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Suppose that a and b are positive integers such that (a+bi)^2 = 3+4i. What is a+bi?

 Jan 29, 2019
 #1
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 (a+bi)^2 = 3+4i

 

(a + bi)^2  =   a^2 + 2abi - b^2

 

Equating coefficients, we have

 

a^2 - b^2  =  3     (1)

2ab = 4    ⇒  ab = 2  ⇒  b = 2/a    (2)

 

Sub (2) into (1)

 

a^2 - (2/a)^2 = 4

 

a^4 - 4  = 3a^2

 

a^4 - 3a^2 - 4 = 0        factor

 

(a^2 - 4) (a^2 + 1) = 0

 

A real  solultion will come from

 

(a^2 - 4) = 0

(a + 2) (a - 2) = 0

Since a is positve, then a = 2

And b = 2/a = 2/2 = 1

 

a + bi   =     2 + i

 

 

cool cool cool

 Jan 29, 2019

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