Expressing complex numbers in the x + iy form

[Logan Land]

Homework Statement

((1-i)/(sqrt2))^42
express in x+iy form

Homework Equations

z1/z1=(r1/r2)e^(i(theta1-theta2))

The Attempt at a Solution

Ive found that (1-i) has r=sqrt2 so since r is sqrt2 and x=1 y=-1 so the angle is 7pi/4
so then I have (sqrt2e^(-i7pi/4)/sqrt2)^42
now from here is where I don't understand where to go to obtain x+iy form since it is raised to the 42 power and I don't have a e on the bottom

 

[Stephen Tashi]

so then I have (sqrt2e^(-i7pi/4)/sqrt2)^42

Don't change the result back to a + bi form until you have finished raising it to the 42 power. Use the formula for (r e^{i \theta})^n = ....

 

[Logan Land]

Don't change the result back to a + bi form until you have finished raising it to the 42 power. Use the formula for (r e^{i \theta})^n = ....

that formula isn't in my book

 

[Stephen Tashi]

that formula isn't in my book

It works just like the formula for (ab^k)^n = ... works for real numbers a,b,k, and integer n.

 

[Logan Land]

It works just like the formula for (ab^k)^n = ... works for real numbers a,b,k, and integer n.

so then ((sqrt2)e^(-i7pi/4))/sqrt2)^42
= 1/e^147ipi/2 ?? doesn't seem like that's correct

 

[Stephen Tashi]

so then ((sqrt2)e^(-i7pi/4))/sqrt2)^42
= 1/e^147ipi/2 ?? doesn't seem like that's correct

You didn't make a correct application of (ab^k)^n = a^n b^{kn} For one thing, you didn't raise a = \sqrt{2} to a power.

 

[Mark44]

Is de Moivre's Formula in your book? http://en.wikipedia.org/wiki/De_Moivre's_formula

$e^{inx} = cos(nx) + i\ sin(nx)$

 

[BvU]

so then ((sqrt2)e^(-i7pi/4))/sqrt2)^42
= 1/e^147ipi/2 ?? doesn't seem like that's correct

Maybe it doesn't seem correct to you, but you really did it right according to me !

(I was wondering why you didn't let the factors sqrt(2) cancel straightaway. Matter of obfuscating notation ?)
$${1-i\over \sqrt 2} = e^{{7\over 4}\pi}$$
Draw a unit circle, consider the x-axis the real axis and the y-axis the imaginary axis, and mark your ${1-i\over \sqrt 2}$

You should see the ${7\over 4}\pi$ (with a plus sign ! *) , the de Moivre thing, and lots more goodies. Play with squaring the number, etc.*) check where $e^{-{7\over 4}\pi}$ is located on that unit circle !