What is ((2-i)^2-i^{15})/(3+i) ?

The solution to ((2-i)^2-i^{15})/(3+i) is 3/5-6/5 i

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simplify ((2-i)^2-i^{15})/(3+i)

simplify

\frac{( 2 - i ) ^{2} - i ^{15}}{3 + i}

\frac{3}{5} - \frac{6}{5} i

Solution steps

\frac{( 2 - i ) ^{2} - i ^{15}}{3 + i}

Multiply by the conjugate

\frac{3 - i}{3 - i}

= \frac{( ( 2 - i ) ^{2} - i ^{15} ) ( 3 - i )}{( 3 + i ) ( 3 - i )}

Simplify

\frac{( ( 2 - i ) ^{2} - i ^{15} ) ( 3 - i )}{( 3 + i ) ( 3 - i )} : \frac{- 12 i + 6}{10}

= \frac{- 12 i + 6}{10}

Rewrite in standard complex form:

\frac{3}{5} - \frac{6}{5} i

= \frac{3}{5} - \frac{6}{5} i

(\frac{9 x ^{2}}{4 y ^{3}}) (\frac{2 y ^{2}}{x})

\frac{cot ( x )}{cos ( x )}

9 (x - 1) - (x - 1)^{2}

\frac{500 ( \frac{1}{2} ) ^{1001}}{1001}

What is ((2-i)^2-i^{15})/(3+i) ?
The solution to ((2-i)^2-i^{15})/(3+i) is 3/5-6/5 i