Question

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

Answer 1

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

Answer 2

nope

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

ok then...(7.28,1.85)

Answer 11

@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 \( \theta \in [0,2\pi) \).

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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.

Answer 18

Do you not have deMoivre's theorem?

Answer 19

yh couldn't we just say\[(r ^{n}<n \theta)\]

Answer 20

yea i do understand that, the coordinates i gave were correct for eulers and polar:

w= 7.28 e^j(1.85 rad)

Answer 21

using eulers you would then take the log?

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

\[\theta=105.95\]

Answer 28

or 1.849 rads

Answer 29

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

Answer 30

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

Answer 31

i know there're 2999 more

Answer 32

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

Answer 33

are you looking for a specific angle ?

Answer 34

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

Answer 35

which angle there're 3000 of them ?

Answer 36

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

Answer 37

no the angles can't exceed 360 degrees

Answer 38

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

Answer 39

you can use this formula to find one angle between the range of 0 to 360
\[r ^{1/n}<(\theta+360(k))/n\]

Answer 40

n= power, k=position of the angle

Answer 41

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