Expressing complex numbers in the x + iy form

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Homework Help Overview

The discussion revolves around expressing the complex number \(\left(\frac{1-i}{\sqrt{2}}\right)^{42}\) in the form \(x + iy\). Participants are exploring the application of polar coordinates and de Moivre's theorem in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to simplify the expression using polar coordinates, noting the modulus and argument of the complex number. Some participants suggest using de Moivre's theorem for raising the expression to a power, while others question the application of formulas and the correctness of intermediate steps.

Discussion Status

Participants are actively discussing the correct application of formulas related to complex numbers. There is a mix of interpretations regarding the simplification process, with some guidance offered on not reverting to rectangular form too early. The discussion is ongoing, with various perspectives on the steps taken.

Contextual Notes

Some participants mention that certain formulas are not included in their textbooks, which may affect their understanding and application of the concepts being discussed. There is also a reference to the unit circle and its relevance to the problem.

Logan Land
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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
 
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Logan Land said:
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].
 
Stephen Tashi said:
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
 
Logan Land said:
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].
 
Stephen Tashi said:
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
 
Logan Land said:
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.
 
Logan Land said:
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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