Question

u = −1 + 4 sqrt(3)
i
Find square root of u using polar form of u

Answer 1

um... im not in collage so i cant help u sorry

Answer 2

@Photon336 @jhonyy9 @vishweshshrimali5 @ganeshie8 @Directrix @Isaiah.Feynman @IrishBoy123 @inkyvoyd @aaronq @Astrophysics @kropot72 @ParthKohli

Answer 3

\(z = -1 + 4 \sqrt{3} ~ i = 7 (-\dfrac{1}{7} + \dfrac{4 \sqrt{3}}{7} ~ i)\)

\( = 7 e ^{ i ( \theta + 2 n \pi )}, \qquad \tan \theta = -4 \sqrt{3}\)

\(z^{1/2} = \sqrt{7} e ^{ i (\frac{\theta}{2} + n \pi )} = \sqrt{7} e ^{ i \frac{\theta}{2} } e^{ i ~ n \pi }\)

\( = \sqrt{7}e ^{ i \frac{\theta}{2} } (-1)^n = \sqrt{7} ~ cis ( \frac{\theta}{2} ) ~ (-1)^n\)

using the double angle formulae, i get \(\cos \frac{\theta}{2} = \sqrt{\dfrac{3}{7}}\) and \(\sin \frac{\theta}{2} = \dfrac{2}{\sqrt{7}}\)

so

\((-1)^n (\ \sqrt{3} + 2 ~ i )\)

or \(\pm (\ \sqrt{3} +2 ~ i )\)

Answer 4

Can you clear my confusion?
The polar form is \[\large\rm 7e^{-1.427i} \]where the theta is calculated by \[\large\rm \theta =tan^{-1}( -4\sqrt{3}) \]
But when I substitute the value of theta in 7cos(theta) to give me -1 it gives me 1. Why is that happening? (Why the polar form is giving a different value when changed into angle form?)

Answer 5

And I am getting Rcos (theta/2) = 2 not square root 3