Question

Compute the modulus and argument of each complex number:
A. sqrt10(cos(5pi/7) + i sin(5pi/7))

Answer 1

Does this follow De Moivre's formula?

Answer 2

yes

Answer 3

De Moivre's Formula states:\((\cos(x) +i\sin(x))^n = \cos(nx)+i\sin(nx)\)

Answer 4

And is your function: \(\sqrt{10((cos(5pi/7) + i sin(5pi/7))}\) ?

Answer 5

yes it is

Answer 6

So first multiply 10 to both cosine and sine.\[\left[\left(10\cos\left(\frac{5\pi}{7}\right)\right) +\left(10i\sin\left(\frac{5\pi}{7}\right)\right)\right]^{1/2}\] Now multiply \(\dfrac{1}{2}\) to the angle measure for both angle measurements. \[10\cos\left(\frac{1}{2}\cdot \frac{5\pi}{7}\right) + 10i\sin\left(\frac{1}{2}\cdot \frac{5\pi}{7}\right)\]

Answer 7

ok

Answer 8

I feel like the power should be multiplied to the 10 as well but I am not sure.

Answer 9

I think it has to be since you're multiplying the whole thing by \[\frac{ 1 }{ 2 }\]

Answer 10

Well the modulus is just a fancy way of saying the length of a complex number from the origin to that point, which really just means find the hypotenuse of a triangle with the real and imaginary parts the legs. The argument is just a fancy way of saying the angle you rotated.

I suggest you don't distribute the 10, instead separate it. Multiplying by a number will only ever scale the length, it can't turn it! It's the real and imaginary parts with cosine and sine that will change the direction. However now in complex numbers, square roots, cube roots, and actually all exponents will rotate the number if it's on a complex number.

I can give you some examples to show you if you like. =)

Answer 11

Please!!!

Answer 12

would you like help ?

Answer 13

demoiver's theorem is valid for rational exponents as well.
[r ( cos t + i sin t )] ^(1/n) = r^(1/n) * ( cos ( t / n + i sin (t / n )