What does #sqrt(3+i)# equal in a+bi form?

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Answer 1 · 2015-11-06T23:44:28

#sqrt(3+i) = (sqrt((sqrt(10)+3)/2)) + (sqrt((sqrt(10)-3)/2)) i#

Suppose #(a+bi)^2 = 3+i#

#(a+bi)^2 = (a^2-b^2)+2abi#

So equating real and imaginary parts we get:

#a^2-b^2 = 3#

#2ab = 1#

Hence #b = 1/(2a)#, which we can substitute into the first equation to get:

#3 = a^2-(1/(2a))^2 = a^2-1/(4a^2)#

Multiply both ends by #4a^2# to get:

#12(a^2) = 4(a^2)^2-1#

So:

#4(a^2)^2-12(a^2)-1 = 0#

#a^2 = (12+-sqrt(12^2+16))/8 = (12+-sqrt(160))/8 = (3+-sqrt(10))/2#

Since #sqrt(10) > 3#, choose the #+# sign to get Real values for #a#:

#a = +-sqrt((sqrt(10)+3)/2)#

#b = +-sqrt(a^2-3) = +-sqrt((sqrt(10)-3)/2)#

where #b# has the same sign as #a# since #b = 1/(2a)#

The principal square root is in Q1 with #a, b > 0#

That is:

#sqrt(3+i) = (sqrt((sqrt(10)+3)/2)) + (sqrt((sqrt(10)-3)/2)) i#

In fact, if #c, d > 0# then we can similarly show:

#sqrt(c+di) = (sqrt((sqrt(c^2+d^2)+c)/2)) + (sqrt((sqrt(c^2+d^2)-c)/2)) i#