Solve 1+4i/4-i

$$\frac { 1 + 4 i } { 4 - i }$$

Evaluate

$$i$$

Solution Steps

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

$$\frac{\left(1+4i\right)\left(4+i\right)}{\left(4-i\right)\left(4+i\right)}$$

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

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

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

$$\frac{\left(1+4i\right)\left(4+i\right)}{17}$$

Multiply complex numbers 1+4i and 4+i like you multiply binomials.

$$\frac{1\times 4+i+4i\times 4+4i^{2}}{17}$$

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

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

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

$$\frac{4+i+16i-4}{17}$$

Combine the real and imaginary parts in 4+i+16i-4.

$$\frac{4-4+\left(1+16\right)i}{17}$$

Do the additions in 4-4+\left(1+16\right)i.

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

Divide 17i by 17 to get i.

$$i$$

Real Part

$$0$$

Solution Steps

Multiply both numerator and denominator of \frac{1+4i}{4-i} by the complex conjugate of the denominator, 4+i.

$$Re(\frac{\left(1+4i\right)\left(4+i\right)}{\left(4-i\right)\left(4+i\right)})$$

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

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

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

$$Re(\frac{\left(1+4i\right)\left(4+i\right)}{17})$$

Multiply complex numbers 1+4i and 4+i like you multiply binomials.

$$Re(\frac{1\times 4+i+4i\times 4+4i^{2}}{17})$$

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

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

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

$$Re(\frac{4+i+16i-4}{17})$$

Combine the real and imaginary parts in 4+i+16i-4.

$$Re(\frac{4-4+\left(1+16\right)i}{17})$$

Do the additions in 4-4+\left(1+16\right)i.

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

Divide 17i by 17 to get i.

$$Re(i)$$

The real part of i is 0.

$$0$$