Solve 3-i/2+i=

$$\frac { 3 - i } { 2 + i } =$$

Evaluate

$$1-i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 2-i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{\left(3-i\right)\left(2-i\right)}{5}$$

Multiply complex numbers 3-i and 2-i like you multiply binomials.

$$\frac{3\times 2+3\left(-i\right)-i\times 2-\left(-i^{2}\right)}{5}$$

By definition, i^{2} is -1.

$$\frac{3\times 2+3\left(-i\right)-i\times 2-\left(-\left(-1\right)\right)}{5}$$

Do the multiplications in 3\times 2+3\left(-i\right)-i\times 2-\left(-\left(-1\right)\right).

$$\frac{6-3i-2i-1}{5}$$

Combine the real and imaginary parts in 6-3i-2i-1.

$$\frac{6-1+\left(-3-2\right)i}{5}$$

Do the additions in 6-1+\left(-3-2\right)i.

$$\frac{5-5i}{5}$$

Divide 5-5i by 5 to get 1-i.

$$1-i$$

Real Part

$$1$$

Solution Steps

Multiply both numerator and denominator of \frac{3-i}{2+i} by the complex conjugate of the denominator, 2-i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{\left(3-i\right)\left(2-i\right)}{5})$$

Multiply complex numbers 3-i and 2-i like you multiply binomials.

$$Re(\frac{3\times 2+3\left(-i\right)-i\times 2-\left(-i^{2}\right)}{5})$$

By definition, i^{2} is -1.

$$Re(\frac{3\times 2+3\left(-i\right)-i\times 2-\left(-\left(-1\right)\right)}{5})$$

Do the multiplications in 3\times 2+3\left(-i\right)-i\times 2-\left(-\left(-1\right)\right).

$$Re(\frac{6-3i-2i-1}{5})$$

Combine the real and imaginary parts in 6-3i-2i-1.

$$Re(\frac{6-1+\left(-3-2\right)i}{5})$$

Do the additions in 6-1+\left(-3-2\right)i.

$$Re(\frac{5-5i}{5})$$

Divide 5-5i by 5 to get 1-i.

$$Re(1-i)$$

The real part of 1-i is 1.

$$1$$