#sqrt{2i}={1+i, -1-i}#
Let us look at some details.
Let #z=sqrt{2i}#.
(Note that #z# are complex numbers.)
by squaring,
#Rightarrow z^2=2i#
by using the exponential form #z=re^{i theta}#,
#Rightarrow r^2e^{i(2theta)}=2i=2e^{i(pi/2+2npi)}#
#Rightarrow {(r^2=2 Rightarrow r=sqrt{2}),
(2theta=pi/2+2npi Rightarrow theta=pi/4+npi):}#
So, #z=sqrt{2}e^{i(pi/4+npi)}#
by Eular's Formula: #e^{i theta}=cos theta +isin theta#
#Rightarrow z=sqrt{2}[cos(pi/4+npi)+isin(pi/4+npi)]#
#=sqrt{2}(pm1/sqrt{2}pm1/sqrt{2}i)=pm1pmi#
I kept the following original post just in case someone needs it.
#(2i)^(1/2)# = #(2)^(1/2)# #(i)^(1/2)#,
#(i)^(1/2)# = -1
#(2i)^(1/2)# = #(2)^(1/2)# x -1
#(2)^(1/2)# = 1.41
#(2i)^(1/2)# = 1.41 x -1 = -1.41
