Solve 12i/1+i

$$\frac { 12 i } { 1 + i }$$

Evaluate

$$6+6i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1-i.

$$\frac{12i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$$

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

$$\frac{12i\left(1-i\right)}{1^{2}-i^{2}}$$

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

$$\frac{12i\left(1-i\right)}{2}$$

Multiply 12i times 1-i.

$$\frac{12i\times 1+12\left(-1\right)i^{2}}{2}$$

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

$$\frac{12i\times 1+12\left(-1\right)\left(-1\right)}{2}$$

Do the multiplications in 12i\times 1+12\left(-1\right)\left(-1\right). Reorder the terms.

$$\frac{12+12i}{2}$$

Divide 12+12i by 2 to get 6+6i.

$$6+6i$$

Real Part

$$6$$

Solution Steps

Multiply both numerator and denominator of \frac{12i}{1+i} by the complex conjugate of the denominator, 1-i.

$$Re(\frac{12i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)})$$

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

$$Re(\frac{12i\left(1-i\right)}{1^{2}-i^{2}})$$

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

$$Re(\frac{12i\left(1-i\right)}{2})$$

Multiply 12i times 1-i.

$$Re(\frac{12i\times 1+12\left(-1\right)i^{2}}{2})$$

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

$$Re(\frac{12i\times 1+12\left(-1\right)\left(-1\right)}{2})$$

Do the multiplications in 12i\times 1+12\left(-1\right)\left(-1\right). Reorder the terms.

$$Re(\frac{12+12i}{2})$$

Divide 12+12i by 2 to get 6+6i.

$$Re(6+6i)$$

The real part of 6+6i is 6.

$$6$$