Determine what is real of complex number z=1/(1-i)+1/(2-i)+1/(1-2i)?

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You need to consider the complex number z as a summation of complex numbers `z_1 = 1/(1-i), z_2 = 1/(2-i), z_3 = 1/(1-2i)` , such that:

`z = z_1 + z_2 + z_3`

You need to multiply the complex number `z_1` by conjugate of `1 - i` , such that:

`z_1 = (1+i)/((1 - i)(1 + i)) => z_1 = (1+i)/(1^2 - i^2)`

Replacing -1 for ` i^2` yields:

`z_1 = (1+i)/(1^2 - (-1)) => z_1 = (1+i)/2`

You need to multiply the complex number `z_2` by conjugate of `2 - i` , such that:

`z_2 = (2+i)/((2+i)(2-i))`

`z_2 = (2+i)/(4 - i^2) => z_2 = (2+i)/5`

You need to multiply the complex number `z_3` by conjugate of `1 - 2i` , such that:

`z_3 = (1+2i)/((1-2i)(1+2i))`

`z_3 = (1+2i)/(1-4i^2) => z_3 = (1+2i)/(1+4) => z_3 = (1+2i)/5`

`z = (1+i)/2 + (2+i)/5 + (1+2i)/5`

Bringing the terms to a common denominator yields:

`z = (5(1 + i) + 2(2 + i + 1 + 2i))/10`

`z = (5 + 5i + 6 + 6i)/10 => z = (11 + 11i)/10`

`z = 11/10 + (11/10)*i`

**Hence, evaluating the real part of the complex number z, yields **`Re(z = z_1 + z_2 + z_3) = 11/10.`

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