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

- anonymous

- katieb

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- anonymous

would it simply be 3000 * (angle of (-2+7i)) ?

- EarthCitizen

nope

- EarthCitizen

\[w=(-2+j7)^{3000}\]

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## More answers

- EarthCitizen

there's two way's to solve this either directly or use de Moivre's theorem, the choice is yours

- anonymous

i havent learned that theorem yet so most likely the direct way

- EarthCitizen

the direct way is log as you can see the power is 3000

- anonymous

yea but we're dealing with complex numbers.. do the same rules apply then

- EarthCitizen

first thing you need to do here is change z into polar form

- EarthCitizen

that is in the form of \[r ^{n}

- anonymous

ok then...(7.28,1.85)

- 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

Then \[ z^{3000} = A^{3000} e^{3000\theta i} \]

- EarthCitizen

yh, in polar form \[r<\theta\]

- JamesJ

(@EC, I see. Not standard notation, but I get it.)

- EarthCitizen

@ JamesJ is Euler's theorem the only way to solve ?

- anonymous

i didnt think it was the only way, there has to be another

- 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

Do you not have deMoivre's theorem?

- EarthCitizen

yh couldn't we just say\[(r ^{n}

- 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

using eulers you would then take the log?

- EarthCitizen

\[e ^{j \theta}= \cos(\theta)+jsin (\theta)\]

- 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

\[(re ^{j \theta})=r ^{n}e ^{jn \theta}=(7.28)^{3000}(\cos(3000\theta)+jsin(3000\theta)\]

- EarthCitizen

to convert to radians\[(\theta \times \pi)/180\]

- EarthCitizen

that's de Moivre's,
\[(r<\theta)^{n}=(r ^{n}

- EarthCitizen

\[\theta=105.95\]

- EarthCitizen

or 1.849 rads

- anonymous

the angle you gave 1.85 rad is only for (-2+7j)^1

- anonymous

but when raised to the 3000 the angle becomes -.765

- EarthCitizen

i know there're 2999 more

- 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

are you looking for a specific angle ?

- anonymous

yeah an angle reduced to the domain between 0 and 2pi. or between 0 and 360 degrees.

- EarthCitizen

which angle there're 3000 of them ?

- anonymous

haha the one between 0 and 360 degrees, there will be only one. the other angles have a greater total degree

- EarthCitizen

no the angles can't exceed 360 degrees

- EarthCitizen

all 3000 of them are within the range of 0 to 360

- 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

n= power, k=position of the angle

- EarthCitizen

http://www.wolframalpha.com/input/?i=%28-2%2Bi7%29^3000 these are all the angles according to wolfy

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