Solve 3-4i/2-i

$$\frac { 3 - 4 i } { 2 - i }$$

Evaluate

$$2-i$$

Solution Steps

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

$$\frac{\left(3-4i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}$$

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

$$\frac{\left(3-4i\right)\left(2+i\right)}{2^{2}-i^{2}}$$

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

$$\frac{\left(3-4i\right)\left(2+i\right)}{5}$$

Multiply complex numbers 3-4i and 2+i like you multiply binomials.

$$\frac{3\times 2+3i-4i\times 2-4i^{2}}{5}$$

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

$$\frac{3\times 2+3i-4i\times 2-4\left(-1\right)}{5}$$

Do the multiplications in 3\times 2+3i-4i\times 2-4\left(-1\right).

$$\frac{6+3i-8i+4}{5}$$

Combine the real and imaginary parts in 6+3i-8i+4.

$$\frac{6+4+\left(3-8\right)i}{5}$$

Do the additions in 6+4+\left(3-8\right)i.

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

Divide 10-5i by 5 to get 2-i.

$$2-i$$

Real Part

$$2$$

Solution Steps

Multiply both numerator and denominator of \frac{3-4i}{2-i} by the complex conjugate of the denominator, 2+i.

$$Re(\frac{\left(3-4i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)})$$

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

$$Re(\frac{\left(3-4i\right)\left(2+i\right)}{2^{2}-i^{2}})$$

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

$$Re(\frac{\left(3-4i\right)\left(2+i\right)}{5})$$

Multiply complex numbers 3-4i and 2+i like you multiply binomials.

$$Re(\frac{3\times 2+3i-4i\times 2-4i^{2}}{5})$$

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

$$Re(\frac{3\times 2+3i-4i\times 2-4\left(-1\right)}{5})$$

Do the multiplications in 3\times 2+3i-4i\times 2-4\left(-1\right).

$$Re(\frac{6+3i-8i+4}{5})$$

Combine the real and imaginary parts in 6+3i-8i+4.

$$Re(\frac{6+4+\left(3-8\right)i}{5})$$

Do the additions in 6+4+\left(3-8\right)i.

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

Divide 10-5i by 5 to get 2-i.

$$Re(2-i)$$

The real part of 2-i is 2.

$$2$$