Engineering Math - Complex Roots and Powers

In summary, the homework equations involve taking x+iy and converting it to polar coordinates. The r comes from the magnitude of the square root of (cos2 + sin2).
  • #1
Tonik
13
0

Homework Statement


Evaluate the following in x+iy form.
i3+i

Homework Equations


i=rei[itex]\Theta[/itex]

The Attempt at a Solution


i=rei[itex]\frac{pi}{2}[/itex]
i3+i = i3*ii
(-i)(ei2[itex]\frac{pi}{2}[/itex])
=-ie-[itex]\frac{pi}{2}[/itex]
=-.2079i

I understand how it all works out except [itex]\frac{pi}{2}[/itex]. I can't figure out how they got [itex]\frac{pi}{2}[/itex] in the first place. Any ideas? Thanks in advance.
 
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  • #2
My editor won't allow greek letters, so in place of theta I'll use T

You know: eiT=cosT + i.sinT

So let T=Pi/2, so that cosT=0, and we have

ei.Pi/2= i

Q: How do you say in words: rei.theta

I'm wondering what the r is part of?
 
  • #3
If [itex]y= i^i[/itex], then [itex]ln(y)= i ln(i)[/itex]. Further, i has modulus r= 1 and argument [itex]\theta= \pi/2[/itex] so that [itex]i= e^{i(\pi/2+ 2k\pi)}[/itex] for k any integer ([itex]e^{2k\pi}= 1[/itex] for all integer k so these are all different ways of writing i). Then [itex]ln(i)= i(\pi/2+ 2k\pi)[/itex] and so [itex]ln(y)= -(\pi/2+ 2k\pi)[/itex].
 
  • #4
NascentOxygen said:
My editor won't allow greek letters, so in place of theta I'll use T

You know: eiT=cosT + i.sinT

So let T=Pi/2, so that cosT=0, and we have

ei.Pi/2= i

Q: How do you say in words: rei.theta

I'm wondering what the r is part of?

The r comes from converting x+iy (rectangular coordinates) to polar coordinates.
x+iy=r(cos[itex]\Theta[/itex]+isin[itex]\Theta[/itex])=rei[itex]\Theta[/itex]
As far as how r is used, I'm unsure. None of the problems I have been given/done have used that r at all.
 
  • #5
NascentOxygen said:
My editor won't allow greek letters, so in place of theta I'll use T

You know: eiT=cosT + i.sinT

So let T=Pi/2, so that cosT=0, and we have

ei.Pi/2= i

Q: How do you say in words: rei.theta

I'm wondering what the r is part of?

HallsofIvy said:
If [itex]y= i^i[/itex], then [itex]ln(y)= i ln(i)[/itex]. Further, i has modulus r= 1 and argument [itex]\theta= \pi/2[/itex] so that [itex]i= e^{i(\pi/2+ 2k\pi)}[/itex] for k any integer ([itex]e^{2k\pi}= 1[/itex] for all integer k so these are all different ways of writing i). Then [itex]ln(i)= i(\pi/2+ 2k\pi)[/itex] and so [itex]ln(y)= -(\pi/2+ 2k\pi)[/itex].

Thank you both for the explanations! You're my heroes.
 
  • #6
Tonik said:
The r comes from converting x+iy (rectangular coordinates) to polar coordinates.
x+iy=r(cos[itex]\Theta[/itex]+isin[itex]\Theta[/itex])=rei[itex]\Theta[/itex]
As far as how r is used, I'm unsure. None of the problems I have been given/done have used that r at all.
I guessed that r was the magnitude, but you confused me by introducing it in (3) for i= and I know i has a magnitude of unity. The square root of (cos2 + sin2) is always 1.

I just wanted to make sure "re" wasn't a combination that I am unfamiliar with.

Well, 5.cosT + 5.i.sinT would equal 5.eiT so that's when it's needed.
 
  • #7
NascentOxygen said:
I guessed that r was the magnitude, but you confused me by introducing it in (3) for i= and I know i has a magnitude of unity. The square root of (cos2 + sin2) is always 1.

I just wanted to make sure "re" wasn't a combination that I am unfamiliar with.

Well, 5.cosT + 5.i.sinT would equal 5.eiT so that's when it's needed.

Yeah, I think you're probably right. Thanks again! :)
 

1. What are complex roots and powers in engineering math?

Complex roots and powers are mathematical concepts that involve imaginary numbers, which are numbers that can be written as a combination of a real number and the imaginary unit, i. In engineering math, these concepts are used to solve equations and analyze systems that involve complex quantities.

2. How are complex roots and powers used in engineering applications?

Complex roots and powers are used in engineering applications to model and analyze systems that involve oscillations, such as electrical circuits, mechanical systems, and signal processing. They are also used in control systems, where the behavior of a system is described by a transfer function with complex coefficients.

3. What is the difference between a complex root and a complex power?

A complex root refers to the solution of an equation that involves complex numbers, while a complex power refers to the result of raising a complex number to a power. Complex roots and powers are related, as the power of a complex number can be written in terms of its roots using De Moivre's theorem.

4. How are complex roots and powers calculated?

To calculate complex roots and powers, the basic rules of complex arithmetic are used, such as the distributive, associative, and commutative properties. To find the roots of a complex number, the quadratic formula or the general formula for finding roots of a polynomial can be used. To raise a complex number to a power, De Moivre's theorem or the polar form of complex numbers can be used.

5. In what engineering fields are complex roots and powers most commonly used?

Complex roots and powers are commonly used in various engineering fields, such as electrical engineering, mechanical engineering, control systems engineering, and signal processing. They are also used in other scientific fields, such as physics and mathematics, to model and analyze systems that involve complex quantities.

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