(1+i)(1-2i)+(1+i)(1+2i)/(1+2i)^2-(1-2i)

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вопрос

First, let's simplify the expression step by step.

Expanding the first part: (1+i)(1-2i)
= 1(1) + 1(-2i) + i(1) + i(-2i)
= 1 - 2i + i - 2i^2
= 1 - 2i + i + 2
= 3 - i

Expanding the second part: (1+i)(1+2i)
= 1(1) + 1(2i) + i(1) + i(2i)
= 1 + 2i + i + 2i^2
= 1 + 2i + i - 2
= -1 + 3i

Now, we substitute these results back into the original expression:

(3 - i + (-1 + 3i)) / (1+2i)^2 - (1-2i)

= (2 + 2i) / (1+2i)^2 - (1-2i)

To simplify further, we need to square the denominator (1+2i)^2:

(1+2i)^2 = (1+2i)(1+2i)
= 11 + 12i + 2i1 + 2i2i
= 1 + 2i + 2i + 4i^2
= 1 + 4i + 4i - 4
= -3 + 8i

Now substitute back:

(2 + 2i) / (-3 + 8i) - (1-2i)

Now, let's simplify this expression further if needed.