Solve 104i/5+i

$$\frac { 104 i } { 5 + i }$$

Evaluate

$$4+20i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 5-i.

$$\frac{104i\left(5-i\right)}{\left(5+i\right)\left(5-i\right)}$$

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

$$\frac{104i\left(5-i\right)}{5^{2}-i^{2}}$$

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

$$\frac{104i\left(5-i\right)}{26}$$

Multiply 104i times 5-i.

$$\frac{104i\times 5+104\left(-1\right)i^{2}}{26}$$

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

$$\frac{104i\times 5+104\left(-1\right)\left(-1\right)}{26}$$

Do the multiplications in 104i\times 5+104\left(-1\right)\left(-1\right). Reorder the terms.

$$\frac{104+520i}{26}$$

Divide 104+520i by 26 to get 4+20i.

$$4+20i$$

Real Part

$$4$$

Solution Steps

Multiply both numerator and denominator of \frac{104i}{5+i} by the complex conjugate of the denominator, 5-i.

$$Re(\frac{104i\left(5-i\right)}{\left(5+i\right)\left(5-i\right)})$$

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

$$Re(\frac{104i\left(5-i\right)}{5^{2}-i^{2}})$$

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

$$Re(\frac{104i\left(5-i\right)}{26})$$

Multiply 104i times 5-i.

$$Re(\frac{104i\times 5+104\left(-1\right)i^{2}}{26})$$

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

$$Re(\frac{104i\times 5+104\left(-1\right)\left(-1\right)}{26})$$

Do the multiplications in 104i\times 5+104\left(-1\right)\left(-1\right). Reorder the terms.

$$Re(\frac{104+520i}{26})$$

Divide 104+520i by 26 to get 4+20i.

$$Re(4+20i)$$

The real part of 4+20i is 4.

$$4$$