find the square root of -3+4i

Asked by nmenghani | 5th Apr, 2013, 08:35: PM

Expert Answer:

Let the square root be a+ib, where a and b are real numbers
 
Hence, (a+ib)*(a+ib) = -3+4i
a^2 + 2iab -b^2 = -3+4i
Comparing real and imaginary parts
a^2  -b^2 = -3
2ab = 4i
a = 2/b
 
Hence, 4/b^2 - b^2 = -3
 
b^4 -3b^2 -4 = 0
 
(b^2 -4)(b^2 +1) =0
Since, b^2 can positive as b is real, hence b^2 = 4
b = +2 or -2
Hence, a = 1 or -1
 
So, the square root would be 1+2i or -1-2i
 
 

Answered by  | 7th Apr, 2013, 07:16: AM

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