Solve 2-i/3+i

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

Evaluate

$$\frac{1}{2}-\frac{1}{2}i=0.5-0.5i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-i.

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

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

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

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

$$\frac{\left(2-i\right)\left(3-i\right)}{10}$$

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

$$\frac{2\times 3+2\left(-i\right)-i\times 3-\left(-i^{2}\right)}{10}$$

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

$$\frac{2\times 3+2\left(-i\right)-i\times 3-\left(-\left(-1\right)\right)}{10}$$

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

$$\frac{6-2i-3i-1}{10}$$

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

$$\frac{6-1+\left(-2-3\right)i}{10}$$

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

$$\frac{5-5i}{10}$$

Divide 5-5i by 10 to get \frac{1}{2}-\frac{1}{2}i.

$$\frac{1}{2}-\frac{1}{2}i$$

Real Part

$$\frac{1}{2} = 0.5$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(2-i\right)\left(3-i\right)}{10})$$

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

$$Re(\frac{2\times 3+2\left(-i\right)-i\times 3-\left(-i^{2}\right)}{10})$$

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

$$Re(\frac{2\times 3+2\left(-i\right)-i\times 3-\left(-\left(-1\right)\right)}{10})$$

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

$$Re(\frac{6-2i-3i-1}{10})$$

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

$$Re(\frac{6-1+\left(-2-3\right)i}{10})$$

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

$$Re(\frac{5-5i}{10})$$

Divide 5-5i by 10 to get \frac{1}{2}-\frac{1}{2}i.

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

The real part of \frac{1}{2}-\frac{1}{2}i is \frac{1}{2}.

$$\frac{1}{2}$$