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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)² = 1 + 2i - 1 = 2i, the original equation becomes (a + bi)² = (1 + i)², 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)² = 2i a² + 2abi - b² = 2i a² - b² = 0, ab = 1 a² - 1/a² = 0 a² = 1/a² a^4 = 1 ? a = ±1 b = 1/a = 1/±1 = ±1 (? 1/1 = 1, 1/-1 = -1) ? a = b = ±1

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