Show Roots of Complex Sine Are Real Valued

[mancini0]

Homework Statement

Hi guys, I am having difficulty with the following problem:

Show the complex sine function has real valued zeros.

Homework Equations

I know that sin(z) = (e^iz - e^-iz) /2i

The Attempt at a Solution

I have to find roots of sin(z), i.e, when the above equation = 0. First I multiplied both sides by 2i, to cancel the 2i in the denominator.

leaving e^iz -e^-iz = 0.

then e^iz = e^-iz .

So e^iz = 1/e^1z

Now I am stuck. I can break it down to trigonometric form, but I'm not sure how that helps...

cos(z) + i sin(z) = 1 / cos(z) + i sin(z)

I think I have to get the above expression into real and imaginary (u and v) parts:
cos(x+iy) +i (sin(x+iy) = 1/[cos(x+iy) + i*sin(x+iy)]

Any guidance would be much appreciated...

Thanks!

 

[Dick]

Factor it. 0=e^(iz)-e^(-iz)=e^(-iz)*(e^(2iz)-1). The first factor is never zero, right?

 

[mancini0]

Hmmm... so the roots are when e^(2iz) - 1 = 0, i.e, when e^(2iz) = 1.

e^(2iz) = e^(iz) * e^(iz) = (cos(z) + isin(z) ) * (cos(z) + isin(z))

1= cos^2(z) - sin^2(z) + i(sin(z)cos(z))

Recognizing that cos^2(z) - sin^2(z) = cos(2z), I can rewrite the above as

1 = cos(2z) + isin(z)cos(z)

Now my intuition tells the above must be real valued, since I feel that whatever I put in for z which satisfies the above equation will cancel the imaginary part. I just don't know how to solve the above equation in the complex plane, i.e, how would I expand cos(z) to cos(x+iy)?

 

[micromass]

Isn't there an easier way to solve e^2iz=1 by simply taking logarithms?

 

[Dick]

Hmmm... so the roots are when e^(2iz) - 1 = 0, i.e, when e^(2iz) = 1.

e^(2iz) = e^(iz) * e^(iz) = (cos(z) + isin(z) ) * (cos(z) + isin(z))

1= cos^2(z) - sin^2(z) + i(sin(z)cos(z))

Recognizing that cos^2(z) - sin^2(z) = cos(2z), I can rewrite the above as

1 = cos(2z) + isin(z)cos(z)

Now my intuition tells the above must be real valued, since I feel that whatever I put in for z which satisfies the above equation will cancel the imaginary part. I just don't know how to solve the above equation in the complex plane, i.e, how would I expand cos(z) to cos(x+iy)?

Work out what e^(2iz) is when z=x+iy. Then set that to 1.

 

[Dick]

Isn't there an easier way to solve e^2iz=1 by simply taking logarithms?

If you happen to know log is multivalued and all that stuff. I think a more basic approach might be better here.

 

[mancini0]

e^2(iz) = e^2i(x+iy) = e^(2ix -2y)
= e^2ix / e^2y since subtraction in the exponents corresponds to division of base.

The above = 1 when e^2ix = e^2y

i.e, when 2ix = 2y, when ix = y

x = -iy

 

[mancini0]

We haven't discussed the complex logarithm, that comes tomorrow! I don't see how my above solution shows sin(z) is always real valued.

 

[Dick]

e^2(iz) = e^2i(x+iy) = e^(2ix -2y)
= e^2ix / e^2y since subtraction in the exponents corresponds to division of base.

The above = 1 when e^2ix = e^2y

i.e, when 2ix = 2y, when ix = y

x = -iy

No, no. The complex function e^z isn't one-to-one. You can't cancel it like that. Use deMoivre on e^(2ix). You know a lot about the REAL sine and cosine functions.

 

[mancini0]

Hmmm, I think it hit me...if x = -iy, then z=x+iy implies z = -iy +iy = 0 is the root.

 

[Dick]

Hmmm, I think it hit me...if x = -iy, then z=x+iy implies z = -iy +iy = 0 is the root.

See my previous post.

 

[mancini0]

Okay, thank you very much for your help thus far.

By De Moivre's Theorem I would take the nth power of the modulus and multiply the argument by n.
But why would I do this on e^2ix? I thought De Moivre's theorem only applies when a complex number is raised to an integer?

 

[SammyS]

[tex]\sin(z)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)\,,\ \text{ where }\ z = x + iy[/tex]

So sin(x)cosh(y) = 0 and cos(x)sinh(y) = 0 .

 

[Dick]

Okay, thank you very much for your help thus far.

By De Moivre's Theorem I would take the nth power of the modulus and multiply the argument by n.
But why would I do this on e^2ix? I thought De Moivre's theorem only applies when a complex number is raised to an integer?

e^(2ix)=cos(2x)+i*sin(2x). That's all I meant. So 1=(cos(2x)+i*sin(2x))*e^(-2y). First notice that the imaginary part has to vanish. What does that tell you about x? Use that to deduce what y must be.