Solve 5+3i/4-i

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

Evaluate

$$1+i$$

Solution Steps

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

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

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

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

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

$$\frac{\left(5+3i\right)\left(4+i\right)}{17}$$

Multiply complex numbers 5+3i and 4+i like you multiply binomials.

$$\frac{5\times 4+5i+3i\times 4+3i^{2}}{17}$$

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

$$\frac{5\times 4+5i+3i\times 4+3\left(-1\right)}{17}$$

Do the multiplications in 5\times 4+5i+3i\times 4+3\left(-1\right).

$$\frac{20+5i+12i-3}{17}$$

Combine the real and imaginary parts in 20+5i+12i-3.

$$\frac{20-3+\left(5+12\right)i}{17}$$

Do the additions in 20-3+\left(5+12\right)i.

$$\frac{17+17i}{17}$$

Divide 17+17i by 17 to get 1+i.

$$1+i$$

Real Part

$$1$$

Solution Steps

Multiply both numerator and denominator of \frac{5+3i}{4-i} by the complex conjugate of the denominator, 4+i.

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

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

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

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

$$Re(\frac{\left(5+3i\right)\left(4+i\right)}{17})$$

Multiply complex numbers 5+3i and 4+i like you multiply binomials.

$$Re(\frac{5\times 4+5i+3i\times 4+3i^{2}}{17})$$

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

$$Re(\frac{5\times 4+5i+3i\times 4+3\left(-1\right)}{17})$$

Do the multiplications in 5\times 4+5i+3i\times 4+3\left(-1\right).

$$Re(\frac{20+5i+12i-3}{17})$$

Combine the real and imaginary parts in 20+5i+12i-3.

$$Re(\frac{20-3+\left(5+12\right)i}{17})$$

Do the additions in 20-3+\left(5+12\right)i.

$$Re(\frac{17+17i}{17})$$

Divide 17+17i by 17 to get 1+i.

$$Re(1+i)$$

The real part of 1+i is 1.

$$1$$