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Posted by DarthWho on November 23, 2011 at 11:45 PM |

Sorry for the huge delay in getting this one up my bad.

anyway:

how fo we calculate a^b when both A and B are complex numbers?

well this is where the complex exponentiation formula comes into play (this is a long one so hold onto your hats!

well first we have the formula:

E+Fi=(A+Bi)^(C+Di)

how do we work that out you ask? well it just so happens that that formula is equal to:

E+Fi=(A^2+B^2)^(C/2)*e^(-D*ARG(A+Bi))*(COS(C*ARG(A+Bi)+D*ln(A^2+B^2)/2)+i*SIN(C*ARG(A+Bi)+D*ln(A^2+B^2)/2))

or to split it up into components

G=(A^2+B^2)^(C/2)*e^(-D*ARG(A+Bi))

E=G*COS(C*ARG(A+Bi)+D*ln(A^2+B^2)/2)

F=G*SIN(C*ARG(A+Bi)+D*ln(A^2+B^2)/2)

I will try to update more regularly in the future;

this sunday: Calculus and why programmers should have some basic understanding of the subject. this will kick off a new seried which will focus on educating self professed math dummies on the mysteries of basic calculus.

Categories: Equation of the Week, Complex Numbers

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EllVutt

9:20 PM on October 19, 2019

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