the form of y...

 

[HallsofIvy]

Homework Statement

Calculate sin(z) = i

Homework Equations

Sin, cos, sinh, cosh exponential formulas?

The Attempt at a Solution

I tried expanding the sin out with exponentials, I think I could take e^iz and substitute that for z, so I'd have:

(z+ z^-1)/2i = i

Maybe I could multiply everything by z and find the roots of that polynomial aaand...?

Looks good to me- except that sin(z)= (z- z^(-1))/2i.

I see nothing wrong with (z-z^{-1})/2i= i then z- z^{-1}= -2.
(z-z^{-1})/2i= i then z- z^{-1}= -2.

Multiplying through by z, z^2- 1= -2z so z^2+ 2z-1= 0. That's kind of easy to solve isn't it?

 

[vela]

(z+ z^-1)/2i = i

Looks like we all missed the fact you should have a minus sign between the two terms for sine.

 

[HallsofIvy]

I went back and edited so I can pretend I didn't miss that!

 

[Lavabug]

Thanks for the replies everyone, totally forgot about this thread. I got the roots $$\omega = -1 \pm \sqrt{2}$$
Now what am I supposed to do with them? I recall seeing a similar problem done in a book and they would just do the following:
$$z = \ln{([-1 \pm \sqrt{2}]e^{i\theta})}$$

Not sure what they did there, or what theta is in this case. Can someone please pick it up where I'm stuck?

 

[Hurkyl]

I got the roots $$\omega = -1 \pm \sqrt{2}$$

That's great! What is $\omega$?

 

[vela]

You have $w=e^{iz}=-1\pm\sqrt{2}$, so solve for z.

 

[Lavabug]

I think I see it now, z would be -i*ln(roots). I'm a bit confused though, feel like I lost track of what I was doing after substituting for w and finding the roots.

 

[Hurkyl]

I think I see it now, z would be -i*ln(roots). I'm a bit confused though, feel like I lost track of what I was doing after substituting for w and finding the roots.

This is one of the most important ideas of mathematics. To solve a problem:

1. Reduce it into an easier problem
2. Solve the easier problem
3. Use the solution to the easier problem to help solve the original problem

Step 1 frequently comes with a recipe for doing step 3 -- in this case, the relationship between z and $\omega$.

 

[Lavabug]

Hi again, thanks for the replies. I think I understand it now, please correct me if I'm wrong: the angle in my solution should be pi (the angle that i forms with y=0 in the complex plane) + 2pi*k, which covers all possible z's that'll satisfy it.

I found a similar problem in a text: cosz = 2, and the solution was also the log of roots*e^i(arg). Where the arg was just 2piK, which I assume refers to an angle of 0 which is where "2" lies. Forgive me if I'm not being faithful to the vocabulary of complex numbers.

 

[vela]

That's not right.

You have two roots, $w_1 = -1+\sqrt{2}$ and $w_2 = -1-\sqrt{2}$. Express those in polar form $w=re^{i\theta}$, then take their logs.

 

[Lavabug]

That's not right.

You have two roots, $w_1 = -1+\sqrt{2}$ and $w_2 = -1-\sqrt{2}$. Express those in polar form $w=re^{i\theta}$, then take their logs.

Ok so I'd have $w = \sqrt{3} e^{i\theta}$
which works for both of the roots(theta is zero + 2piK because they lie on y=0 on the first quadrant).

But how am I going to put the roots in my solution? I'm not seeing the connection.

 

[vela]

Try again. Note that both roots are purely real. There's no imaginary part. Also, for a positive real number, the argument is 0, but here, both roots aren't positive.