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

$$\frac { ( 1 - i - 3 ) ^ { 2 } } { 1 - i + i }$$

Evaluate

$$3+4i$$

Solution Steps

Subtract 3 from 1-i to get -2-i.

$$\frac{\left(-2-i\right)^{2}}{1-i+i}$$

Calculate -2-i to the power of 2 and get 3+4i.

$$\frac{3+4i}{1-i+i}$$

Add 1-i and i to get 1.

$$\frac{3+4i}{1}$$

Anything divided by one gives itself.

$$3+4i$$

Real Part

$$3$$

Solution Steps

Subtract 3 from 1-i to get -2-i.

$$Re(\frac{\left(-2-i\right)^{2}}{1-i+i})$$

Calculate -2-i to the power of 2 and get 3+4i.

$$Re(\frac{3+4i}{1-i+i})$$

Add 1-i and i to get 1.

$$Re(\frac{3+4i}{1})$$

Anything divided by one gives itself.

$$Re(3+4i)$$

The real part of 3+4i is 3.

$$3$$