Expert: Scott A Wilson - 5/11/2009

Question
uhm i think i was not cleat but i know all about the Pythagorean thing
and the 0-90(x,y) / 90-180 (-x,y) / 180-270 (-x,-y) / 270/360 (x,-y)

thats middle school stuff but i meant that i wanted you to rexplain how you use them to get the square of complex number i mean ok i know that tangent is the opposite/adjacent
also i know about the radian and degree conversion still i dont understand how to use them for complex number root


Answer
To compute the root of a complex number, it is easy to do in polar coordinates.

In x-y coordinates, the x axis measures the distance horizontally and the y axis measures the direction vertically.

In polar coordinates, r measures how far from the origin and Θ measures the angle upwards from the positive x direction.

To do a simple problem, lets square 1+i.  
The answer would be (1+i)(1+i) = 1 + i + i + i² = 1 + 2i - 1 = 2i.

That's not that hard, but 1+i in polar coordinates...
Well, the graph would put it at x=1, y=1
(since 1 in the y direction is i).  
This makes r² = x²+y² and tan(Θ) = y/x.
From here, r = √(1+1) = √2 and tan(Θ) = 1/1 = 1.
That makes Θ be 45°.

To square a number in polar coordinates, if the number is (r, Θ),
is (r², 2Θ).  So the number would be (√2², 2*45°) = (2, 90°).
Since 90° is straight up, and y is for i, that is 2i.
This is just so the fact it works can be seen.

Now (1+i) seems easy to square, and maybe easy to cube,
but what about (1+i)^8?  Not so hard in polard coordinates.
The number is 1 + i = (√2, 45°).

Taking it to the 8th gives √2^8 = 2^4 = 16 for r.
For the new Θ, it is 8*45° = 360°.  Note that 360° is the same as 0°.
Using this, we can say that (1+i)^8 = (16, 0°),
which is merely 16.

Using this method, it can be seen that there are 3 cube roots,
4 4th roots, 5 5th roots, 6 6th roots, etc.

What are the 5 roots of 32?  (It's not just 2).
In polar coordinates, 32 is (32, 0°) = (32, 360°) = (32, 720°) =
(32, 1080°) = (32, 1440°).

Now to take the 5th root, find the 5th root of 32 and divide all the angles by 5.  It can be seen that the 5th root of 32 is 2.
The angles turn out to be 0°, 72°, 144°, 216°, and 288°.

Or what about the fourth root of 28,561?
Well, 13*13 = 169, 13 * 169 = 2,197, and 13 * 2,197 = 28,561.
This known, it would be easy to answer 13.
However, there are 3 other roots - they are 13i, -13, and - 13i.
Take any of these to the 4th and 2,197 is gotten.
Maybe you can see that.

Well, what about the 5th root of 371,293.  That's also 13, you know.
Yet now I say, there are 4 other roots.  That's where polar coordinates come up again.  Put the number in polar coordinates and compute the 5 roots by adding 360°, 2*360°, 3*360°, and 4*360°.

If each of the angles is divide by 5, we can see that the angles are 0, 1, 2, 3, and 4, each times 72°.

So they would be
(13,0°), (13,72°), (13,144°), (13,216°), and (13,288°).

To take polar numbers to the 5th, take r to the 5th and multiply the angles by 5.  That gives us (371,293, 0°) for each of them, since eahc of them would give a multiple of 360°, which is a full circle.

Now just try and compute those 5 fifth roots in (x,y) coordinates.
I don't know how to do it.

Basically, in polar coordinates on the x=real, y=imaginary scale,
almost any power can be computed.

For example, take (3 + (√720)i)^(4/3).
Well, you just need to compue (3^3 + 3*3^2*√720i + ... fooey!
Convert it to polar first.
In polar, r = √(3² + 720) = √729 = 27.
tan(Θ) = √720 / 3.
Now to take it to the 4/3 power, the cube root would be
r=27^(1/3) = 3 and Θ = arctan(√720/3).
To the fourth, we would then get r^4 = 81 and 4Θ = 4arctan(√720/3).
Now the angle is a bit messy, but the number was not that hard to find.  Using Excel, Θ = 8.94° and 4Θ = 35.78°.  Note that Θ was kept in extended form before I multiplied by 4, so it looks a little off.  This is because Θ is really 8.94427191...

Note that this power stuff makes sense in polar coordinates.
Just think of a negative number.  Its really that positive of that number for r and 180° for Θ.  Square it and you get r² and Θ = 360°,
which is the same as (r², 0°).  You've probably known for a long time that squaring a negative left you with a positive, but now you know why in polar version.

Do you understand?

Even though I have dealt with numbers in polar, it seems hot to me!