What is log_{e}((1-i/2)/(-1-i/2)) ?

The solution to log_{e}((1-i/2)/(-1-i/2)) is ln(1-i/2)-ln(-1-i/2)

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simplify log_{e}((1-i/2)/(-1-i/2))

simplify

lo g_{e} (\frac{1 - \frac{i}{2}}{- 1 - \frac{i}{2}})

ln (1 - \frac{i}{2}) - ln (- 1 - \frac{i}{2})

Solution steps

lo g_{e} (\frac{1 - \frac{i}{2}}{- 1 - \frac{i}{2}})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

lo g_{e} (\frac{1 - \frac{i}{2}}{- 1 - \frac{i}{2}}) = lo g_{e} (1 - \frac{i}{2}) - lo g_{e} (- 1 - \frac{i}{2})

= lo g_{e} (1 - \frac{i}{2}) - lo g_{e} (- 1 - \frac{i}{2})

Apply log rule:

lo g_{e} (x) = ln (x)

= ln (1 - \frac{i}{2}) - lo g_{e} (- 1 - \frac{i}{2})

Apply log rule:

lo g_{e} (x) = ln (x)

= ln (1 - \frac{i}{2}) - ln (- 1 - \frac{i}{2})

2 lo g_{b} (w - 3) - 6 lo g_{b} (w + 3)

- lo g_{10} (25 \cdot 1 0^{- 12})

- ln (\frac{13}{3})

ln (2 - x) + ln (2 + x)

ln ∣ sin (x) ∣ + ln ∣ csc (x) - cot (x) ∣

What is log_{e}((1-i/2)/(-1-i/2)) ?
The solution to log_{e}((1-i/2)/(-1-i/2)) is ln(1-i/2)-ln(-1-i/2)