# Homework #10

## 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?

• A pole is where the formula asks you to divide by zero.

-2 is not a pole, because $X(-2) = \frac{1}{1-\frac{1}{5(-2)}}+\frac{(-2)}{2}\frac{1}{1-\frac{(-2)}{2}} = \frac{10}{11}-\frac{1}{2}$

3. Question from a student:

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

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 $a$.

• What to do after I find the value of a and b?

• If $z_1$ and $z_2$ are solutions of a quadratic polynomial $z^2 + z + 1-i$, then that means: $z^2 + z + 1-i = (z-z_1)(z-z_2)$

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)