The polar form of a complex number is #Re^(itheta)#, where #R# is the number's *modulus* (its distance from #0#) and #theta# is the angle formed by the positive real axis and the number's vector on the complex plane.

We have a nice way of converting to polar coordinates by using Euler's formula: #e^(itheta) = cos(theta)+isin(theta)#. Thus, if we can find and factor out #R#, we can find (theta) from the remaining number.

In this case, we will first find #2+i# in polar form, and then apply the power of #1/2#.

To find #R#, we find the number's modulus: #|a+bi| = sqrt(a^2+b^2)#

#|2+i| = sqrt(2^2+1^2) = sqrt(5)#

#=> 2+i = sqrt(5)(2/sqrt(5)+1/sqrt(5)i)#

So, we have #cos(theta) = 2/sqrt(5)# and #sin(theta)=1/sqrt(5)#

As #arccos(2/sqrt(5))=arcsin(1/sqrt(5))~~0.4636# is not one of the "nice" angles, we'll leave it in that form. For ease of use, let's let #theta_0 = arccos(2/sqrt(5))# and write that for the remainder of the problem.

Proceeding, we now have

#2+i = sqrt(5)(cos(theta_0)+isin(theta_0))#

By Euler's formula, this gives us

#2+i = sqrt(5)e^(itheta_0)#

Note that we can add #2pii# in any integer multiple without changing the value due to the cyclic nature of #sin(theta)# and #cos(theta)#. This will become relevant once we take the root.

#2+i = sqrt(5)e^(i(theta_0+2pin))#

Finally, we take a power of #1/2# to get

#sqrt(2+i) = (sqrt(5)e^(itheta_0))^(1/2)#

#=5^(1/4)e^(i(theta_0+2pin)/2)#

#=5^(1/4)e^(i(theta_0/2+pin))#