What is (3i^{30}-i^{19})/(2i-1) ?

The solution to (3i^{30}-i^{19})/(2i-1) is 1+i

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simplify (3i^{30}-i^{19})/(2i-1)

simplify

\frac{3 i ^{30} - i ^{19}}{2 i - 1}

1 + i

Solution steps

\frac{3 i ^{30} - i ^{19}}{2 i - 1}

Rewrite

2 i - 1

in standard complex form:

- 1 + 2 i

= \frac{3 i ^{30} - i ^{19}}{- 1 + 2 i}

Multiply by the conjugate

\frac{- 1 - 2 i}{- 1 - 2 i}

= \frac{( 3 i ^{30} - i ^{19} ) ( - 1 - 2 i )}{( - 1 + 2 i ) ( - 1 - 2 i )}

Simplify

\frac{( 3 i ^{30} - i ^{19} ) ( - 1 - 2 i )}{( - 1 + 2 i ) ( - 1 - 2 i )} : \frac{5 i + 5}{5}

= \frac{5 i + 5}{5}

Rewrite in standard complex form:

1 + i

= 1 + i

(x + 4) (x)

- - 6

\frac{125 x ^{8} y ^{6}}{125 x y ^{5}}

\frac{2}{( \frac{5}{x - 1} ) + 3}

(1 + r)^{3}

What is (3i^{30}-i^{19})/(2i-1) ?
The solution to (3i^{30}-i^{19})/(2i-1) is 1+i