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

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

Evaluate

$$-3-4i$$

Solution Steps

Multiply both numerator and denominator of \frac{3-i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Do the multiplications in \frac{\left(3-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

$$\left(\frac{2-4i}{2}\right)^{2}$$

Divide 2-4i by 2 to get 1-2i.

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

Calculate 1-2i to the power of 2 and get -3-4i.

$$-3-4i$$

Real Part

$$-3$$

Solution Steps

Multiply both numerator and denominator of \frac{3-i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Do the multiplications in \frac{\left(3-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

$$Re(\left(\frac{2-4i}{2}\right)^{2})$$

Divide 2-4i by 2 to get 1-2i.

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

Calculate 1-2i to the power of 2 and get -3-4i.

$$Re(-3-4i)$$

The real part of -3-4i is -3.

$$-3$$