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

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

Evaluate

$$1-7i$$

Solution Steps

Calculate 3-i to the power of 2 and get 8-6i.

$$\frac{8-6i}{1+i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 1-i.

$$\frac{\left(8-6i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$$

Do the multiplications in \frac{\left(8-6i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

$$\frac{2-14i}{2}$$

Divide 2-14i by 2 to get 1-7i.

$$1-7i$$

Real Part

$$1$$

Solution Steps

Calculate 3-i to the power of 2 and get 8-6i.

$$Re(\frac{8-6i}{1+i})$$

Multiply both numerator and denominator of \frac{8-6i}{1+i} by the complex conjugate of the denominator, 1-i.

$$Re(\frac{\left(8-6i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)})$$

Do the multiplications in \frac{\left(8-6i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

$$Re(\frac{2-14i}{2})$$

Divide 2-14i by 2 to get 1-7i.

$$Re(1-7i)$$

The real part of 1-7i is 1.

$$1$$