# Expressing complex numbers in the x + iy form

## 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 dont understand where to go to obtain x+iy form since it is raised to the 42 power and I dont 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 $(r e^{i \theta})^n = ...$.

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 isnt in my book

Stephen Tashi
that formula isnt in my book

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

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 ?? doesnt seem like thats correct

Stephen Tashi
so then ((sqrt2)e^(-i7pi/4))/sqrt2)^42
= 1/e^147ipi/2 ?? doesnt seem like thats 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.

BvU
Homework Helper
so then ((sqrt2)e^(-i7pi/4))/sqrt2)^42
= 1/e^147ipi/2 ?? doesnt seem like thats 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 !