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Grade: 11
        
find the square root of the complex number:
(3+4i)/(3-4i)
15 days ago

Answers : (1)

Arun
16262 Points
							
Dear student
 
I am thinking that you are asking square root of any one.
 
Set the square root of this equal to a + bi. Square both sides. Simplify. Then compare the real and imaginary parts separately and solve for a and b. 

√(3 + 4i) = a + bi 
3 + 4i = (a+bi)^2 
3 + 4i = a^2 + 2abi + (bi)^2 
3 + 4i = a^2 + 2abi - b^2 
So: 
3 = a^2 - b^2, and 4i = 2abi, which means 
3 = (a^2 - b^2) and 2 = ab 

Replace a = 2/b in the first equation to get 
3 = (2/b)^2 - b^2 
3 = (4 / b^2) - b^2 
3b^2 = 4 - b^4 
b^4 + 3b^2 - 4 = 0 
(b^2 + 4)(b^2 - 1) = 0 

So b^2 = -4, 1. We're only looking for real values of b, so b = ±1. This means a = 2/b = ±2. So 2 + i and -2 - i are two solutions. 
15 days ago
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