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Schrodinger
 4 years ago
This involves De Moivre's Theorem, heads up. So, given a number that you're supposed to find all of the complex roots for, in this case, the cube roots of 1000, i'm failing to understand where an instructional i'm on suddenly gets this:
"The number 1000, or 1000 + 0i, is a complex number written in rectangular form. The polar notation for a complex number a + b*i is r(cos[theta] + i*sin[theta]), where r is the absolute value of the complex number and theta is the angle in standard position whose terminal side passes through the point (a.b).
(cont'd below)
Schrodinger
 4 years ago
This involves De Moivre's Theorem, heads up. So, given a number that you're supposed to find all of the complex roots for, in this case, the cube roots of 1000, i'm failing to understand where an instructional i'm on suddenly gets this: "The number 1000, or 1000 + 0i, is a complex number written in rectangular form. The polar notation for a complex number a + b*i is r(cos[theta] + i*sin[theta]), where r is the absolute value of the complex number and theta is the angle in standard position whose terminal side passes through the point (a.b). (cont'd below)

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Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0"...The complex number 1000, or 1000 + 0i, is shown below in polar form. 1000(cos [0 degrees] + i sin [0 degrees])." How did they get to theta being zero, all of a sudden? I just don't understand how they got the angle, at all. I'm assuming it has to do with 0i?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first off to find the cube roots of 1000 is identical to finding the cube roots of 1 and then multiplying the result by 10

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not concerned with that at the moment, mainly just the angle part.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the angle is 0 because in the complex plane the real numbers live on the \(x\) axis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352818446497:dw

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know what that means, I just don't understand the significance of that statement at all. xP

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0(I'm selfteaching, so i'm not totally aware of everything that's being talked about.)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok you want the angle right? and the way you make an angle for unit circle trig is you start here dw:1352818720308:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0meaning that in the complex plane, all positive real numbers make an angle of 0 and all negative real numbers have an angle of \(\pi\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is also of course true that \(\cos(0)=1\) and \(\sin(0)=0\) so for any positive number say \(1000\) you can write \[1000=1000\left(\cos(0)+i\sin(0)\right)\] in a way it is a rather silly thing to write, except that you are about to use it to find cube roots

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0I'll have to look up whatever the "complex plane" is, but that can't be true, in the way that i'm typically thinking of it. And positive real numbers of what? How do numbers by themselves simply form an angle? I can imagine a number being represented as a vector just forming a line, but numbers themselves don't generate an angle, unless that's basically what you mean when you say all positive real numbers produce an angle of zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0also notice that the representation is not unique, since it is also true that \[\cos(2\pi)=1\] and \[\sin(2\pi)=0\] meaning \[1000=1000\left(\cos(2\pi)+i\sin(2\pi)\right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the real line lives inside the complex plane

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0I don't quite get it yet, but that's just me. Thanks for your help so far.

phi
 4 years ago
Best ResponseYou've already chosen the best response.0Here is a little background http://www.khanacademy.org/math/algebra/complexnumbers/v/complexnumberspart1

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks. Sorry, I would reply promptly but for some reason, Openstudy causes my computer to start randomly tripping balls as hard as possible.

phi
 4 years ago
Best ResponseYou've already chosen the best response.0This little app shows the complex plane http://www.khanacademy.org/math/algebra/complexnumbers/e/the_complex_plane notice that you can use polar coordinates (r (distance) and theta (angle from the xaxis) to define a point on the plane, rather than x+ iy (rectangular form) the distance r (use pythagoras) is sqrt( x^2+ y^2) the angle is the arc tangent of (y/x)

Schrodinger
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, this helps a lot. I thought that was potentially how coordinates would be plotted in the imaginary plane, but I didn't want to assume. Thanks!
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