# Homework Help: Express Imaginary number in standard form.

1. Apr 23, 2007

### linuxux

1. The problem statement, all variables and given/known data

Need to see if what im doing is right, here is the problem:

z= 4i/(-1+i), express z^20 in standard form.

so,

z= 4i/(-1+i) = 2-2i

z^20 = 2*sqrt(2) x cis[20(-pi/4)] = 2*sqrt(2) x cis(-5pi)
= 2*sqrt(2) x [cos(-5pi) + i sin(-5pi)]
= 2*sqrt(2) x [-1 + 0i]
= 2*sqrt(2) x [-1] = -2048

so the final answer is only a real number of magnitude -2048+0i. is this correct? thanks.

Last edited: Apr 24, 2007
2. Apr 23, 2007

### Dick

Nope. Try again. cos(-5pi)=-1 and the magnitude of z is 2*sqrt(2). You can do better.

3. Apr 24, 2007

### linuxux

4. Apr 24, 2007

### mezarashi

Okay, how exactly it it get from

2*sqrt(2) x [-1] to this number: -2048

5. Apr 24, 2007

### linuxux

it shouldn't, thats a mistake, I was following an example with a negative exponent thus explaining why i thought it should be raised to the positive 20. but, z^20 does have to be change to z^1, so the equation will then go from this (i think):

z^20 = [2*sqrt(2)]^(20) x cis[(20)(-pi/4)]

to this:
z = [2*sqrt(2)]^(1/20) x cis[(1/20)(-pi/4)]

is that right?

Last edited: Apr 24, 2007
6. Apr 24, 2007

### linuxux

wait, looking at my above post, i see where i have gone wrong, I'll try to correct it...

7. Apr 24, 2007

### linuxux

z = [2*sqrt(2)] x cis[(-pi/4)]
= [2*sqrt(2)] * [cos(-pi/4) + i sin(-pi/4)]
= [2*sqrt(2)] * [sqrt(2)/2 + i -sqrt(2)/2]
= 2 - 2i

so the answer is z^20 = 2 - 2i in standard form. is that right? (doesnt seem right, thats what i started with...)

Last edited: Apr 24, 2007
8. Apr 24, 2007

### mezarashi

I think you're confusing yourself and confusing me as well. Are you trying to find z^(20) or z^(-20)?

I was previously only highlighting that $$(2 \sqrt{2})^{20}$$ is NOT equal to 2048.

Here's a good old wiki page to refresh your memory:
http://en.wikipedia.org/wiki/De_Moivre's_formula#Applications

Last edited: Apr 24, 2007
9. Apr 24, 2007

### linuxux

Yes, now i am confused, but the question is right:

z= 4i/(-1+i), express z^20 in standard form.

lets try again,

we are raising everything in z to the power of 20,

z^20 = [2*sqrt(2)]^20 x cis[20(-pi/4)] = [2*sqrt(2)]^20 x cis(-5pi)

(note how large the number [2*sqrt(2)]^20, there is no reasonable way to write it other than how i have just now, this seem odd...)

now,

[2*sqrt(2)]^20 x cis(-5pi) = [2*sqrt(2)]^20 x [cos(-5pi)+i sin(-5pi)]

is this correct?...

next,

[2*sqrt(2)]^20 x [cos(-5pi)+i sin(-5pi)] = [2*sqrt(2)]^20 x [1 + 0i]

is this correct?...

Last edited: Apr 24, 2007
10. Apr 24, 2007

### mezarashi

I agree with you up to:

z^20 = [2*sqrt(2)]^20 x [cos(-5pi)+i sin(-5pi)]

From here, it should be easy to see that cos(-5pi) = -1 and sin(-5pi) = 0. Problem solved. Note that [2*sqrt(2)]^20 = 2^30. Remember that one integer in our computer uses 4 bytes, that's 2^32.

11. Apr 24, 2007

### linuxux

thanks.

12. Apr 24, 2007

### HallsofIvy

No, this should be (2*sqrt(2))^20 cis(-5pi)
[/quote] = 2*sqrt(2) x [cos(-5pi) + i sin(-5pi)]
= 2*sqrt(2) x [-1 + 0i]
= 2*sqrt(2) x [-1] = -2048