What is (i^{30}(5-i))/(-1+i) ?

The solution to (i^{30}(5-i))/(-1+i) is 3+2i

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simplify (i^{30}(5-i))/(-1+i)

simplify

\frac{i ^{30} ( 5 - i )}{- 1 + i}

3 + 2 i

Solution steps

\frac{i ^{30} ( 5 - i )}{- 1 + i}

Multiply by the conjugate

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

= \frac{i ^{30} ( 5 - i ) ( - 1 - i )}{( - 1 + i ) ( - 1 - i )}

Simplify

\frac{i ^{30} ( 5 - i ) ( - 1 - i )}{( - 1 + i ) ( - 1 - i )} : \frac{6 + 4 i}{2}

= \frac{6 + 4 i}{2}

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

= \frac{6}{2} + \frac{4}{2} i

Simplify

\frac{6}{2} + \frac{4}{2} i : 3 + 2 i

= 3 + 2 i

- 245

- 224

- 216

1 + 4 (0.5)^{2}