What does #sqrt(3+i)# equal in a+bi form?
1 Answer
Explanation:
Suppose
#(a+bi)^2 = (a^2-b^2)+2abi#
So equating real and imaginary parts we get:
#a^2-b^2 = 3#
#2ab = 1#
Hence
#3 = a^2-(1/(2a))^2 = a^2-1/(4a^2)#
Multiply both ends by
#12(a^2) = 4(a^2)^2-1#
So:
#4(a^2)^2-12(a^2)-1 = 0#
From the quadratic formula we get:
#a^2 = (12+-sqrt(12^2+16))/8 = (12+-sqrt(160))/8 = (3+-sqrt(10))/2#
Since
#a = +-sqrt((sqrt(10)+3)/2)#
#b = +-sqrt(a^2-3) = +-sqrt((sqrt(10)-3)/2)#
where
The principal square root is in Q1 with
That is:
#sqrt(3+i) = (sqrt((sqrt(10)+3)/2)) + (sqrt((sqrt(10)-3)/2)) i#
In fact, if
#sqrt(c+di) = (sqrt((sqrt(c^2+d^2)+c)/2)) + (sqrt((sqrt(c^2+d^2)-c)/2)) i#