What is ((1-3i)^2)/(i-1) ?

The solution to ((1-3i)^2)/(i-1) is 1+7i

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simplify ((1-3i)^2)/(i-1)

simplify

\frac{( 1 - 3 i ) ^{2}}{i - 1}

1 + 7 i

Solution steps

\frac{( 1 - 3 i ) ^{2}}{i - 1}

Rewrite

i - 1

in standard complex form:

- 1 + i

= \frac{( 1 - 3 i ) ^{2}}{- 1 + i}

Multiply by the conjugate

\frac{- 1 - i}{- 1 - i}

= \frac{( 1 - 3 i ) ^{2} ( - 1 - i )}{( - 1 + i ) ( - 1 - i )}

Simplify

\frac{( 1 - 3 i ) ^{2} ( - 1 - i )}{( - 1 + i ) ( - 1 - i )} : \frac{14 i + 2}{2}

= \frac{14 i + 2}{2}

Rewrite in standard complex form:

1 + 7 i

= 1 + 7 i

3 2^{- \frac{3}{5}}

(4 x + 5)^{2} = 5 x^{2} + 4 x

lo g_{3} (x) = 2

32 - \frac{64}{3} + \frac{32}{5}

m^{2} - 6 m + 5