Solve 10i/3+i

$$\frac { 10 i } { 3 + i }$$

Evaluate

$$1+3i$$

Solution Steps

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

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

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

$$\frac{10i\left(3-i\right)}{3^{2}-i^{2}}$$

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

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

Multiply 10i times 3-i.

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

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

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

Do the multiplications in 10i\times 3+10\left(-1\right)\left(-1\right). Reorder the terms.

$$\frac{10+30i}{10}$$

Divide 10+30i by 10 to get 1+3i.

$$1+3i$$

Real Part

$$1$$

Solution Steps

Multiply both numerator and denominator of \frac{10i}{3+i} by the complex conjugate of the denominator, 3-i.

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

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

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

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

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

Multiply 10i times 3-i.

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

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

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

Do the multiplications in 10i\times 3+10\left(-1\right)\left(-1\right). Reorder the terms.

$$Re(\frac{10+30i}{10})$$

Divide 10+30i by 10 to get 1+3i.

$$Re(1+3i)$$

The real part of 1+3i is 1.

$$1$$