Question 512605
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|\frac{5\ +\ 4i}{7\ -\ 5i}\right|]


First rationalize your denominator and then worry about the absolute value.


Multiply your fraction by the number 1 in the form of the conjugate of the denominator divided by itself.  If you have a complex number *[tex \Large \alpha\ +\ \beta{i}], then *[tex \Large \alpha\ -\ \beta{i}] is the conjugate.  So you want to multiply your fraction by *[tex \Large \frac{7\ +\ 5i}{7\ +\ 5i}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{5\ +\ 4i}{7\ -\ 5i}\right)\left(\frac{7\ +\ 5i}{7\ +\ 5i}\right)]


Foil the numerator and recall that the product of conjugates is the difference of two squares.  Don't forget that *[tex \Large i^2\ =\ -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{35\ +\ 25i\ +\ 28i\ +\ (-20)}{49\ -\ (-25)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15\ +\ 53i}{74}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15}{74}\ +\ \frac{53}{74}i]


Then recall that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |a\ +\ bi|\ =\ \sqrt{a^2\ +\ b^2}]


You can do the rest of the arithmetic.


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