1. ## Complex number question

The question:
Suppose both c and $\displaystyle (1 + ic)^5$ are real (c != 0). Show that c = $\displaystyle \pm \sqrt{5 \pm 2\sqrt{5}}$

How do I solve this? I substituted the value of c into the complex number with the intent to find the magnitude and angle, but I'm not sure how to handle the $\displaystyle \pm$. Is it two different complex numbers? Am I attempting this correctly? Thanks.

2. You can use the Binomial Theorem to expand $\displaystyle \displaystyle (1 + ic)^5$...

$\displaystyle \displaystyle (1 + ic)^5 = \sum_{r = 0}^n{{n\choose{r}}1^{n-r}(ic)^r}$

$\displaystyle \displaystyle = 1 + 5ic + 10(ic)^2 + 10(ic)^3 + 5(ic)^4 + (ic)^5$

$\displaystyle \displaystyle = 1 + 5ic -10c^2 - 10ic^3 + 5c^4 + ic^5$

$\displaystyle \displaystyle = 1 - 10c^2 + 5c^4 + (5c - 10c^3 + c^5)i$.

Since you know that $\displaystyle \displaystyle (1 + ic)^5$ is real, that means the imaginary part is $\displaystyle \displaystyle 0$.

So $\displaystyle \displaystyle 5c - 10c^3 + c^5 = 0$.

You should now be able to solve for $\displaystyle \displaystyle c$.

3. Solving quartics isn't exactly my speciality. :/

What's the best way to tackle $\displaystyle c(5 - 10c^2 + c^4) = 0$? Or can I just rewrite the given value of 'c' since I know that's going to be the solution anyway? :P

4. Originally Posted by Glitch
Solving quartics isn't exactly my speciality. :/

What's the best way to tackle $\displaystyle c(5 - 10c^2 + c^4) = 0$? Or can I just rewrite the given value of 'c' since I know that's going to be the solution anyway? :P
You already know that $\displaystyle \displaystyle c \neq 0$, so $\displaystyle \displaystyle c^4 - 10c^2 + 5 = 0$.

If you let $\displaystyle \displaystyle C = c^2$ then the equation becomes $\displaystyle \displaystyle C^2 - 10C + 5 = 0$.

I'm sure you can solve this for $\displaystyle \displaystyle C$, then use that to solve for $\displaystyle \displaystyle c$.

5. Oh dang. I forgot that trick. Sometimes I wonder if I'm a descendant of Homer J Simpson.

6. Originally Posted by Glitch
Oh dang. I forgot that trick. Sometimes I wonder if I'm a descendant of Homer J Simpson.

If you were Lisa or Maggie you'd be alright :P

7. Originally Posted by Prove It
If you were Lisa or Maggie you'd be alright :P
See, I can't even mock myself correctly! :P

8. Originally Posted by Glitch
See, I can't even mock myself correctly! :P
Of course, that's assuming that Lisa and Maggie are really his kids... Now that I think about it, isn't Maggie's real dad Kang the alien? Hahahaha

9. Originally Posted by Prove It
Of course, that's assuming that Lisa and Maggie are really his kids... Now that I think about it, isn't Maggie's real dad Kang the alien? Hahahaha

Kinky.

10. Originally Posted by Glitch

Kinky.
Even moreso, considering the aliens are mormons >< HAHAHA