what is √(1+i)?

i+1 = √2 e^i(pi/4)

Therefore

√(1+i) = (i+1)^(1/2) = (√2)^(1/2) e^i(pi/8)

To put in the form x+iy use re^(it) = rcos(t) + i r sin(t).

Bear in mind that every complex number has two square roots separated by 180° on a circle centered at the origin.

First off, there's two square roots, just like √-1 = i or -i [since (-i)^2 = (-1)^2 i^2 = -1]. So you'll have to say which square root you mean. Secondly, it's best to think of these problems in terms of the complex plane, with the x-axis being the real axis and the y-axis being the imaginary axis. Complex number multiplication in the complex plane just adds angles and multiplies magnitudes (ask your teacher for specifics if this isn't enough for you). Hence finding square roots or cube roots or whatever just reduces to finding complex numbers with the correct angle and magnitude. That's what Mathmom's answer is really doing (though she essentially arbitrarily picked one answer over the other), and that's what'll allow you to solve these problems on your own in general.

You can't put the answer to this into decimal form. It's what is called a complex number -- an expression containing real and imaginary numbers. ("i" is the square root of negative 1, which is called an "imaginary" number. It has no decimal representation.)

The way you have written it is about as simple as you're going to be able to get it.

√(1+i) = √[√2(√2 / 2 + i √2 / 2)] = 2^(1 / 4) √e^(i pi / 4) = 2^(1 / 4) e^(i pi / 8) = 2^(1 / 4) cos(pi / 8) + i 2^(1 / 4)sin(pi / 8)