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How do you simplify complex numbers to a negative exponent? - Answers

Lets try z^-1 which is 1/z. Now z=x+iy then 1/z = 1/(x+iy). This is equal to (x-iy)/((x+iy)(x-iy)) and is equal to (x-iy)/(xx+yy). The real part is u=x/(xx+yy) and the imaginary part is -y/(xx+yy). Don't forget that i squared is -1 when working out (x+iy)(x-iy). Get a good program like Mathmatica as higher exponents get real time consuming and will cause brain damage! Netflyer



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How do you simplify complex numbers to a negative exponent? - Answers

https://math.answers.com/basic-math/How_do_you_simplify_complex_numbers_to_a_negative_exponent

Lets try z^-1 which is 1/z. Now z=x+iy then 1/z = 1/(x+iy). This is equal to (x-iy)/((x+iy)(x-iy)) and is equal to (x-iy)/(xx+yy). The real part is u=x/(xx+yy) and the imaginary part is -y/(xx+yy). Don't forget that i squared is -1 when working out (x+iy)(x-iy). Get a good program like Mathmatica as higher exponents get real time consuming and will cause brain damage! Netflyer



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https://math.answers.com/basic-math/How_do_you_simplify_complex_numbers_to_a_negative_exponent

How do you simplify complex numbers to a negative exponent? - Answers

Lets try z^-1 which is 1/z. Now z=x+iy then 1/z = 1/(x+iy). This is equal to (x-iy)/((x+iy)(x-iy)) and is equal to (x-iy)/(xx+yy). The real part is u=x/(xx+yy) and the imaginary part is -y/(xx+yy). Don't forget that i squared is -1 when working out (x+iy)(x-iy). Get a good program like Mathmatica as higher exponents get real time consuming and will cause brain damage! Netflyer

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      Lets try z^-1 which is 1/z. Now z=x+iy then 1/z = 1/(x+iy). This is equal to (x-iy)/((x+iy)(x-iy)) and is equal to (x-iy)/(xx+yy). The real part is u=x/(xx+yy) and the imaginary part is -y/(xx+yy). Don't forget that i squared is -1 when working out (x+iy)(x-iy). Get a good program like Mathmatica as higher exponents get real time consuming and will cause brain damage! Netflyer

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