# Complex number question

• Mar 11th 2011, 06:08 PM
Glitch
Complex number question
The question:
Suppose both c and $(1 + ic)^5$ are real (c != 0). Show that c = $\pm \sqrt{5 \pm 2\sqrt{5}}$

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 $\pm$. Is it two different complex numbers? Am I attempting this correctly? Thanks.
• Mar 11th 2011, 06:15 PM
Prove It
You can use the Binomial Theorem to expand $\displaystyle (1 + ic)^5$...

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

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

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

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

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

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

You should now be able to solve for $\displaystyle c$.
• Mar 11th 2011, 08:43 PM
Glitch
Solving quartics isn't exactly my speciality. :/

What's the best way to tackle $c(5 - 10c^2 + c^4) = 0$? Or can I just rewrite the given value of 'c' since I know that's going to be the solution anyway? :P
• Mar 11th 2011, 11:23 PM
Prove It
Quote:

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

What's the best way to tackle $c(5 - 10c^2 + c^4) = 0$? 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 $\displaystyle c \neq 0$, so $\displaystyle c^4 - 10c^2 + 5 = 0$.

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

I'm sure you can solve this for $\displaystyle C$, then use that to solve for $\displaystyle c$.
• Mar 11th 2011, 11:50 PM
Glitch
Oh dang. I forgot that trick. Sometimes I wonder if I'm a descendant of Homer J Simpson.

http://fishbowlinventory.files.wordp...er-simpson.jpg
• Mar 12th 2011, 12:23 AM
Prove It
Quote:

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

http://fishbowlinventory.files.wordp...er-simpson.jpg

If you were Lisa or Maggie you'd be alright :P
• Mar 12th 2011, 12:26 AM
Glitch
Quote:

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
• Mar 12th 2011, 12:32 AM
Prove It
Quote:

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
• Mar 12th 2011, 12:37 AM
Glitch
Quote:

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

http://www.thorninpaw.com/u/htdocs/t..._and_kodos.jpg
Kinky.
• Mar 12th 2011, 01:20 AM
Prove It
Quote:

Originally Posted by Glitch

Even moreso, considering the aliens are mormons >< HAHAHA