Very simple complex powers problem

  • Thread starter Aziza
  • Start date
  • Tags
    Complex
In summary, the complex number (-1)^i, when written in x+iy form, can have multiple values depending on the value of k. However, the book only provides the principal value when k=0, but your solution is more complete and takes into account all possible values. This is due to the multivalued nature of exponentiation to a complex power.
  • #1
Aziza
190
1
Write the following in x+iy form: (-1)^i

my solution:
(-1)i = (ei(π+2πk))i = e-(π+2πk)

However, my book just states the answer as e-(π) ...but these are not the same...
 
Physics news on Phys.org
  • #2
For integer [itex]k[/itex], sure they are. Any full turn around the unit circle doesn't change the value.
 
  • #3
Muphrid said:
For integer [itex]k[/itex], sure they are. Any full turn around the unit circle doesn't change the value.

but ei2πk and e2πk are different...the first one equals 1 but the second does not...

i mean look just plug in k=1 and k=2...you get e^-3 and e^-5, which are not the same
 
  • #4
Oh, okay, I see what's going on. Yeah, in this case, you may be more correct than the book is. Someone with more recent experience than I in complex analysis may know something more specific to this, though.
 
  • #5
Muphrid said:
Oh, okay, I see what's going on. Yeah, in this case, you may be more correct than the book is. Someone with more recent experience than I in complex analysis may know something more specific to this, though.

oh alright. idk my book is doing this same mistake for each and every related problem. so i was thinking that i must be doing something wrong. but maybe they are just giving the answer when k=0...although they never specify this...
 
  • #6
Aziza said:
Write the following in x+iy form: (-1)^i

my solution:
(-1)i = (ei(π+2πk))i = e-(π+2πk)

However, my book just states the answer as e-(π) ...but these are not the same...

Your answer is more complete than the book's. Exponentiation to a complex power can give a multivalued result. The book just gave the principal value, which, if you're asked for a single answer, is the usual value we take.
 
  • #7
Aziza said:
Write the following in x+iy form: (-1)^i

my solution:
(-1)i = (ei(π+2πk))i = e-(π+2πk)

However, my book just states the answer as e-(π) ...but these are not the same...
(Those π's are hard to read !)

Sure enough.

Give WolframAlpha, [itex]\displaystyle \left(e^{\pi} \right)^{1/i}[/itex] or [itex]\displaystyle \left(e^{-3\pi} \right)^{1/i}\,,[/itex] and it gives -1.

Of course, [itex]\displaystyle e^{-3\pi}\approx 0.0000806995175703\,,[/itex] [itex]\displaystyle e^{-\pi}\approx 0.04321391826377226\,,[/itex] [itex]\displaystyle e^{\pi}\approx 23.140692632779263\,,[/itex]

So it looks like [itex]\displaystyle (-1)^i=e^{(\text{odd integer multiple of }\pi)}\ .[/itex]
 

FAQ: Very simple complex powers problem

What is a "Very simple complex powers problem"?

A "Very simple complex powers problem" is a mathematical problem that involves raising a complex number to a certain power. Complex numbers are numbers that have both a real and imaginary component, and they are represented in the form a + bi, where a and b are real numbers and i is the imaginary unit. Raising a complex number to a power involves multiplying it by itself a certain number of times.

How do you solve a "Very simple complex powers problem"?

To solve a "Very simple complex powers problem", you can use De Moivre's theorem, which states that for any complex number z = r(cosθ + isinθ), the nth power of z is equal to r^n(cos(nθ) + isin(nθ)). This allows you to easily calculate the result of raising a complex number to a power without having to carry out multiple multiplications.

What are some real-world applications of "Very simple complex powers problems"?

"Very simple complex powers problems" have many real-world applications, particularly in fields such as electrical engineering and physics. They can be used to model and analyze the behavior of electrical circuits, as well as in the study of oscillations and waves.

What are the common challenges in solving "Very simple complex powers problems"?

One common challenge in solving "Very simple complex powers problems" is understanding the concept of complex numbers and how they behave when raised to a power. Another challenge may be keeping track of the different steps involved in solving the problem, such as using De Moivre's theorem or converting between polar and rectangular forms of complex numbers.

Are there any tips for solving "Very simple complex powers problems" more efficiently?

Yes, to solve "Very simple complex powers problems" more efficiently, it is important to have a strong understanding of complex numbers and the properties of their powers. It can also be helpful to practice converting between polar and rectangular forms and familiarizing yourself with De Moivre's theorem. Additionally, breaking the problem down into smaller steps and keeping organized notes can also make the process more efficient.

Similar threads

Replies
1
Views
1K
Replies
1
Views
868
Replies
17
Views
2K
Replies
13
Views
2K
Replies
12
Views
2K
Replies
18
Views
3K
Back
Top