Solve 2-i/4+3i

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

Evaluate

$$\frac{1}{5}-\frac{2}{5}i=0.2-0.4i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 4-3i.

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

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

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

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

$$\frac{\left(2-i\right)\left(4-3i\right)}{25}$$

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

$$\frac{2\times 4+2\times \left(-3i\right)-i\times 4-\left(-3i^{2}\right)}{25}$$

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

$$\frac{2\times 4+2\times \left(-3i\right)-i\times 4-\left(-3\left(-1\right)\right)}{25}$$

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

$$\frac{8-6i-4i-3}{25}$$

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

$$\frac{8-3+\left(-6-4\right)i}{25}$$

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

$$\frac{5-10i}{25}$$

Divide 5-10i by 25 to get \frac{1}{5}-\frac{2}{5}i.

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

Real Part

$$\frac{1}{5} = 0.2$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(2-i\right)\left(4-3i\right)}{25})$$

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

$$Re(\frac{2\times 4+2\times \left(-3i\right)-i\times 4-\left(-3i^{2}\right)}{25})$$

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

$$Re(\frac{2\times 4+2\times \left(-3i\right)-i\times 4-\left(-3\left(-1\right)\right)}{25})$$

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

$$Re(\frac{8-6i-4i-3}{25})$$

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

$$Re(\frac{8-3+\left(-6-4\right)i}{25})$$

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

$$Re(\frac{5-10i}{25})$$

Divide 5-10i by 25 to get \frac{1}{5}-\frac{2}{5}i.

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

The real part of \frac{1}{5}-\frac{2}{5}i is \frac{1}{5}.

$$\frac{1}{5}$$