Solve 20i/3+i

$$\frac { 20 i } { 3 + i }$$

Evaluate

$$2+6i$$

Solution Steps

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

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

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

$$\frac{20i\left(3-i\right)}{3^{2}-i^{2}}$$

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

$$\frac{20i\left(3-i\right)}{10}$$

Multiply 20i times 3-i.

$$\frac{20i\times 3+20\left(-1\right)i^{2}}{10}$$

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

$$\frac{20i\times 3+20\left(-1\right)\left(-1\right)}{10}$$

Do the multiplications in 20i\times 3+20\left(-1\right)\left(-1\right). Reorder the terms.

$$\frac{20+60i}{10}$$

Divide 20+60i by 10 to get 2+6i.

$$2+6i$$

Real Part

$$2$$

Solution Steps

Multiply both numerator and denominator of \frac{20i}{3+i} by the complex conjugate of the denominator, 3-i.

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

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

$$Re(\frac{20i\left(3-i\right)}{3^{2}-i^{2}})$$

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

$$Re(\frac{20i\left(3-i\right)}{10})$$

Multiply 20i times 3-i.

$$Re(\frac{20i\times 3+20\left(-1\right)i^{2}}{10})$$

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

$$Re(\frac{20i\times 3+20\left(-1\right)\left(-1\right)}{10})$$

Do the multiplications in 20i\times 3+20\left(-1\right)\left(-1\right). Reorder the terms.

$$Re(\frac{20+60i}{10})$$

Divide 20+60i by 10 to get 2+6i.

$$Re(2+6i)$$

The real part of 2+6i is 2.

$$2$$