Homework #10

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15 thoughts on “Homework #10

  1. For question 2 a where you use partial fraction to go from (Z/Z-i)(Z/Z+1+i), would you be able to clarify how you found those coefficients A and B? Is there a good way to do partial fractions when you have complex numbers in the denominator? I apologize if this seems simple, I may be missing something basic but I cant get the same answer.

    • To get the coefficients,
      \frac{z}{z-i}\frac{z}{z+1+i} = A\frac{z}{z-i} + B\frac{z}{z+1+i}
      and compare the numerators, as usual. You should get:
      A+B=1 and (1+i)A-i B=0
      Another way to write the answer is A=\frac{i}{1+2i} and B=\frac{1+i}{1+2i}
      ( to get the numbers in the answers sheet, multiply by \frac{1-2i}{1-2i} )

  2. For question 1a, I find 2>|z|>1/5. I understand that it’s an annulus, but I can’t see why -2 and -1/5 are not included for pole. Can you explain?

  3. Question from a student:

    For question 2a, how do we factor the complex denominator? Is there a specific method or just guessing?

    Answer:
    Guessing isn’t too bad for 2a, but there’s a more systematic way if you prefer:

    One method to factor z^2+z+1-i is to substitute z=a+ib, and then solve for real part and imaginary part equal zero. That is:
    a^2-b^2+a+1=0 and 2 a b + b - 1=0

    To solve these two equations is then tricky.

    Rewrite the second equation as (2a + 1)b = 1, which inspires us to rewrite the first equation using the pattern (2a+1):
    a^2-b^2+a+1 = \frac{1}{4}(2a+1)^2-b^2+\frac{3}{4}

    We can then subsitute together and get:
    \frac{1}{4}\frac{1}{b^2}-b^2+\frac{3}{4}=0
    4\left(b^2\right)^2 - 3\left(b^2\right) - 1=0

    This is a quadratic equation with solutions:
    \left(b^2\right) = \frac{3 \pm \sqrt{9+16}}{8} = 1 \text{   or   } -\frac{1}{4}
    Since b is real, it must be b^2 = 1.

    To finish, plug in b = \pm 1 and find the values of a.

  4. Is a mistake in question 5b? The roots that have been found might not be correct. Could you please check that and confirm?! The roots I found are z_1=1+\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} and z_2=1-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}.

  5. Is the answer to 5a missing a ‘z’ in the numerator? What happened to the ‘z’ that was being multiplied by x(z)?

  6. For 2 b is there a short way to find a, b, and c in the PFE? Multiplying out all the terms is really a big mess, and the residue method doesn’t seem to work because of the ‘z’s in the numerator. Thanks.

  7. For Q5b
    The roots for (z^2 – 2 Z +1 – i) could be (1 +/- sqrt(i)) as well, right?
    because (z – 1 – sqrt(i)) * (z – 1 + sqrt(i)) = (z^2 – 2 Z +1 – i)

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