Real Analysis – Finding Zeros of Two Equations

calculuscomplex numberscomplex-analysisreal-analysis

Consider the equations
1+\frac{1}{z^k}=0 \quad\mbox{and}\quad 1+\frac{1}{z^k}+\frac{1}{(z+1)^k}=0,

where $k$ is a positive integer $\ge 4$. I would like to show for instance that the number of zeros in $\frac{\pi}{2}\lt \mbox{Arg}(z) \lt \frac{2\pi}{3}$ is the same for both equations. I'm not sure if this is true, I'm basing this on numerical evidence.

Of course the first equation is equivalent to $z^k=-1$, so all the zeros lie on the unit circle and it is easy to explicitely compute the zeros. The problem is with the second equation, there can be some zeros outside the unit circle, so the condition $\frac{\pi}{2}\lt \mbox{Arg}(z) \lt \frac{2\pi}{3}$ is really important here. Intuitively what I think happens is that in this region the term $1/(z+1)^k$ is small for big $k$, but when $\mbox{Arg}(z)$ is close to $2\pi/3$ this is not always the case (in that case $|1/(z+1)^k|\approx 1$, so it does contribute to the second equation).

Is there a way to get past the issue near $2\pi/3$ and show that the two equations have the same number of zeros and moreover the solutions are close to each other? (Again this is from numerical evidence)

Best Answer

The two equations don't have the same number of solution in the given sector for every $k$. Define $(I)$ as equation $z^k+1=0$ and $(II)$ as equation $(1+z^k)(1+z)^k+z^k=0$. For $k=8$, numerically $(I)$ has $1$ solutions and $(II)$ has no solutions in sector $\frac{\pi}{2} < \text{Arg}(z) < \frac{2\pi}{3}$. If you denote $N_1,N_2$ as the solutions of $(I),(II)$ respectively in the given sector, then for $k\le 50$, (all of them are even :o )

$$ \begin{array}{c|c|c} \hline k& N_1& N_2\\ \hline 8& 1& 0\\ \hline 10& 0& 1\\ \hline 14& 1& 0\\ \hline 18& 1& 2\\ \hline 20& 2& 1\\ \hline 32& 3& 2\\ \hline 34& 2& 3\\ \hline 38& 3& 2\\ \hline 42& 3& 4\\ \hline \end{array} $$

Now coming to magnitude of the solutions, all solutions to the right of $x=-1/2$ line are indeed close to each other. This is illustrated in the image below for $k=195$. enter image description here Addendum: The following matlab code was used to produce this image, it only works till $k=195$:

syms z
k = 195
eqn1 = z^k==-1;
eqn2 = (z^k+1)*(1+1/z)^k == -1;
%Solve both the equations 
Root1=vpasolve(eqn1);  Root2=vpasolve(eqn2);
%Scatter plot the solutions
hold on
%Real and imaginary axis
xline(0); yline(0);
[n1,m1] = size(Root1);
%Make lines connecting root to origin 
for j=1:n1
line([0 real(Root1(j))],[0 imag(Root1(j))])
[n2,m2] = size(Root2);
for j=1:n2
line([0 real(Root2(j))],[0 imag(Root2(j))],'Color','red')
hold off
axis equal

Observe that some zeroes of equation $II$ are vertically stacked near the vertical line $x=-1/2$. This is also observed for smaller values of $k$, enter image description here

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