Expressing complex numbers in the x + iy form

In summary, the conversation discusses finding the expression (1-i)^42 in x+iy form. The formula z1/z1=(r1/r2)e^(i(theta1-theta2)) is mentioned and the attempt at a solution involves finding the value of (sqrt2e^(-i7pi/4)/sqrt2)^42. It is suggested to use the formula (r e^{i \theta})^n = ... to solve this problem. De Moivre's formula is also referenced and it is ultimately determined that the expression is equal to 1/e^147ipi/2.
  • #1
Logan Land
84
0

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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  • #2
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].
 
  • #3
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
 
  • #4
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].
 
  • #5
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
 
  • #6
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.
 
  • #8
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 !
 

What is the x + iy form of a complex number?

The x + iy form of a complex number is also known as the rectangular form. In this form, the real part of the complex number is represented by the x value, and the imaginary part is represented by the iy value.

Why is it important to express complex numbers in the x + iy form?

Expressing complex numbers in the x + iy form allows us to easily perform calculations such as addition, subtraction, multiplication, and division. It also helps in understanding the geometric interpretation of complex numbers on the complex plane.

How do you convert a complex number from the x + iy form to the polar form?

To convert a complex number from the x + iy form to the polar form, we can use the formulas r = √(x^2 + y^2) and θ = tan^-1 (y/x). The magnitude of the complex number is represented by r, and the angle with the positive real axis is represented by θ.

What is the conjugate of a complex number in the x + iy form?

The conjugate of a complex number in the x + iy form is the complex number with the same real part, but with a negative imaginary part. For example, the conjugate of 3 + 2i is 3 - 2i.

How do you add or subtract complex numbers in the x + iy form?

To add or subtract complex numbers in the x + iy form, we simply add or subtract the real and imaginary parts separately. For example, (2 + 3i) + (4 + 5i) = (2 + 4) + (3 + 5)i = 6 + 8i.

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