I usually don't comment on these types of videos but man! The way you explain things is so friendly and extremely easy to understand. You're an excellent teacher!
Great video, thanks so much. I was wondering what the modulus would be if you had to find something like the cube root of a number. Is it 3? Would you still divide theta by 3?
I have never reply to any comment(s) on youtube before...but I wanted to reply to this one. It works for your question. 1st you need to find the 5th root of 32...which is 2....2nd you need to divide 150 by 5 which is 30...3rd you need to divide 360 by 5 which is 72. Then you build all of your new roots from the modules of the 5th root of 2...So, your 1st of 5 modules will start like this... 2[cos(30) + i sin(30)], and then you add 72 to the rest of the modules 2[cos(102) + i sin(102)]...etc. etc. I hope this helps....good luck....
@@justsaynototv8366 hey ! how about the question only given z = -8 then find all 6th roots. My k(0) is 2[cos 30 + isin 30 , then k(1) is 2[cos 90 + isin 90] ........ etc + 60 is this correct???
This is clean, but I have to ask - why are we doing this? I have to solve this for some math exams but there's nothing afterwards, just "do it to do it".
8 years later, still saving students from grey areas, nothing but respect for you sir
I usually don't comment on these types of videos but man! The way you explain things is so friendly and extremely easy to understand. You're an excellent teacher!
Thanks! :^D
saved me 10 years after publishing. Thanks, i have my midterm for calculus tomorrow.
This saved my life. I couldn't find a single source online that explained it this well. My college prof sucks lol, thanks!
FINALLY!!! Someone who actually shows in straightforward terms how to apply De Moivre's theorem in order to obtain the nth root of a complex number!
Yes!!!! My jaw dropped on the floor when he simplified the thinking and approach without diluting the content. This was amazing
11 years ago but still gold ❤
This is so clean. So awesome. Highly recommended to any uni student.
That was awesome! I was being taught the hard way! Dividing it by 3 at every step (finding the cube roots)! THANKS!
Man, you explained this really well. Thanks for the help!
love you man, this is so much simpler to understand thanksssssssss!!!
Oh gosh you saved my high school life thank you soooo much🤍🤍
It's 1am and I have an exam in the afternoon. Thanks for coming in clutch
You... you are an amazing human being. Thank you!!!
Thanks! :^D
Great video, thanks so much. I was wondering what the modulus would be if you had to find something like the cube root of a number. Is it 3? Would you still divide theta by 3?
bless
your
soul
Thanks for the help on my Diff Eq test that is in less than an hour.
Incredible! Amazing.
Thnx. I hve understood this clearly.
Thanks 11 years later!
Thank you for this. 🙏🏽💯
danke
Very Helpful video,
Thank you so much
Great 👍👍👍
This technique is op !
THANKS!!!!!
fucking awesome bro.Thank you.
Doesn't work for a particular equation I tried to solve
32[cos150 + isin150]
5th root
I have never reply to any comment(s) on youtube before...but I wanted to reply to this one. It works for your question. 1st you need to find the 5th root of 32...which is 2....2nd you need to divide 150 by 5 which is 30...3rd you need to divide 360 by 5 which is 72. Then you build all of your new roots from the modules of the 5th root of 2...So, your 1st of 5 modules will start like this... 2[cos(30) + i sin(30)], and then you add 72 to the rest of the modules 2[cos(102) + i sin(102)]...etc. etc. I hope this helps....good luck....
@@justsaynototv8366 hey ! how about the question only given z = -8 then find all 6th roots. My k(0) is 2[cos 30 + isin 30 , then k(1) is 2[cos 90 + isin 90] ........ etc + 60 is this correct???
thanks!
thankyou!!!
nice man
Thx
ayo whats k tho
❤️
This is clean, but I have to ask - why are we doing this? I have to solve this for some math exams but there's nothing afterwards, just "do it to do it".
Bruh 👁️👄👁️ 😂