Convert the rectangular coordinate (-8,i (square root of 3) into polar coordinates.
How do I do this?? I tried every way I knew of to convert the y coordinate (i (square root of three)) to polar form, but it didn't work. I think that the polar form of the x coordinate is square root of 61. Thank you!!
Think of the unit circle (or any circle centered at (0,0) for that matter). At any angle θ,
tanθ = y/x
r^2 = x^2+y^2
Complex number a+bi is plotted in the x-y plane, so
r^2 = 8^2+√3^2 = 64+3
r = √67
tanθ = √3/-8
That is in QIV (where x is positive and y is negative), so
θ = -12.22° or 0.213 radians
In polar coordinates, that is, of course,
(√67, -0.213)
To the person who solved this, you forgot the importance of the I and to include it in your answer
To convert the given rectangular coordinate (-8, i √3) into polar coordinates, you can follow these steps:
Step 1: Find the magnitude or radius (r) of the coordinate.
The magnitude or the distance from the origin to the point is given by r = √(x^2 + y^2), where (x,y) represents the rectangular coordinates. In this case, x = -8 and y = i √3.
Substituting the values, we get:
r = √((-8)^2 + (i √3)^2)
= √(64 - 3)
= √61
So, the magnitude or radius (r) is √61.
Step 2: Find the angle (θ) of the coordinate.
The angle θ is given by the equation tanθ = y/x. In this case, x = -8 and y = i √3.
Substituting the values, we get:
tanθ = (i √3) / -8
To simplify this expression:
Multiply both the numerator and denominator by the conjugate of the denominator, which is (8 + i √3):
tanθ = (i √3 * (8 + i √3)) / (-8 * (8 + i √3))
tanθ = (8i √3 - √3) / (-64 + 3i √3)
To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator, which is (-64 - 3i √3):
tanθ = [(8i √3 - √3) * (-64 - 3i √3)] / [(-64)^2 - (3i √3)^2]
tanθ = (-512i √3 + 24 + 192i - 8i^2 * 3) / (4096 + 27 * 3)
tanθ = (-536i √3 + 24 + 192i) / 4125
tanθ = -536i √3 / 4125 + 24 / 4125 + 192i / 4125
Simplifying further:
tanθ = (-16i √3 / 125) + (8 / 1375) + (64i / 275)
So, the angle θ of the coordinate is tanθ = (-16i √3 / 125) + (8 / 1375) + (64i / 275), which can also be denoted as approximately 1.136 + 0.367i.
Therefore, the polar coordinate of (-8, i √3) is (√61, 1.136 + 0.367i).