Complex Analysis - Value of imaginary part.

[NewtonianAlch]

Homework Statement

Suppose both c and (1 + ic)$^{5}$ are real (c $\neq$ 0).
Show that c = ± $\sqrt{5 ± 2\sqrt{5}}$
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.

 

[micromass]

What did you get after expanding it out?

 

[NewtonianAlch]

After expanding and simplifying somewhat:

1 + 5ic - 10c$^{2}$ - 10ic$^{3}$ + 5c$^{4}$ + ic$^{5}$

 

[micromass]

OK, that value is supposed to be real. So the terms with i should vanish. What does that give you?

 

[NewtonianAlch]

1 - 10c$^{2}$ + 5c$^{4}$...

Ah, so now I just factorise and solve?

 

[micromass]

1 - 10c$^{2}$ + 5c$^{4}$...

What do you mean by this?

 

[NewtonianAlch]

What do you mean by this?

5c^4 - 10c^2 + 1 = 0

 

[micromass]

5c^4 - 10c^2 + 1 = 0

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

 

[NewtonianAlch]

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.

 

[SammyS]

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.

 

[NewtonianAlch]

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.

 

[SammyS]

After expanding and simplifying somewhat:

1 + 5ic - 10c$^{2}$ - 10ic$^{3}$ + 5c$^{4}$ + ic$^{5}$

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$^{3}$+ ic$^{5}$

Factoring out ci gives:

5 - 10c$^{2}$+ c$^{4}$

 

[NewtonianAlch]

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.

 

[micromass]

When is a+bi real??

 

[NewtonianAlch]

When is a+bi real??

When b = 0.

 

[micromass]

When b = 0.

OK, so when is $$1 + 5ic - 10c^2 - 10ic^3 + 5c^4 + ic^5$$ real?

 

[NewtonianAlch]

I am tempted to say when c = 0, but c $\neq$ 0 according to the question.

 

[NewtonianAlch]

Oh hang on, I separated them out.

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

 

[micromass]

Note that

$$1 + 5ic - 10c^2 - 10ic^3 + 5c^4 + ic^5=(1 - 10c^2 + 5c^4)+ i(5c - 10c^3 + c^5)$$

Do you see it now?

 

[micromass]

Oh hang on, I separated them out.

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

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

$$c^5-10c^3+5c=0$$

Now you can factor this equation.

 

[NewtonianAlch]

Hmm, that's an interesting question. Thanks for your help. I'm going to try that out now!