Expressing complex numbers in the x + iy form

  • #1
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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
 
  • #2
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 [itex] (r e^{i \theta})^n = ... [/itex].
 
  • #3
Don't change the result back to a + bi form until you have finished raising it to the 42 power. Use the formula for [itex] (r e^{i \theta})^n = ... [/itex].
that formula isn't in my book
 
  • #4
that formula isn't in my book

It works just like the formula for [itex] (ab^k)^n = ...[/itex] works for real numbers [itex] a,b,k, [/itex] and integer [itex] n [/itex].
 
  • #5
It works just like the formula for [itex] (ab^k)^n = ...[/itex] works for real numbers [itex] a,b,k, [/itex] and integer [itex] n [/itex].

so then ((sqrt2)e^(-i7pi/4))/sqrt2)^42
= 1/e^147ipi/2 ?? doesn't seem like that's correct
 
  • #6
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 [itex] (ab^k)^n = a^n b^{kn} [/itex] For one thing, you didn't raise [itex] a = \sqrt{2} [/itex] to a power.
 
  • #8
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 !
 

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