Complex numbers - polar to cartesian question?

Question

Say one was given the following complex number in cartesian form, (2)-(2isqrt3) » a - bi. And one was told to transform it into polar form » rcis (theta).
Would one take the degree as 150, or -60. taking 150, will give the answer 4cis(5/6pi) and taking 60 will give 4cis(-1/3pi).

So which degree should one take, and why so?

The angle: why doesn't the angle start at 0, and then move from the 1st to 2nd to 3rd then 4th quadrant. why does the angle start at 0 and go the opposite way, from the1st quadrant right to the 4th and cause a negative angle.

Answer 1

2 - i 2√3 lies in 4 th quadrant.
tan θ = - (2√3 / 2) = - √3
θ = - 60° = 300° = 5π / 3 radians
r² = 2² + 12
r² = 16
r = 4
2 - i 2√3 = 4 (cos 5π/3 + i sin 5π/3 )

Answer 2

The sq. root of an imaginary form is yet yet another imaginary form. First be conscious that (a+bi)^2 = a^2 +2abi -b^2 = a^2-b^2 + (2ab)i so as an social gathering, (2+i)^2 = 4 +4i -a million) =3+4i subsequently the sqrt of three+4i is two+i and 2-a million with the aid of shown fact that complicated numbers constantly are attainable in conjugate pairs. So enable's see a thank you to instruct that the sqrt of three+4i is somewhat 2 + i and 2 - i remember (a+bi)^2= a^2 -b^2 + (2ab)i So we would desire to coach that a^2 - b^2 = 3 and 2ab = 4 placed b= 2/.a into first equation getting a^2-4/a^4=3 a^4 -3a^2 -4 =0This is a quadratic in a^2 that would desire to be sparkling up by factoring giving a^2 = 4 or -a million. Reject the -a million as we want a real answer so x^2 = 4 and a = 2 or -2 b=2/a so b= a million or -a million subsequently the sqrt of three+=4i is two+i or 2-i desire this helped.

Answer 3

the degree is equal to: -artan(2sqrt3/2)
im not sure wat that equals but if u hav a calculator to solve that then that is ur angle.

it is because u find theta by solving artan(b/a) , for a-bi
and because 2-2isqrt3 is in the 4th quardrant (bottom right) u need to multiply the answer by negative 1 to get the correct angle. just one of those rules u need to kno, depending on what quadrant the line is in the rule u need to use is different.

Answer 4

If x + iy is turned into r( cos(t) + i sin(t) ), then the real and imaginary parts have to correspond. Hence:
x = r cos(t)
y = r sin(t)
and
r^2 = x^2 + y^2
r is taken to be positive.

Hence:
2 = r cos(t) .........(1)
-2sqrt(3) = r sin(t) .........(2)
Squaring and adding:
r^2(cos^2(t) + sin^2(t)) = 4 + 12 = 16
r^2 = 16
r = 4

Substituting for r in (1) and (2):
2 = 4cos(t)
cos(t) = 1/2

-2sqrt(3) = 4sin(t)
sin(t) = -sqrt(3) / 2.

As both sin(t) and cos(t) are known, their signs dictate that the angle has to be in the 4th quadrant.
Hence -60deg is your answer, and not 150deg.