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# Finding the fourth root?

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Find the four  fourth roots of -3 + 4i and express the roots in polar coordinates.

chilledz3non  May 29, 2014

#1
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If we have a general point (x, y) in Cartesian coordinates, the polar form is (R, Θ) where R = √(x2 + y2) and

Θ = tan-1(y/x).

And if we are in the complex plane, the point z = x + iy is z = Re in polar coordinates.

Your point is z = -3 + 4i  so, in polar form it is z = Re where R = 5 and Θ = tan-1(4/-3) (note that this is an angle in the 2nd quadrant, because the x-value is negative while the y-value is positive). Because you can add any multiple of 2pi (360°) to Θ without changing the value of e, we can also write z = rei(Θ+2pi*k) where k is an integer.

To find the fourth root of a complex number in polar form we simply take the fourth root of r and divide the angle by 4.  That is: z1/4 = r1/4ei(θ+2pi*k)/4.  So  z1/4 = 51/4ei(tan-1(4/-3)+2pi*k).

We can let k = 1, 2, 3 and 4 to get the four roots.

We can write this as z1/4 = re,  where r = 51/4 and θ = tan-1(4/-3) + 2pi*k.

The result of doing this is summarised in the image below:

Alan  May 29, 2014
#1
+27229
+5

If we have a general point (x, y) in Cartesian coordinates, the polar form is (R, Θ) where R = √(x2 + y2) and

Θ = tan-1(y/x).

And if we are in the complex plane, the point z = x + iy is z = Re in polar coordinates.

Your point is z = -3 + 4i  so, in polar form it is z = Re where R = 5 and Θ = tan-1(4/-3) (note that this is an angle in the 2nd quadrant, because the x-value is negative while the y-value is positive). Because you can add any multiple of 2pi (360°) to Θ without changing the value of e, we can also write z = rei(Θ+2pi*k) where k is an integer.

To find the fourth root of a complex number in polar form we simply take the fourth root of r and divide the angle by 4.  That is: z1/4 = r1/4ei(θ+2pi*k)/4.  So  z1/4 = 51/4ei(tan-1(4/-3)+2pi*k).

We can let k = 1, 2, 3 and 4 to get the four roots.

We can write this as z1/4 = re,  where r = 51/4 and θ = tan-1(4/-3) + 2pi*k.

The result of doing this is summarised in the image below:

Alan  May 29, 2014
#2
+564
0

Thank you Alan!

chilledz3non  May 29, 2014