Toronto Math Forum

Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the  given annulus: $z^3-3z+1$ in ${1<|z|<2}$
Answer: $f(z)=z^3-3z+1$
On the circle $|z|=1$
$$|z^3+1| \leq |z^3|+|1|=2$$
$$|-3z| = |3z| = 3$$
So $|z^3+1|<|-3z|$. By Roche's Theorem, $f(z)$ and $g(z)=-3z$ have the same number of zeroes with $|z|=1$.
Since $g(z)$ has one zero with $|z|=1$, we know that $f(z)$ has one zero with $|z|=1$.
On the circle $|z|=2$
$$|z^3|=8$$
$$|-3z+1| \leq |3z|+|1| = 7$$
So $|-3z+1|<|z^3|$. By Roche's Theorem, $f(z)$ and $g(z)=z^3$ have the same number of zeroes with $|z|=2$.
Since $g(z)$ has three zeroes with $|z|=1$, we know that $f(z)$ has three zeroes with $|z|=1$.
$$3-1=2$$
Therefore, $f(z)=z^3-3z+1$ has two zeroes in ${1<|z|<2}$.

Question: Find the radius of convergence of the given power series.
$$\sum^\infty _{k=0} \frac{(k!)^2}{(2k)!}(z-2)^k$$
Answer: By ratio test, we have that:
$$\frac{1}{R} = lim_{k\rightarrow \infty} |\frac{\frac{((k+1)!)^2}{(2(k+1))!}}{\frac{(k!)^2}{(2k)!}}|$$
$$= lim_{k\rightarrow \infty} |\frac{\frac{(k+1)^2(k)^2(k-1)^2...}{(2k+2)(2k+1)(2k)...}}{\frac{(k)^2(k-1)^2...}{(2k)(2k-1)...}}|$$
$$= lim_{k\rightarrow \infty} |\frac{(k+1)^2}{(2k+2)(2k+1)}|$$
$$= lim_{k\rightarrow \infty} \frac{k+1}{4k+2}$$
By L’Hospital’s Rule, we know that
$$\frac{1}{R} = lim_{k\rightarrow \infty} \frac{k+1}{4k+2} = lim_{k\rightarrow \infty} \frac{1}{4} = \frac{1}{4}$$
Therefore, we conclude that $R=4$.

Questions:
Write the equation of the perpendicular bisector of the line segment joining $-1+2i$ and $1-2i$.
Answer:
Let $z = x + iy$.
Since $z$ is an arbitrary point on the perpendicular bisector of the line segment joining $-1+2i$ and $1-2i$, we know that the distance between $z$ and $-1+2i$ is same as the distance between $z$ and $1-2i$.
Then, we can calculate that:
$$|z-(-1+2i)| = |z - (1-2i)|$$
$$|(x + iy)-(-1+2i)| = |(x + iy) - (1-2i)|$$
$$|(x + 1)+(y-2)i| = |(x - 1) + (y + 2)i|$$
$$\sqrt{(x+1)^2 + (y-2)^2} = \sqrt{(x - 1)^2 + (y + 2)^2}$$
$$(x+1)^2 + (y-2)^2 = (x - 1)^2 + (y + 2)^2$$
$$x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 - 2x + 1 + y^2 + 4y + 4$$
$$4x = 8y$$
$$y = \frac{1}{2} x$$
Then, we write the equation in complex number notation:
$$Re((\frac{1}{2} + i)z) = 0$$