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What is the square root of #2i#?

1 Answer

#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