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March 12th, 2011, 06:48 AM   #1
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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.
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March 12th, 2011, 10:56 AM   #2
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Re: Number and sets

From 2ab=2 substitute a in for b (because a=b). So we get 2a*a=2.
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March 12th, 2011, 11:06 AM   #3
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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
...
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March 12th, 2011, 10:38 PM   #4
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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.
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March 12th, 2011, 11:00 PM   #5
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(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
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