Complex Q (1 Viewer)

[krypticlemonjuice]

Can someone please explain this whole question to me please

I don't understand why in the solution they added pi/3 instead of pi/6

 

[tywebb]

A angle of a revolution is 2π so they add 2π/6=π/3

 

[howcanibesmarter]

lol why r all the questions from nsb past papers

 

[Average Boreduser]

lol why r all the questions from nsb past papers

bc they fr that easy. (makes everyone feel like they have less of a skill issue obv)

 

[howcanibesmarter]

bc they fr that easy. (makes everyone feel like they have less of a skill issue obv)

wat

 

[tywebb]

i. The given vertex is 2e^πi/6. Hence the other 5 are z=2e^(2k+1)πi/6, k=±1, ±2, -3.

ii. z⁴=16e^2πki/3, k=0, ±1 giving the vertices of an equilateral triangle circumscribed by a circle with radius 16. Hence Area(S)=(3x16²/2)sin(2π/3)=192√3 (since the area of a regular n-gon circumscribed by a circle of radius r is (nr²/2)sin(2π/n)).

This is more efficient than the attached NSB solution which takes a whole page!.

 

[tywebb]

And if you don't know where (nr²/2)sin(2π/n) comes from, I assume you know about (1/2)absinC, yeah? Just apply it to this:


in the circle like this:

And just as an aside, you can use the area of a regular apeirogon (regular polygon with sides) to prove the area of a circle:

 

[krypticlemonjuice]

i. The given vertex is 2e^πi/6. Hence the other 5 are z=2e^(2k+1)πi/6, k=±1, ±2, -3.

ii. z⁴=16e^2πki/3, k=0, ±1 giving an equilateral triangle circumscribed by a circle with radius 16. Hence Area(S)=(3x16²/2)sin(2π/3)=192√3 (since the area of a regular n-gon circumscribed by a circle of radius r is (nr²/2)sin(2π/n)).

This is more efficient than the attached NSB solution which takes a whole page!.

Hmmm, why did you let n = 3?

 

[tywebb]

Hmmm, why did you let n = 3?

For a triangle, n=3

 

[krypticlemonjuice]

For a triangle, n=3
View attachment 38871

Ahhh, thank you