Using de Moivre's Theorem - Example (2) : ExamSolutions Maths Revision Tutorials

Example question of using de Moivre's theorem
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[@helenday5031]

This site has for me repeatedly brought clarity to things that are terrribly complicated . imo that is what great teaching is often about.

[@bostangpalaguna228]

Steps:
1. Draw the argand diagram to get some sense and to get arg(Z)
2. Look for mod(Z) and turn into euler's form
3. Apply De Moivre's

[@joemcmillian4179]

The video content is so excellent, congratulations

[@pstabali7910]

Thank you very much.

[@morganabbotts]

Fantastic.

[@howardkunda8545]

Excellent work

[@al-anoud-123]

Thank you

[@trilliondreams9048]

How to do get arg(z) as -π/6 from π/6

[@stevenkaystevekayelu5823]

You opened my mind sir!! Thank you🙏

[@-bismarck]

How do you calulate the arg(z) cuz I have another way to calculate it
Cos= 3½÷2
Sin=-1÷2 by that you know it is (pi÷6) and it is from the
(4th quarter) wich means the arg (z) law is
arg (z) = 2pi - theta
By that
arg (z) = 2pi - pi÷6 = 11pi÷6

[@dhruvchaulkar8208]

Great

[@JonBudinas]

Could you also have the Arg(z) as 11/6pi?

[@ExamSolutions_Maths]

You could have but generally it is accepted that the arg should lie in the range -pi to pi.

[@ahmadmhmd5465]

Is that important to rearrange the the theta

[@tikidorsisa5987]

Good

[@josephmusic2067]

nice nice nice

[@niazgulhazarboz9043]

Hi sir, shouldn't the arg be positive as the original value was negative and the congegate makes it negative so -(-) =+ ?

[@ExamSolutions_Maths]

Hi, I am confused by your question as I am not working with the conjugate of z

[@bredahkendi1959]

He is right. He has used the argand diagram which means he has considered the sign already. If you use -1|√3 you get the same answer

[@kitindah04bush55]

This is so easily explained

[@rianmasoud420]

☺🙏

[@lorri31]

just where did pi over 6 come from?..i dont get it on my calculator..plz help);

[@SummerLangereis]

You just need to know that from the top of your head :p I usually do: cos0= x/r = root of 3 over 2. So 0 is 1/6 pi (which is the same as pi over 6). Because it's in the 4 quadrant it's negative so - pi over 6.

[@rudrakaushik5674]

Hey, is this a mistake or am I confused? I was wondering how you converted 256(cos2/3pi + isin2/3pi) to 256 (-1/2 + isqrt3/2)
should it not be that sqrt -sqrt3/2 is the real part and the 1/2 is the imaginary due to its position in the unit circle.

[@Lighkdk]

yo rudra

[@rudrakaushik5674]

@@Lighkdk wsgud bro 🤣

[@rudrakaushik5674]

@@Lighkdk wassup bro🤣

[@nimrakhan2744]

yhan y ki value to -1 h to apne +1 kuew li h?

[@TWENTYSEVENSTORIES]

Great Question.

[@sureshaadnaik9687]

Hii sir (√3-i)^7 upon (I+1)^10 solve the example

[@deoasutai]

Thank you