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Expressing complex numbers in the x + iy form

  • Thread starter Logan Land
  • Start date
  • #1

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
 

Answers and Replies

  • #2
Stephen Tashi
Science Advisor
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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 isnt in my book
 
  • #4
Stephen Tashi
Science Advisor
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that formula isnt 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 ?? doesnt seem like thats correct
 
  • #6
Stephen Tashi
Science Advisor
7,017
1,237
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 [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
BvU
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2019 Award
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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 !
 

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