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 March 12th, 2011, 06:48 AM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Number and sets Given (a+bi)^2=2i. Find all the real values of a and b. The solution (a+bi)^2=2i a^2-b^2+2abi=2i Equating real parts: a^2-b^2=0 a^2=b^2 a=+- b Equating the imaginary part: 2ab=2 ab=1 Please explain to me about this part below. I am confused. Thank you. When a=b, a^2=1 (I still don't understand how to get this) a=+-1 b=+-1 When a=-b, -a^2=1 a^2=-1 (rejected) because a should be a real number The real values of a and b are 1 and 1, -1 and -1. March 12th, 2011, 10:56 AM #2 Senior Member   Joined: Nov 2010 From: Staten Island, NY Posts: 152 Thanks: 0 Re: Number and sets From 2ab=2 substitute a in for b (because a=b). So we get 2a*a=2. March 12th, 2011, 11:06 AM #3 Global Moderator   Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Number and sets You found that ab = 1. Then when a = b, we rewrite the above as a(a) = 1 after substituting a in place of b. a^2 = 1 ... March 12th, 2011, 10:38 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,927 Thanks: 2205 Using (1 + i)� = 1 + 2i - 1 = 2i, the original equation becomes (a + bi)� = (1 + i)�, so a + bi = �(1 + i). If a and b are real, equating real and imaginary parts gives a = b = �1. March 12th, 2011, 11:00 PM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 (a + bi)� = 2i a� + 2abi - b� = 2i a� - b� = 0, ab = 1 a� - 1/a� = 0 a� = 1/a� a^4 = 1 ? a = �1 b = 1/a = 1/�1 = �1 (? 1/1 = 1, 1/-1 = -1) ? a = b = �1 Tags number, sets Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mid Advanced Statistics 0 July 22nd, 2013 07:07 AM tbillion Applied Math 17 September 5th, 2012 06:13 AM Jamsisos Advanced Statistics 1 June 21st, 2012 01:11 PM hoyy1kolko Algebra 4 March 15th, 2011 06:44 AM mAraujo Real Analysis 2 July 26th, 2009 12:56 PM

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