Question 1007384
Find the two square roots for the following complex number. 
Write your answers in standard form:  2i
<pre>
Let the answer be a complex number 
a+bi with a,b real. Then

(a+bi)Č = 2i

aČ+2abi+bČiČ = 2i

aČ+2abi+bČ(-1) = 2i

aČ+2abi-bČ=2i

Equate real parts and equate imaginary
parts:

aČ-bČ = 0   2abi = 2i
aČ = bČ       ab = 1
              
So the only possible real solutions
are (a,b) = (1,1) and (a,b) = (-1,-1)

So the two square roots of 2i are

1+i and -1-i    <--answers 

Checking:

(1+i)Č = 1+2i+iČ = 1+2i+(-1) = 2i

Edwin</pre>