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

In summary, the student attempted to solve for sin(z) using exponentials, but ran into an error. They found another solution that involved using the log of roots*e^i(arg).
  • #1
Lavabug
866
37

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...?
 
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  • #2
Hi Lavabug! :smile:

sin(z) = Im(…) ? :wink:
 
  • #3
Lavabug said:

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?
 
  • #4
Lavabug said:
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=eiz, 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.
 
  • #5
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?
 
  • #6
I looked on Wiki and found this

0b7d3b2f55de29dee74d89e3125ccd77.png
 
Last edited by a moderator:
  • #7
So you equate:
[tex]
\sin x\cosh y+i\cos x\sinh y=i
[/tex]
This tells you the form of x and the form of y...
 
  • #8
Lavabug said:

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?
 
Last edited by a moderator:
  • #9
Lavabug said:
(z+ z^-1)/2i = i
Looks like we all missed the fact you should have a minus sign between the two terms for sine. :redface:
 
  • #10
I went back and edited so I can pretend I didn't miss that!
 
  • #11
Thanks for the replies everyone, totally forgot about this thread. I got the roots [tex]\omega = -1 \pm \sqrt{2} [/tex]
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:
[tex]z = \ln{([-1 \pm \sqrt{2}]e^{i\theta})}[/tex]

Not sure what they did there, or what theta is in this case. Can someone please pick it up where I'm stuck?
 
  • #12
Lavabug said:
I got the roots [tex]\omega = -1 \pm \sqrt{2} [/tex]
That's great! What is [itex]\omega[/itex]?
 
  • #13
You have [itex]w=e^{iz}=-1\pm\sqrt{2}[/itex], so solve for z.
 
  • #14
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.
 
  • #15
Lavabug said:
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 [itex]\omega[/itex].
 
  • #16
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.
 
  • #17
That's not right.

You have two roots, [itex]w_1 = -1+\sqrt{2}[/itex] and [itex]w_2 = -1-\sqrt{2}[/itex]. Express those in polar form [itex]w=re^{i\theta}[/itex], then take their logs.
 
  • #18
vela said:
That's not right.

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

Ok so I'd have [itex] w = \sqrt{3} e^{i\theta}[/itex]
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.
 
  • #19
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.
 

1. What are complex numbers?

Complex numbers are numbers that contain both a real and an imaginary component. They are typically written in the form a + bi, where a is the real part and bi is the imaginary part with the imaginary unit i.

2. How do you find the sine of a complex number?

To find the sine of a complex number, we can use Euler's formula, which states that sin(z) = (e^(iz) - e^(-iz)) / (2i). This formula allows us to calculate the sine of any complex number, including sin(z) = i.

3. What is the geometric interpretation of sin(z) = i?

The geometric interpretation of sin(z) = i is a point on the imaginary axis of the complex plane. This point has a magnitude of 1 and is located at an angle of 90 degrees from the origin.

4. How do you graph sin(z) = i on the complex plane?

To graph sin(z) = i on the complex plane, we can plot the point (0,1) on the imaginary axis. This represents the solution to the equation sin(z) = i.

5. What are the applications of complex numbers in real life?

Complex numbers have many applications in various fields such as engineering, physics, and economics. They are used to model and solve problems involving alternating currents, oscillations, and quantum mechanics. They also have practical uses in signal processing, image processing, and financial forecasting.

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