- #1
neu
- 225
- 3
I am a 3rd Physics student and this is news to me!
apparantly root of i is exp(iPi/2))^1/2
Why?
Of course by i I mean Root of minus 1
apparantly root of i is exp(iPi/2))^1/2
Why?
Of course by i I mean Root of minus 1
-2a2 is not equal to 1, for real a.
how do we know that the sqrt of i can be denoted as a+bi?
how do we know that the sqrt of i can be denoted as a+bi?
how do we know that the sqrt of i can be denoted as a+bi?
Well, perhaps he knew that the complex numbers are algebraically closed and so did know that the square root of i is a complex number.a + bi is the general form of any complex number. This said, stupidmath made the assumption that the square root can be expressed as complex number. He didn't know whether or not this is the case first, but lucky him, it is!
Well, perhaps he knew that the complex numbers are algebraically closed and so did know that the square root of i is a complex number.
If you did not know it, it would have made the assumption unsure, not necessarily false.