What is ln(((-2i)^2)/(1+i)) ?

The solution to ln(((-2i)^2)/(1+i)) is 2ln(2)+2ln(i)-ln(1+i)

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simplify ln(((-2i)^2)/(1+i))

simplify

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

2 ln (2) + 2 ln (i) - ln (1 + i)

Solution steps

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

Apply log rule:

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

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

= ln ((- 2 i)^{2}) - ln (1 + i)

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln ((- 2 i)^{2}) = 2 ln (- 2 i)

= 2 ln (- 2 i) - ln (1 + i)

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

2 ln (- 2 i) = 2 ln (2) + 2 ln (i)

= 2 ln (2) + 2 ln (i) - ln (1 + i)

lo g_{4} (\frac{5}{30})

1 - lo g_{2} (5) - lo g_{2} (x + 1) + 3 lo g_{2} (x)

- \frac{1}{2} ln ∣ y + 1 ∣ + \frac{1}{2} ln ∣ y - 1 ∣

ln (\frac{e ^{- θ} θ ^{x}}{x !})

What is ln(((-2i)^2)/(1+i)) ?
The solution to ln(((-2i)^2)/(1+i)) is 2ln(2)+2ln(i)-ln(1+i)