E EvoRevolution Member Joined Jan 25, 2009 Messages 123 Gender Male HSC 2009 Jan 28, 2009 #1 Without using any series expansions, prove (√3+i)^n + (√3 - i)^n is real. Last edited: Jan 28, 2009
jet Banned Joined Jan 4, 2007 Messages 3,148 Gender Male HSC 2009 Jan 28, 2009 #2 You need to retype that equation to make sense. You're square roots are either in the wrong place or missing an argument.
You need to retype that equation to make sense. You're square roots are either in the wrong place or missing an argument.
E EvoRevolution Member Joined Jan 25, 2009 Messages 123 Gender Male HSC 2009 Jan 28, 2009 #3 opps soz
L lolokay Active Member Joined Mar 21, 2008 Messages 1,015 Gender Undisclosed HSC 2009 Jan 28, 2009 #4 just put each in polar (mod/arg) form
jet Banned Joined Jan 4, 2007 Messages 3,148 Gender Male HSC 2009 Jan 28, 2009 #5 sqrt(3) + i = cos(π/6) + isin(π/6) Similarly, sqrt(3) - i = cos(-π/6) + isin(-π/6) Hence (sqrt(3) + i)n + (3 - i)n = cos(nπ/6) + isin(nπ/6) + cos(-nπ/6) + isin(-nπ/6) =cos(nπ/6) + isin(nπ/6) + cos(nπ/6) - isin(nπ/6) = 2cos(nπ/6) Hence the expression is real.
sqrt(3) + i = cos(π/6) + isin(π/6) Similarly, sqrt(3) - i = cos(-π/6) + isin(-π/6) Hence (sqrt(3) + i)n + (3 - i)n = cos(nπ/6) + isin(nπ/6) + cos(-nπ/6) + isin(-nπ/6) =cos(nπ/6) + isin(nπ/6) + cos(nπ/6) - isin(nπ/6) = 2cos(nπ/6) Hence the expression is real.
E EvoRevolution Member Joined Jan 25, 2009 Messages 123 Gender Male HSC 2009 Jan 28, 2009 #6 ohh i understand thanks for the help simple lol