Finding the number of roots using Rouché's Theorem

2021-10-10 05:13:19 | English, Korean

Translated with the help of ChatGPT and Google Translator

Yesterday, while watching YouTube, I found the following very interesting video. It was a video explaining how to find the number of roots within a certain region of a complex function using Rouché's Theorem.



To briefly explain Rouché's Theorem explained in this video, it is as follows.

"If f(z)>g(z)|f(z)|>|g(z)| always holds true on the boundary D\partial D of a region DD, then f(z)f(z) and f(z)insideDThenumberofrootsof+g(z)f(z) inside D The number of roots of +g(z) is the same."

And when f>g|f|>|g| always holds, ff is said to be dominant over gg.

Using this, when trying to find the number of roots of a complex polynomial function h(z)h(z) within a certain region DD, h(z)h(z) is dominant, and the simple term ff and the non-dominant term $ We can easily solve the problem by dividing by g$.


Let's follow the example from the video above. We want to find the number of roots of the function h(z)=z5+3z2+1h(z)=z^5+3z^2+1 where D={z1<z<2}D = \set{z|1<|z|<2}.

To do so, set the area D1={zz<1},D2={zz<2}D_1 = \{z | |z| <1\}, D_2 = \{z ||z|<2\}, then subtract the number of roots in D1D_1 and D1\partial D_1 from the number of roots in D2D_2. .

First, regarding the number of roots inside each region, in D1\partial D_1, z=1|z|=1, so z5=1,3z2=3|z^5|=1, |3z^2|=3, so z5+1z5+1=2<3z2|z^5+1 |\leq|z^5|+1=2<|3z^2| holds true.

Therefore, hh can be divided by f(z)=3z2,g(z)=z5+1f(z)=3z^2, g(z)=z^5+1.

Therefore, the number of roots of h(z)h(z) inside D1D_1 is 2, the same as the number of roots of f(z)f(z). (middle root at z=0z=0)

At this time, you should be careful that only the number of roots is the same, but the values of the roots are not the same. Therefore (in general) hh has no roots at z=0z=0.

Similarly, D2={zz=2}\partial D_2 = \{ z| |z|=2\} Above, z5=32>3z2+1=4>3z2+1|z^5|=32>|3z^2|+1=4>|3z^2+1|, so $f(z)=z^5, g( You can divide hh by z)=3z^2+1$.

Therefore, the number of roots of h(z)h(z) inside D2D_2 is 5, the same as the number of roots of f(z)f(z).

Finally, above D1={zz=1}\partial D_1 = \{z||z|=1\}, we just need to show that hh has no roots.

This also seemed to be a very simple method in the video above, as shown below.

If a root such as z=1|z|=1 exists, then

z5+3z2+1=03z2=z513z2=z5+13=z5+1<z5+1=23<2z^5+3z^2+1=0\\ \therefore 3z^2=-z^5-1\\ \therefore |3z^2| = |z^5+1|\\ \therefore 3 = |z^5+1| < |z^5| + 1 = 2\\ \therefore 3<2

must be established. Obviously, this is false, so there is no such thing as z=1|z|=1 such that z5+3z2+1=0z^5+3z^2+1=0.

In fact, this cannot but be established since f(z)f(z) is dominant over g(z)g(z) on D1\partial D_1. If a root exists on the boundary, h=f+g=0h=f+g = 0, so f=gf=-g will hold, and therefore f=g|f|=|g|, which means ff is not dominant.

Therefore, the number of roots inside D2D_2 - the number of roots on D1\partial D_1 - the number of roots inside D1D_1 = 502=35-0-2= 3.

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