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Alexis1994_13

  • 4 years ago

1/(x+yi) ,make it like a+bi please. Step by step.

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  1. Alexis1994_13
    • 4 years ago
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    x/(x^2+y^2)-yi/(x^2+y^2) is that right ?

  2. darthsid
    • 4 years ago
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    \[\frac{1}{x+iy} = \frac{1}{x+iy} \times \frac{x-iy}{x-iy}\] now, multiplying the denominators and using the rule: \[ (m+n) times (m-n) = m^2 - n^2 \] we get: \[\frac{1}{x+iy} = \frac{x-iy}{x^2-(iy)^2}\] \[\frac{1}{x+iy} = \frac{x-iy}{x^2-i^2y^2}\] \[\frac{1}{x+iy} = \frac{x-iy}{x^2-(-1)y^2}\] \[\frac{1}{x+iy} = \frac{x-iy}{x^2+y^2}\] \[\frac{1}{x+iy} = \frac{x}{x^2+y^2} + i\frac{-y}{x^2+y^2}\]

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