Rouche's Theorem with Example | Complex Analysis #23

[@sethlovelace862]

Great video! I did note one error: at minute 2:50, you claim that |g(z)| = |7-2z^3| = |7|+|-2z^3|, when instead it should be |g(z)| = |7-2z^3|

[@TheMathCoach]

Ohh my mistake, you are completely right! Thanks for keeping me in check, these kinds of errors always make it through somehow. I will pin this comment so more people can see it and I also added some lines about it in the description to make it crystal clear. Thanks again for the feedback and the kinds words, I really appreciate it! Lastly welcome to my channel, I hope you will find it useful :)

[@matloose]

You don't know how much you've helped me

[@TheMathCoach]

Now I know, and I'm happy to help!

[@ingiford175]

So when doing rational root tests, we can toss out any number greater then 2. Interesting.

[@TheMathCoach]

not sure how you got to that from the video, but more power to you if you did!

[@ingiford175]

@@TheMathCoachWell if all the roots have a | z | less than 2, then any rational roots will be under that value. Of course it may be faster to try all the rational roots then do this, but I find finding other ways to 'limit' search spaces of polynomial functions interesting.

[@sivatvs]

If only I had seen this video a couple of weeks ago... ;) very clear explanation....all the best...

[@TheMathCoach]

Thanks for the kind words and good luck on the rest of the exams :) Do you have any specific feedback on the video, so I can keep improving? It would be greatly appreciated ^^

[@sivatvs]

Perhaps you can give another example with a slight variation in the same video....and you can post more videos on different topics in Complex Analysis...

[@TheMathCoach]

Thanks, I will keep that in mind! Complex Analysis is the course I'm focusing on at the moment, uploaded a video about Trigonometric Equations some hours ago and will probably come out with a video about Cauchy's equations in one week, just have to add the description and edit the tags. Thanks again for the feedback and have a nice day!

[@Nightfold]

So concise and clear!

[@TheMathCoach]

Happy you liked it and that you find more of my content useful for you!

[@PunmasterSTP]

Another phenomenal video! Now I know how to appRouche these types of problems...

[@TheMathCoach]

That was the whole point, so I take that as a giant win :)

[@PunmasterSTP]

@@TheMathCoach I'm glad, and this video really helped me to zero in on things and get the (p)whole picture!

[@devashish890]

Thank you Sir , for crystal clear explanation

[@TheMathCoach]

I'm glad you liked it and thanks for the kind words :)

[@AA-le9ls]

You don't need something as powerful as Rouche's theorem to solve the problem mentioned in the second half of this video. To use that theorem here is an overkill, like killing mosquitoes with cannons.

[@TheMathCoach]

I like to hunt with the big guns! but what method would you use instead? I'm interested to know :)

[@krisnendubiswas3374]

It would be better if you had added some more examples with theorem though thanks for the hard work!

[@TheMathCoach]

Happy to help and thanks for the feedback - it is noted!

[@xxjason813]

good job dude!!! very helpful

[@TheMathCoach]

Glad you liked it dude!! :)

[@damnsalacious]

this is really helpful, thank you!

[@TheMathCoach]

Ohh, I'm really glad to hear that! What did you think about the speed of the video, too fast, too slow? If you don't mind me asking, I always appreciate feedback :)

[@gnydnnk8384]

Is it generally difficult/tedious to find how many number of roots in a specific quadrant? I had a few tutorial questions which asked for how many roots in each quadrant.

[@TheMathCoach]

It is a bit more difficult but still works, you just have to choose your simple closed contour C smart. If you for example would like to determine the the number of roots in the first quadrant, when we might do something as the following. We can't draw a simple closed contour which includes all of the first quadrant so we have to manage with a sector of it: Let C go along the on the real axis from 0 to R, then the circular arc |z|=R in the first quadrant from R to Ri, then back to 0 on the positive imaginary axis. This way we have been able to create a simple closed contour C which satisfies rouches theorem and the theorem can be used. Also by increasing R we can include more and more of the first quadrant. Then it is just a question about using Rouches Theorem and testing what happens when you increase R and let it approach infinity, in most cases you will find that it is sufficient to only test R values in between 1-10, since the result can probably be generalized from that result and will therefor not change for bigger values of R. I hope this answered your question and just let me know if you got any follow up questions. Best regards, TheMathCoach

[@richasharma4295]

Sir , please solve more examples, where contour is like , 1

[@TheMathCoach]

I will add that one to my collection of suggestions for future videos. The idea for these kind of examples are the same as the example above, you just have to do it two times one for each circle. For example if you know that a function have 5 zeros inside the bigger circle and then only 2 zeros inside the smaller circle then you can draw the conclusion that the function most have 3 zeros in between the two circles. Here you have an example from the exam I wrote when I took the course. gyazo.com/c3f860125c68203a75fc800c7627ddf1

[@richasharma4295]

TheMathCoach Now , doubt is clear , Thank you sir , for help.

[@TheMathCoach]

No problem, I'm happy to help :)

[@fordtimelord8673]

Thank you.

[@TheMathCoach]

Happppy to help dear doctor who

[@metallsnubben]

At an exam today I misread my own (allowed) notes and thought |f(z)| had to be bigger on _and_ _inside_ C. Made it SO much harder (...impossible, even) for myself for no goddamn reason lol

[@TheMathCoach]

haha, that sounds like real _pain_ and I can imaging everything became much harder by doing that! Happens to us all mate

[@davidjohnson-my6sr]

You might be mistaken at 3mins. |g (z)| = |7-2z^3| =/= |7| + |-2z^3|... you mightve meant the Triangle Inequality.

[@TheMathCoach]

Ohh yes indeed, you are completly right! My results is the result of using the triangle inequality, but I completly forgot to mention it will adding the audio and even used "=" instead of the "inequality sign". Thank you so much for alerting me, I'm glad that I have people like you keeping me in check. I will add some lines about this mistakes in the video description :)

[@lakhanpaul1458]

demirts of rouches theorem???

[@TheMathCoach]

I'm not sure what you mean.

[@shivanshirathore8060]

Sir please btayeye ki hmesa ek hi term kyu lete hai as f(z)

[@TheMathCoach]

Hello, I don't understand what you are saying, sorry.

[@geniusmathematics9123]

Grt sir

[@TheMathCoach]

No problem mate

[@ankurtalukdar174]

👍

[@TheMathCoach]

Thumbsup!