I'm trying to show the cube roots of i Apparently there are 3, which I have gotten, (3)^(1/2)/2 + i/2 for 0pi -(3)^(1/2)/2 + i/2 for 2pi 0 - i for 4pi when I try to take 6pi though I get in this case r = 1 so, z = e^((pi/2 + 6pi)/3)i = e^(13pi/6)i apparently it is suppose to repeat for 6pi but that doesn't seem the case, this is driving me crazy can anyone show me where I'm making my mistake. Apparently, z^(1/3) = r^(1/3) * e^(i*(a/3 + 2*pi)) = r^(1/3) * e^(i*a/3) where a = pi/2 This algebra just seems wrong http://www.wolframalpha.com/input/?i=i^%281%2F3%2
@nincompoop
\[e^\frac{\frac{\pi}{2}+ 6\pi}{3}\]
omg I am dizzy already just looking at it
ya it's wrong… wrong link of wolfram alpha
so, \[e^{\frac{\frac{\pi}{2} + \frac{12\pi}{2}}{\frac{3}{1}}}\]
\[e^{\frac{13\pi}{2(3)}}\]
why would there be cube roots of i? i is what?
|dw:1368949070126:dw|
cube of i = -√-1
i^(1/3) or \[0 + \sqrt[3]{i}\]
cube of i is -i
I'm looking for the cube root no the cube
not*
can you see a problem in my algebra?
13pi/6 is not one of the roots of the cube root of i
im going to just go to sleep and wait for someone to come on that knows this stuff :\ thanks for trying to help I appreciate it
can you rewrite your expression using the equation toolbar or tex commands ?
\[e^\frac{\frac{\pi}{2} + 6\pi}{3}\] = \[e^\frac{\frac{\pi}{2} + \frac{12\pi}{2}}{3}\] = \[e^{\frac{\pi}{2(3)} + \frac{12\pi}{2(3)}}\] = e^(13pi/6)
which was confirmed by wolfram maybe I'm doing something wrong but I arrived at the other roots just fine for 0pi, 2pi and 4pi
ya I think there's something wrong … so your e has the exponent (2pi/2+6pi)/3?
stupid equation is so small in my screen, I couldn't tell…
pi/2 not 2pi/2, but otherwise yes
You must be making a mistake... the fact the 13pi/6 is greater than 2pi is a pretty big clue (the angle should be less than 2pi) this may help: http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Roots.aspx I'm too tired to check your work now but I might come back to this tomorrow. The angle of i is pi/2, so your exponents should have angles \[\Large \frac{ 1 }{ 3 } \times \frac{ \pi }{ 2 }\], \[\Large \frac{ 1 }{ 3 } \times \frac{ \pi }{ 2 } + \frac{ 2 \pi }{ 3 }\], \[\Large \frac{ 1 }{ 3 } \times \frac{ \pi }{ 2 } + \frac{ 4 \pi }{ 3 }\],
Since you add 2*k*pi/n, where k =0, 1, 2... and n is the 3 from the cube root.
Wait.. why are you trying to find four roots? "Apparently there are 3, which I have gotten, (3)^(1/2)/2 + i/2 for 0pi -(3)^(1/2)/2 + i/2 for 2pi 0 - i for 4pi when I try to take 6pi though I get" That's your mistake right there... there is no 6pi, you found all three roots.
oh so you only go up to 4pi? I thought you continued until it repeated
that doesn't make sense
No... you continue until k = n. You can't find four roots of a cube root...
thank you for explaining that lol, makes sense now ugh ha
It makes perfect sense. A square root has two roots. A cube root has three roots. A fourth root has four roots.
yeah I dont know what I was thinking
Thank you soooo much
haha no prob :)
No... you continue until k = n. That should say until k=n-1, since k starts from zero.
right :) thanks for being clear
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