Solve De Moivre's Theorem for z^10

[jwxie]

Homework Statement

Let z = 1 − i. Find z^10.

Homework Equations

The Attempt at a Solution

|z|= \sqrt{(1)^{2}+{-1}^2}
a=|z|cos(\theta ), b=|z|sin(\theta )
and we find that \theta = -pi/4 or 7*pi/4 (Accoridng to this, it is in the fourth quadrant )

[PLAIN]http://dl.dropbox.com/u/14655573/latex/dem.PNG

Why do we add 2*pi at the end? And moreover, if I use 7*pi/4, I notice that the cosine will vanish to 1.2e-12 close to zero, but are we really allowed to ignore that vanish value, and instead write -32i ?

Thanks!

 

[Char. Limit]

Homework Statement

Let z = 1 − i. Find z^10.

Homework Equations

The Attempt at a Solution

|z|= \sqrt{(1)^{2}+{-1}^2}
a=|z|cos(\theta ), b=|z|sin(\theta )
and we find that \theta = -pi/4 or 7*pi/4 (Accoridng to this, it is in the fourth quadrant )

[PLAIN]http://dl.dropbox.com/u/14655573/latex/dem.PNG

Why do we add 2*pi at the end? And moreover, if I use 7*pi/4, I notice that the cosine will vanish to 1.2e-12 close to zero, but are we really allowed to ignore that vanish value, and instead write -32i ?

Thanks!

How are you getting 1.2 * 10^-12? That sounds like a calculator rounding error for cos(7π/4 * 10), which IS equal to 0.

 

[SteamKing]

There is no requirement to add 2pi when expanding z^n according to DeMoivre. Line 3 in the attachment evaluates to the correct result. The small quantity you obtained when evaluating the cosine is due to the approximation your calculator uses to calculate values for this function. Remember, cos(x) = 0 for all values of x = (2k+1)* pi/2 and k is any integer.

 

[jwxie]

Hi Char. Limit.
I guess you are right. That's what I am wondering too... since that would definitely give zero. Sometime I do stuff on the calculator just to simplify things.

and thanks SteamKing. You both are right.
So I guess the whole purpose of adding 2pi is just to simplify the expression. (actually -5pi/2 will be fine).

TI-83 I am using. Interesting, huh?

 

[Ray Vickson]

Homework Statement

Let z = 1 − i. Find z^10.

Homework Equations

The Attempt at a Solution

|z|= \sqrt{(1)^{2}+{-1}^2}
a=|z|cos(\theta ), b=|z|sin(\theta )
and we find that \theta = -pi/4 or 7*pi/4 (Accoridng to this, it is in the fourth quadrant )

[PLAIN]http://dl.dropbox.com/u/14655573/latex/dem.PNG

Why do we add 2*pi at the end? And moreover, if I use 7*pi/4, I notice that the cosine will vanish to 1.2e-12 close to zero, but are we really allowed to ignore that vanish value, and instead write -32i ?

Thanks!

|z| = sqrt(2) and arg(z) = -45 degrees. We have |z^10| = 2^5 = 32 and the angle of z^10 is = 10*arg(z) = -450 degrees. In the standard interval [-180,180] degrees, this angle is -90 degrees, because 450 mod (360) = 90. So arg(z^10) = -90 degrees, hence z^10 = -32i.

RGV

 

[HallsofIvy]

Adding 2\pi, which obviously doesn't change the angle, to -5\pi/2 changes the form to -\pi/2] which is easily seen to be "straight down".

 

[ehild]

There is an easy way (without angles) to get the result of this problem:

(1-i)¹⁰=((1-i)²)⁵,

As i*i=-1,

(1-i)²=-2i,

so (1-i)¹⁰=(-2)⁵ i⁵=-32i

ehild