Question 1171500
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Let *[tex \Large z\ =\ x\ +\ yi] where *[tex \Large x\ \in\ \mathbb{R}] and *[tex \Large y\ \in\ \mathbb{R}].  *[tex \Large |z|\ =\ x^2\ +\ y^2\ = 1].  *[tex \Large \overline{z}\ = \ x\ -\ yi] by definition.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ -\ yi}{1}\,\cdot\,\frac{x\ +\ yi}{x\ + \ yi}\ =\ \frac{x^2\ +\ y^2}{x\ +\ yi}\ =\ \frac{1}{x\ +\ yi}\ =\ \frac{1}{z}]


Same logic for *[tex \Large \overline{w}]

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
I > Ø
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