Expert: Sherman D. - 5/26/2009

Question
What is the square root of i + 1?  i can't find any good explanations on the web.

Thanks,

Daniel

Answer
sqrt(i + 1) or sqrt(1 + i)

a^2 + 2abi + b^2i^2

a^2 + 2abi - b^2

(a^2 - b^2) + 2abi

a^2 - b^2 = 1
2abi = i

2ab = 1
ab = (1/2)
b = (1/(2a))

a^2 - (1/(2a))^2 = 1
a^2 - (1/(4a^2)) = 1
(4a^4 - 1)/(4a^2) = 1
4a^4 - 1 = 4a^2
4a^4 - 4a^2 - 1 = 0

using the quadratic formula

a^2 = (4 +/- sqrt(16 + 16))/8
a^2 = (4 +/- 4sqrt(2))/8
a^2 = (1 +/- sqrt(2))/2

we don't need to use the negative value.

a^2 = (1/2)(1 + sqrt(2))

a = sqrt((1 + sqrt(2))/2)

b = 1/(2(sqrt((1 + sqrt(2))/2)))

a = about 1.0987 or -1.0987
b = about .4551 or -.4551

ANS :
about
1.0987 + .4551i
-1.0987 - .4551i

in short the answer is ±(1.0987 + .4551i)

info found at http://staff.jccc.edu/swilson/complex/sqrootsimag.htm

if you go to www.quickmath.com, click on Solve and type in

x = sqrt(sqrt(-1) + 1)

tell it to solve for x, then click solve.

keep in mind that sqrt values have both a positive and negative, so although quickmath.com only gives you 1 value, there is actually 2 values.