A complex number will, in general, have _____ fifth complex roots. Help fill in the blank? :o
Whatever the number complex root, that's how many there are. 5 5th complex, 4 4th complex, 3 3rd complex, etc.
Oh okay that's easy! :)
how do you find the fourth root of \[-1+i \sqrt{3}\] ?
I'm not positive, but I think we need it in polar form first. At least I know how to do it from polar form, lol.
You remember how to put it in polar form? xD
NO D: >.< :(
Forgot since last night xD Bleh. Well, we need r(costheta + isintheta). r comes from the same old pythagorean theorem thing. \[r = \sqrt{a ^{2}+b ^{2}}\]. A is your negative 1 and b is your sqrt(3) So we can start off by getting r :3
So it'd be 1+ 3 = 4 ?
Square root of 3 Squared is 3 ?
Right. So you would just have sqrt of 4 which is 2 xD
So r = 2 :P Now to get the angle, we need to say that \[\tan \theta = \frac{ b }{ a }\]
so, \[4\sqrt{2}\].. b is \[i \sqrt{3} ?\]
Whoa, wait xD Not sure where 4rt2 came from o.o a = -1 b = \[\sqrt{3}\] no i included \[r = \sqrt{(-1)^{2}+\sqrt{3}^{2}} = \sqrt{4} = 2\] This is where we are so far xD Now since a = -1 and b = sqrt 3, we need to say: \[\tan \theta =\frac{ \sqrt{3} }{ -1 }\]
ohh, okkay !, which is \[-\sqrt{3} ??\] D:
Lol, yes xD No need to freak out, that one is correct :P So now, tangent is negative in the 2nd and 4th quadrants. So if our original rectangular says \[-1 + i \sqrt{3}\], then which quadrant is that,2 or 4?
2 ?
Yep :p The rectangular form basically says left one and up rt(3) xD So now, in the 2nd quadrant, any idea where tan(theta) = -sqrt(3)?
1 quadrant ?
._o
No, no, no, it's in the 2nd quadrant xD That's what I was just basically asking :P |dw:1376165910322:dw| Because tangent is the division of y/x, the only way it can be negative is if it's in a quadrant where either x or y, but not both, are negative. So on that circel, you can see that quadrant 2 and quadrant 4 are the places where you ciuld have a negative tangent.
If you're still alive xD
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