Complex Numbers (How to find the nth root) : ExamSolutions Maths Video Tutorials
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Tutorial on finding the nth root when dealing with complex numbers.
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For 1st Quad arg = Theta
------2nd Quad arg = Pi - Theta
------3rd Quad arg = Theta - Pi
------4th Quad arg = -(Theta)
We have to consider different arguements for diff quads.
In this case our theta was in 3rd quadrant so we used theta - pi
which resulted in -5pi/6 .
you beautiful human
You helped me out.... thank buddy
thankyouuu
Thanks so much this write up was more useful than the video
Thanks a lot! I've been wondering why he did that cause the Arg was already between -pi & pi at first, Again Thanks!
Just start with k=0, then k=1 and so on. You don't really need to do k=-1 etc as k=1, k=2, k=3 etc will start to repeat the roots eventually.
thank you so much, you've made me actually understand complex numbers, and they're no longer complex to me
life saver!
This is one of the best videos ever! keep the good work and please never stop making those, i feel that am really learning with you!
i like how you interpret question
how is the angle (5pi / 6) if both terms on complex # are negative? Should it not be in the 3rd quadrant ?
8:52 - reason why not k = 3, it 31pi/24 which exceeds pi radient. Which is why it took -1
great method of teaching... awsm clearity of complex concepts !!! thnx
You're welcome.
For this type, yes.
thank you so much teacher... your lecture helped me at exact 1 hour before my calculus exam...
@@tuneese2561 rah
Is there a range for theta because in exam questions they always add pi if the initial theta value is negative
Brilliant!
Cool
Thanks a ton Sir!
You are always welcome.
Just made it simpler😢 thanks
in the last line of working where did the power of 1/4 go for the trig part go ?
sorry sir, how to find value of k that we use?
good lecture
BEST
9:45 why does theta have to be less than pi?
how did u get the 1over 4 for the n?
Because we are finding z to the power of 4
Sir, Can we use seven pi upon six instead of - five pi upon six?
i also had 7 pi on 6 and iwas doubting
I was wondering too
is it a trick to know with which values of k i must start? thanks really apprieciated ;p
oller bin You definitely won’t find this useful now but for other viewers, I’m pretty sure you start from 0 and go to a k value 1 less than the power of z
Anurag Reggie ooo thank you
Oller bin how did your exams go?
It should be an integer. Your answers could well be fluky.
Why do we need to include the 2kπ?
In my text book they work in polar form, is this method easier?
.
How did you get the -5pi/6
I want to know too
what do you mean by " 31pi/24 exceeds pie radian"? thanks
+Brunin G3 neri Pi radian = 180 degrees. He is just saying that 31/24pi is bigger than 24/24pi. (The angle is bigger than 180 degrees, going anticlockwise from the x axis). As the range of the principle argument is 0 to pi then 31/24pi exceeds the range we want
do you have a video for drawing shapes with more than 3 vertices on youtube, I don't know how I would go about it!
Sorry, no.
@@ExamSolutions_Maths :(
Is this FP1 or FP2/3??
+Moezph. Depends on what board you are doing.
+ExamSolutions I checked your website, it's edexcel FP2. I'm studying FP1 this year. Great videos sir!
+Moezph. Okay - good luck
how come tan inverse of 8/8.root 3 gives you -5pai/6 in the third quad ,am confused you made a mistake
Based on the complex number, we landed in the 3rd quadrant.
As mentioned in the video, the Arg Z must be between -pi and pi radians, so you need to find the angle from the positive Re axis, to the reference line (red line) by turning in a clockwise direction (giving a negative angle).
As we found the acute angle to be pi/6, we got 5pi/6 by doing pi - pi/6.
@@ExamSolutions_Maths cool story bro
Are you god?
Hardly!
@@ExamSolutions_Maths lol I love you so much.
How k = -1 ?
So what's the final answer? Sorry if this is a dumb question
The solutions are the different values of Z when k =-1,0,1,2
Urdu me
Speaking urdu
Your lectures are veey useless sir very hard and u dont 3xplain how you get values you just write them out if we know then we qont be here obviously all the comments is someone confused pls do better
Complex numbers?
More like simple numbers. :)
speaking for Hindi