I understand how the complex numbers are closed under addition, subtraction, multiplication, and division. But how do you prove they are closed under exponentiation to a complex number like e^(ix) without already assuming euler's formula e^ix = cosx + isinx. The whole video is based on the assumption that e^(ix) would still be a complex number, but how do you justify that?
@pritampadhan59773 жыл бұрын
Sir please provide long video with detail explanation 😍😍😍😍😍
@groverlife40383 жыл бұрын
yeah eddie wooo nice math stuff
@vishnutheunique16703 жыл бұрын
What is answer of int 0 to 2pi. {(acost)^2+(-bsint)^2} ^1/2 dt Where a= 0.3870 b=0.3788 Can anyone help me out?
@legalfictionnaturalfact39693 жыл бұрын
this looks like one of those situations that may be simplified by using tau instead of pi?
@jasonzheng83883 жыл бұрын
this was a question on my linear test from a few months ago😂
@aashsyed12773 жыл бұрын
ANY BODY FROM BLACKPENREDPEN?
@mychevysparkevdidntcatchfi14893 жыл бұрын
if x=i, ke^(i*i)=k/e = k(cos(i)+isin(i)). So real numbers are actually complex. ;-)
@RedBar3D3 жыл бұрын
Complex numbers are by definition just ordered pairs (a, b) of real numbers a, b. We call a its real part and b its imaginary part. When we write x + iy for a complex number, what we actually mean is the pair (x, y), where x and y are real numbers. Now, for a given real number x, you may think of it as the complex number (x, 0). That is as the complex number with real part x and imaginary part 0.
@kingtrades50153 жыл бұрын
Kings Trades
@samurai20683 жыл бұрын
We studied these things in std XI
@florianwollmuetze24123 жыл бұрын
Pretty sure my math exam failed. Watching this to calm myself with simpler math