anonymous
  • anonymous
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
Mathematics
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anonymous
  • anonymous
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
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
would it simply be 3000 * (angle of (-2+7i)) ?
EarthCitizen
  • EarthCitizen
nope
EarthCitizen
  • EarthCitizen
\[w=(-2+j7)^{3000}\]

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EarthCitizen
  • EarthCitizen
there's two way's to solve this either directly or use de Moivre's theorem, the choice is yours
anonymous
  • anonymous
i havent learned that theorem yet so most likely the direct way
EarthCitizen
  • EarthCitizen
the direct way is log as you can see the power is 3000
anonymous
  • anonymous
yea but we're dealing with complex numbers.. do the same rules apply then
EarthCitizen
  • EarthCitizen
first thing you need to do here is change z into polar form
EarthCitizen
  • EarthCitizen
that is in the form of \[r ^{n}
anonymous
  • anonymous
ok then...(7.28,1.85)
JamesJ
  • JamesJ
@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 \( \theta \in [0,2\pi) \).
JamesJ
  • JamesJ
Then \[ z^{3000} = A^{3000} e^{3000\theta i} \]
EarthCitizen
  • EarthCitizen
yh, in polar form \[r<\theta\]
JamesJ
  • JamesJ
(@EC, I see. Not standard notation, but I get it.)
EarthCitizen
  • EarthCitizen
@ JamesJ is Euler's theorem the only way to solve ?
anonymous
  • anonymous
i didnt think it was the only way, there has to be another
JamesJ
  • JamesJ
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.
JamesJ
  • JamesJ
Do you not have deMoivre's theorem?
EarthCitizen
  • EarthCitizen
yh couldn't we just say\[(r ^{n}
anonymous
  • anonymous
yea i do understand that, the coordinates i gave were correct for eulers and polar: w= 7.28 e^j(1.85 rad)
anonymous
  • anonymous
using eulers you would then take the log?
EarthCitizen
  • EarthCitizen
\[e ^{j \theta}= \cos(\theta)+jsin (\theta)\]
anonymous
  • anonymous
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
EarthCitizen
  • EarthCitizen
\[(re ^{j \theta})=r ^{n}e ^{jn \theta}=(7.28)^{3000}(\cos(3000\theta)+jsin(3000\theta)\]
EarthCitizen
  • EarthCitizen
to convert to radians\[(\theta \times \pi)/180\]
EarthCitizen
  • EarthCitizen
that's de Moivre's, \[(r<\theta)^{n}=(r ^{n}
EarthCitizen
  • EarthCitizen
\[\theta=105.95\]
EarthCitizen
  • EarthCitizen
or 1.849 rads
anonymous
  • anonymous
the angle you gave 1.85 rad is only for (-2+7j)^1
anonymous
  • anonymous
but when raised to the 3000 the angle becomes -.765
EarthCitizen
  • EarthCitizen
i know there're 2999 more
anonymous
  • anonymous
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
EarthCitizen
  • EarthCitizen
are you looking for a specific angle ?
anonymous
  • anonymous
yeah an angle reduced to the domain between 0 and 2pi. or between 0 and 360 degrees.
EarthCitizen
  • EarthCitizen
which angle there're 3000 of them ?
anonymous
  • anonymous
haha the one between 0 and 360 degrees, there will be only one. the other angles have a greater total degree
EarthCitizen
  • EarthCitizen
no the angles can't exceed 360 degrees
EarthCitizen
  • EarthCitizen
all 3000 of them are within the range of 0 to 360
EarthCitizen
  • EarthCitizen
you can use this formula to find one angle between the range of 0 to 360 \[r ^{1/n}<(\theta+360(k))/n\]
EarthCitizen
  • EarthCitizen
n= power, k=position of the angle
EarthCitizen
  • EarthCitizen
http://www.wolframalpha.com/input/?i=%28-2%2Bi7%29^3000 these are all the angles according to wolfy

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