Complex Analysis - Value of imaginary part.

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Homework Help Overview

The discussion revolves around a complex analysis problem where participants are tasked with demonstrating that both \( c \) and \( (1 + ic)^{5} \) are real, given that \( c \neq 0 \). The original poster attempts to show that \( c = \pm \sqrt{5 \pm 2\sqrt{5}} \) and later suggests another method involving trigonometric values.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the expansion of the expression and the conditions under which the imaginary parts vanish. There are attempts to factor and simplify the resulting polynomial, with some questioning the correctness of coefficients and terms. Others explore the implications of setting the imaginary part to zero.

Discussion Status

The discussion is ongoing, with participants actively engaging in verifying calculations and clarifying the conditions for the expression to be real. There is a recognition of the need to factor the polynomial derived from the imaginary components, and some participants express confusion about the coefficients involved.

Contextual Notes

Participants are working under the constraint that \( c \neq 0 \) and are exploring the implications of this condition on the realness of the expression. There is also a focus on ensuring that the derived equations align with the original problem statement.

NewtonianAlch
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Homework Statement



Suppose both c and (1 + ic)[itex]^{5}[/itex] are real (c [itex]\neq[/itex] 0).
Show that c = ± [itex]\sqrt{5 ± 2\sqrt{5}}[/itex]
Now use another method to show that either c = ± tan 36◦ or c = ± tan 72◦


The Attempt at a Solution



I expanded it out, but I'm not entirely too sure how to solve this for c. Also, solving for c with wolfram gives c = i - which is not correct as c equals something else entirely.
 
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What did you get after expanding it out?
 
After expanding and simplifying somewhat:

1 + 5ic - 10c[itex]^{2}[/itex] - 10ic[itex]^{3}[/itex] + 5c[itex]^{4}[/itex] + ic[itex]^{5}[/itex]
 
OK, that value is supposed to be real. So the terms with i should vanish. What does that give you?
 
1 - 10c[itex]^{2}[/itex] + 5c[itex]^{4}[/itex]...

Ah, so now I just factorise and solve?
 
NewtonianAlch said:
1 - 10c[itex]^{2}[/itex] + 5c[itex]^{4}[/itex]...

What do you mean by this?
 
micromass said:
What do you mean by this?

5c^4 - 10c^2 + 1 = 0
 
NewtonianAlch said:
5c^4 - 10c^2 + 1 = 0

Why should that be true?? Does that force (1+ic)5 to be real?
 
I'm not too sure, if we solved for c this way, and substitute back into the original equation, then it should return a real value; i.e. no i component

I just tried it in Maple and it returns a value with no i component.
 
  • #10
NewtonianAlch said:
5c^4 - 10c^2 + 1 = 0

You switched some coefficients around.

That should be c4 - 10c2 + 5 = 0, if post #3 is correct.

Completing the square should help with factoring.
 
  • #11
SammyS said:
You switched some coefficients around.

That should be c4 - 10c2 + 5 = 0, if post #3 is correct.

Completing the square should help with factoring.

I can't see how those values switched around.

In the post #3, c4 had a coefficient of 5.
 
  • #12
NewtonianAlch said:
After expanding and simplifying somewhat:

1 + 5ic - 10c[itex]^{2}[/itex] - 10ic[itex]^{3}[/itex] + 5c[itex]^{4}[/itex] + ic[itex]^{5}[/itex]

NewtonianAlch said:
I can't see how those values switched around.

In the post #3, c4 had a coefficient of 5.
Here are the terms with 1 .

5ic - 10ic[itex]^{3}[/itex]+ ic[itex]^{5}[/itex]

Factoring out ci gives:

5 - 10c[itex]^{2}[/itex]+ c[itex]^{4}[/itex]
 
  • #13
I am really confused now. I thought that since we're only interested in the real terms here, any terms with the i component can just be canceled out; hence leaving us with all the real terms, which is why I said

5c^4 - 10c^2 + 1

I'm not sure what you mean by here are the terms with 1.
 
  • #14
When is a+bi real??
 
  • #15
micromass said:
When is a+bi real??


When b = 0.
 
  • #16
NewtonianAlch said:
When b = 0.

OK, so when is [tex]1 + 5ic - 10c^2 - 10ic^3 + 5c^4 + ic^5[/tex] real?
 
  • #17
I am tempted to say when c = 0, but c [itex]\neq[/itex] 0 according to the question.
 
  • #18
Oh hang on, I separated them out.

When c^5 - 10c^3 +5c = 0?
 
  • #19
Note that

[tex]1 + 5ic - 10c^2 - 10ic^3 + 5c^4 + ic^5=(1 - 10c^2 + 5c^4)+ i(5c - 10c^3 + c^5)[/tex]

Do you see it now?
 
  • #20
NewtonianAlch said:
Oh hang on, I separated them out.

When c^5 - 10c^3 +5c = 0?

Yes! So, you're given that [itex](1+ic)^5[/itex] is real. You've shown that this is equivalent to saying that

[tex]c^5-10c^3+5c=0[/tex]

Now you can factor this equation.
 
  • #21
Hmm, that's an interesting question. Thanks for your help. I'm going to try that out now!
 

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