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

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

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

simplify

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

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

Solution steps

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

Rewrite

1 - i - i

in standard complex form:

1 + (- 1 - 1) i

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

Subtract the numbers:

- 1 - 1 = - 2

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

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

2000 (1 + 0.26)^{15}

\frac{3 x - 1}{x + 2} - \frac{x - 2}{x - 1}

(- 2 x y^{4})^{4}

4 x^{4} - 16 x^{3} - 180 x^{2}

2 \cdot 32

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