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:

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

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