What is ((1+2i)/(1-2i)+(2+i)/(1+2i))*1-i ?

The solution to ((1+2i)/(1-2i)+(2+i)/(1+2i))*1-i is 1/5-4/5 i

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simplify ((1+2i)/(1-2i)+(2+i)/(1+2i))*1-i

simplify

(\frac{1 + 2 i}{1 - 2 i} + \frac{2 + i}{1 + 2 i}) \cdot 1 - i

\frac{1}{5} - \frac{4}{5} i

Solution steps

(\frac{1 + 2 i}{1 - 2 i} + \frac{2 + i}{1 + 2 i}) \cdot 1 - i

Apply rule:

a \cdot 1 = a

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

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

\frac{1 + 2 i}{1 - 2 i} = \frac{4 i - 3}{5}

\frac{2 + i}{1 + 2 i} = \frac{4 - 3 i}{5}

= \frac{4 i - 3}{5} + \frac{4 - 3 i}{5} - i

Join

\frac{4 i - 3}{5} + \frac{4 - 3 i}{5} - i : \frac{i + 1}{5} - i

= \frac{i + 1}{5} - i

Join

\frac{i + 1}{5} - i : \frac{- 4 i + 1}{5}

= \frac{- 4 i + 1}{5}

Rewrite in standard complex form:

\frac{1}{5} - \frac{4}{5} i

= \frac{1}{5} - \frac{4}{5} i

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ln (ln (x^{2}))

512 + 27 a^{9}

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What is ((1+2i)/(1-2i)+(2+i)/(1+2i))*1-i ?
The solution to ((1+2i)/(1-2i)+(2+i)/(1+2i))*1-i is 1/5-4/5 i