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.

$$\frac{3\times 2+3i+2i+i^{2}}{5}$$

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

$$\frac{3\times 2+3i+2i-1}{5}$$

Do the multiplications in 3\times 2+3i+2i-1.

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

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.

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

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

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

Do the multiplications in 3\times 2+3i+2i-1.

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

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.

$$1$$