Complex Analysis - Value of imaginary part.

In summary, the conversation discusses how to prove that c can equal ± √(5 ± 2√5) and how to use another method to show that c can equal ± tan 36° or ± tan 72°, given that both c and (1+ic)^5 are real. The method involves expanding and simplifying the equation to eliminate the terms with i, and then factoring to solve for c.
  • #1
NewtonianAlch
453
0

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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  • #2
What did you get after expanding it out?
 
  • #3
After expanding and simplifying somewhat:

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

Ah, so now I just factorise and solve?
 
  • #6
NewtonianAlch said:
1 - 10c[itex]^{2}[/itex] + 5c[itex]^{4}[/itex]...

What do you mean by this?
 
  • #7
micromass said:
What do you mean by this?

5c^4 - 10c^2 + 1 = 0
 
  • #8
NewtonianAlch said:
5c^4 - 10c^2 + 1 = 0

Why should that be true?? Does that force (1+ic)5 to be real?
 
  • #9
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!
 

Related to Complex Analysis - Value of imaginary part.

1. What is the value of the imaginary part in complex analysis?

In complex analysis, the value of the imaginary part refers to the coefficient of the imaginary unit i in a complex number. It is written as bi, where b is a real number and i is the imaginary unit (i.e. √-1).

2. How is the imaginary part related to the real part in complex analysis?

In complex analysis, the imaginary part is related to the real part through the complex conjugate. The complex conjugate of a complex number a + bi is a - bi, where a and b are real numbers. The real part and the imaginary part are also used to calculate the modulus (absolute value) and argument (angle) of a complex number.

3. Can the imaginary part of a complex number be zero?

Yes, the imaginary part of a complex number can be zero. In fact, when the imaginary part is zero, the complex number is a real number. This is because a complex number with zero imaginary part can be written as a + 0i, which is equivalent to just a.

4. How is the imaginary part used to represent complex numbers geometrically?

In complex analysis, complex numbers are often represented geometrically on a complex plane, with the real part as the x-coordinate and the imaginary part as the y-coordinate. This allows for a visual representation of complex numbers and operations such as addition, subtraction, and multiplication.

5. How does the value of the imaginary part affect the behavior of a complex function?

The value of the imaginary part can greatly affect the behavior of a complex function. For example, when the imaginary part of a complex number is large, the function may exhibit oscillatory behavior. Additionally, the location of poles and zeros (the roots and singularities) in the complex plane can be determined by the value of the imaginary part.

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