Complex Number Problem

[cmkluza]

The problem statement is as follows:

Consider w = z / (z² + 1) where z = x + iy, y ≠ 0 and z² + 1 ≠ 0.
Given that Im(w) = 0, show that |z| = 1

I've attempted expanding the function and trying to go from there:
w = (x + iy) / ((x + iy)² + 1) = (x + iy) / (x² + 2xiy - y² + 1) = [(x + iy)(x - iy)²] / [(x² + 2xiy - y² + 1)(x - iy)²] = (x³ + 3x²iy + 3xy² + iy³) / (3x² - 2xiy + y²)

I'm then getting stuck trying to get the imaginary numbers out of the denominator.
Once I do that, I wouldn't imagine it would be overly difficult to find the Im(w) and from there try to see what |z| = √(x² + y²) is.

(Also, as a side note, could anyone tell me how to make my math look more "math-like" in these forums? I hate this typed out appearance, as it lacks structure, and is more difficult to read)

Any help with this will be greatly appreciated, thanks!

 

[Steven G]

The problem statement is as follows:

Consider w = z / (z² + 1) where z = x + iy, y ≠ 0 and z² + 1 ≠ 0.
Given that Im(w) = 0, show that |z| = 1

I've attempted expanding the function and trying to go from there:
w = (x + iy) / ((x + iy)² + 1) = (x + iy) / (x² + 2xiy - y² + 1) = [(x + iy)(x - iy)²] / [(x² + 2xiy - y² + 1)(x - iy)²] = (x³ + 3x²iy + 3xy² + iy³) / (3x² - 2xiy + y²)

I'm then getting stuck trying to get the imaginary numbers out of the denominator.
Once I do that, I wouldn't imagine it would be overly difficult to find the Im(w) and from there try to see what |z| = √(x² + y²) is.

(Also, as a side note, could anyone tell me how to make my math look more "math-like" in these forums? I hate this typed out appearance, as it lacks structure, and is more difficult to read)

Any help with this will be greatly appreciated, thanks!

w = (x + iy) / ((x + iy)² + 1) = (x + iy) / (x² + 2xiy - y² + 1) = (x+iy)/(x²- y² + 1-2xyi). Now multiply top and bottom by the complex conjugate of the denominator. This should help

 

[cmkluza]

w = (x + iy) / ((x + iy)² + 1) = (x + iy) / (x² + 2xiy - y² + 1) = (x+iy)/(x²- y² + 1-2xyi). Now multiply top and bottom by the complex conjugate of the denominator. This should help

Thank you very much for your help! I'm afraid that I'm not quite getting where to go next. I'm not used to seeing something with a complex number in the denominator with four terms. I'd like to say the complex conjugate is just 1 + 2xyi, but it seems more likely that it's x² - y² + 1 + 2xyi, however multiplying either of these numbers in the denominator will still leave imaginary numbers in the denominator. I apologize for not understanding where you're going with this, but could you give me another hint?

Thanks!

 

[Ishuda]

The problem statement is as follows:

Consider w = z / (z² + 1) where z = x + iy, y ≠ 0 and z² + 1 ≠ 0.
Given that Im(w) = 0, show that |z| = 1

I've attempted expanding the function and trying to go from there:
w = (x + iy) / ((x + iy)² + 1) = (x + iy) / (x² + 2xiy - y² + 1) = [(x + iy)(x - iy)²] / [(x² + 2xiy - y² + 1)(x - iy)²] = (x³ + 3x²iy + 3xy² + iy³) / (3x² - 2xiy + y²)

I'm then getting stuck trying to get the imaginary numbers out of the denominator.
Once I do that, I wouldn't imagine it would be overly difficult to find the Im(w) and from there try to see what |z| = √(x² + y²) is.

(Also, as a side note, could anyone tell me how to make my math look more "math-like" in these forums? I hate this typed out appearance, as it lacks structure, and is more difficult to read)

Any help with this will be greatly appreciated, thanks!

First, for more 'math like' forms, I assume you mean something like
\(\displaystyle w = \frac{z}{z^2 + 1}\)
That is accomplished with the following

[tex]w = \frac{z}{z^2 + 1}[/tex]

That is, a string of letters between the bracketed tex and /tex tag pair. Formatting the text is discussed at
ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf
which also has a couple of other links for more detail.

To the problem: The denominator is, as you indicated,
x² + 2xiy - y² + 1 = (x² - y² + 1) + i (2 x y) = a + i b
where not both a and b are zero. If we multiply denominator and numerator by a - i b we have
\(\displaystyle w = \frac{z}{z^2 + 1} = \frac {(x + i y) (a - i b)}{a^2+b^2} = \frac{a x + b y + i (a y - b x)}{a^2+b^2}\)
So, the imaginary part of w is given by
Imag(w) = c (a y - b x)
where
\(\displaystyle c = \frac{1}{a^2+b^2}\)
Now, if Imag(w) is zero then, since c is not zero
\(\displaystyle y = \frac{b}{a} x\)
So, use that to see if you can show
|z²|=x² + y² =1
That is, just substitute in for a and b and see where it leads.

 

[cmkluza]

First, for more 'math like' forms, I assume you mean something like
\(\displaystyle w = \frac{z}{z^2 + 1}\)
That is accomplished with the following

[tex]w = \frac{z}{z^2 + 1}[/tex]

That is, a string of letters between the bracketed tex and /tex tag pair. Formatting the text is discussed at
ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf
which also has a couple of other links for more detail.

To the problem: The denominator is, as you indicated,
x² + 2xiy - y² + 1 = (x² - y² + 1) + i (2 x y) = a + i b
where not both a and b are zero. If we multiply denominator and numerator by a - i b we have
\(\displaystyle w = \frac{z}{z^2 + 1} = \frac {(x + i y) (a - i b)}{a^2+b^2} = \frac{a x + b y + i (a y - b x)}{a^2+b^2}\)
So, the imaginary part of w is given by
Imag(w) = c (a y - b x)
where
\(\displaystyle c = \frac{1}{a^2+b^2}\)
Now, if Imag(w) is zero then, since c is not zero
\(\displaystyle y = \frac{b}{a} x\)
So, use that to see if you can show
|z²|=x² + y² =1
That is, just substitute in for a and b and see where it leads.

Thank you very much for your help! As it ends up, I'll need a lot more work with complex numbers before I'm proficient. Also, thanks for showing me how to write down things in a more neat fashion, it'll surely help me in the future!