Question
Root of iota plus root of minus iota
Answers
Steve
I assume by iota you mean <i>i</i>, which is √-1
√i = (1+i)/√2
√-i = (1-i)/√2
add them to get 2/√2 = √2
√i = (1+i)/√2
√-i = (1-i)/√2
add them to get 2/√2 = √2
Adarsh
Let root iota &root -iota = y
Now square on both side we get i - i+ 2 i^1/2 -i^1/2
= 2 [-i^2]^1/2
= 2
As y^2 = 2
Then y 2^1/2
Now square on both side we get i - i+ 2 i^1/2 -i^1/2
= 2 [-i^2]^1/2
= 2
As y^2 = 2
Then y 2^1/2