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If z= (2+7i)
and w = z^3000
whats the angle theta of w in the region [0,2pi)
I want to know how to compute this. I can easily plug it into wolfram but it doesnt explain the steps
 2 years ago
 2 years ago
If z= (2+7i) and w = z^3000 whats the angle theta of w in the region [0,2pi) I want to know how to compute this. I can easily plug it into wolfram but it doesnt explain the steps
 2 years ago
 2 years ago

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tdabboudBest ResponseYou've already chosen the best response.0
would it simply be 3000 * (angle of (2+7i)) ?
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
\[w=(2+j7)^{3000}\]
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
there's two way's to solve this either directly or use de Moivre's theorem, the choice is yours
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
i havent learned that theorem yet so most likely the direct way
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
the direct way is log as you can see the power is 3000
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
yea but we're dealing with complex numbers.. do the same rules apply then
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
first thing you need to do here is change z into polar form
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
that is in the form of \[r ^{n}<n \theta\]
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
@EC, that's not polar form for complex numbers. You want to write \[ z = Ae^{i\theta} \] for some positive number \( A \) which is the magnitude of \( z \), and some angle \( \theta \in [0,2\pi) \).
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
Then \[ z^{3000} = A^{3000} e^{3000\theta i} \]
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
yh, in polar form \[r<\theta\]
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
(@EC, I see. Not standard notation, but I get it.)
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
@ JamesJ is Euler's theorem the only way to solve ?
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
i didnt think it was the only way, there has to be another
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
It's far away the fastest. Unless you use some version of it, you would need to multiply that out explicitly. The whole point of the exercise with such a ridiculous exponent is to make you use some polar form.
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
Do you not have deMoivre's theorem?
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
yh couldn't we just say\[(r ^{n}<n \theta)\]
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
yea i do understand that, the coordinates i gave were correct for eulers and polar: w= 7.28 e^j(1.85 rad)
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
using eulers you would then take the log?
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
\[e ^{j \theta}= \cos(\theta)+jsin (\theta)\]
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
I am still lost... one of my problems we did they simply took the exponent to the front and multiplied it by the angle the problem was (3+j)^4 so they did: 4(2.8198 rad) = 11.27 rad... you can then subtract 2pi to get to the correct domain. b but i tried this and its not working correctly
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
\[(re ^{j \theta})=r ^{n}e ^{jn \theta}=(7.28)^{3000}(\cos(3000\theta)+jsin(3000\theta)\]
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
to convert to radians\[(\theta \times \pi)/180\]
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
that's de Moivre's, \[(r<\theta)^{n}=(r ^{n}<n \theta)\]
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
the angle you gave 1.85 rad is only for (2+7j)^1
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
but when raised to the 3000 the angle becomes .765
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
i know there're 2999 more
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
yeah thats the problem I am having i found the angle you gave me easily, but how can you find that angle when raised to the 3000. would it be 3000 times that angle like we have done previously? then just reduce it to 0,2pi
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
are you looking for a specific angle ?
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
yeah an angle reduced to the domain between 0 and 2pi. or between 0 and 360 degrees.
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
which angle there're 3000 of them ?
 2 years ago

tdabboudBest ResponseYou've already chosen the best response.0
haha the one between 0 and 360 degrees, there will be only one. the other angles have a greater total degree
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
no the angles can't exceed 360 degrees
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
all 3000 of them are within the range of 0 to 360
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
you can use this formula to find one angle between the range of 0 to 360 \[r ^{1/n}<(\theta+360(k))/n\]
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
n= power, k=position of the angle
 2 years ago

EarthCitizenBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=%282%2Bi7%29^3000 these are all the angles according to wolfy
 2 years ago
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