Question 469997
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5}{3\ -\ 4i}]


Five principles required.


1.  Anything divided by itself is 1, i.e. *[tex \LARGE \frac{a}{a}\ =\ 1\ \forall\ a\ \in\ \mathbb{C}]


2.  1 is the multiplicative identity, i.e. *[tex \LARGE a\ \cdot\ 1\ =\ a\ \forall\ a\ \in\ \mathbb{C}]


3.  The product of a pair of conjugate binomials is the difference of 2 squares, i.e. *[tex \LARGE (a\ +\ b)(a\ -\ b)\ =\ a^2\ -\ b^2]


4.  *[tex \LARGE i^2\ =\ -1]


Combining 3 and 4, we can show that *[tex \LARGE (a\ +\ bi)(a\ -\ bi)\ =\ a^2\ +\ b^2]


5.  Multiplication of a complex number by a real number is defined as: *[tex \LARGE  c(a\ +\ bi)\ =\ ac\ +\ bci]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5}{3\ -\ 4i}]


Apply principles 1 and 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{5}{3\ -\ 4i}\right)\left(\frac{3\ +\ 4i}{3\ +\ 4i}\right)]


Apply principles 3, 4, and 5 and the definition of multiplication of fractions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15\ +\ 20i}{25}]


Simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{5}\ +\ \frac{4i}{5}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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