Solve 4-3i/2+i

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

Evaluate

$$1-2i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide 5-10i by 5 to get 1-2i.

$$1-2i$$

Real Part

$$1$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide 5-10i by 5 to get 1-2i.

$$Re(1-2i)$$

The real part of 1-2i is 1.

$$1$$