   Complex Number Reciprocals (Multiplicative Inverses), approached Algebraically by
Video Lecture 10 of 26
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### Lecture Description

Complex Analysis Video #10 (Complex Arithmetic, Part 10). Multiplicative Inverses of Complex Numbers approached Algebraically.

Details: Review modulus and argument properties of multiplication. What would it mean for a + b*I to have a multiplicative inverse x + y*i such that (a + b*i)*(x + y*i) = 1? Derive it using a system of equations in the unknowns x & y. Solve via substitution assuming that "a" is not zero. Find a general formula. Have Mathematica check it using Simplify and ComplexExpand. What if a = 0? Still get the same answer if b is nonzero. The only time it doesn't work is if both "a" and "b" equal zero. Find multiplicative inverse of z1 = 5 + 7*i. Use the derived formula to get it. How should we divide z1 = 5 + 7i by z3 = 2 + 3i? Same as z1*(1/z3)...could also do as z1/z3 (the answer is 31/13-i/13).

### Course Description

This is a mini crash course providing all you need to know to understand complex numbers, and study Complex Analysis. Mathematica is used to help visualize the complex plane.