Solve 3i(1-i)/1+i

$$\frac { 3 i ( 1 - i ) } { 1 + i }$$

Evaluate

$$3$$

Solution Steps

Multiply 3i times 1-i.

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

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

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

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

$$\frac{3+3i}{1+i}$$

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

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

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

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

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

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

Multiply complex numbers 3+3i and 1-i like you multiply binomials.

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

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

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

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

$$\frac{3-3i+3i+3}{2}$$

Combine the real and imaginary parts in 3-3i+3i+3.

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

Do the additions in 3+3+\left(-3+3\right)i.

$$\frac{6}{2}$$

Divide 6 by 2 to get 3.

$$3$$

Real Part

$$3$$

Solution Steps

Multiply 3i times 1-i.

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

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

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

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

$$Re(\frac{3+3i}{1+i})$$

Multiply both numerator and denominator of \frac{3+3i}{1+i} by the complex conjugate of the denominator, 1-i.

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

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

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

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

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

Multiply complex numbers 3+3i and 1-i like you multiply binomials.

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

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

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

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

$$Re(\frac{3-3i+3i+3}{2})$$

Combine the real and imaginary parts in 3-3i+3i+3.

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

Do the additions in 3+3+\left(-3+3\right)i.

$$Re(\frac{6}{2})$$

Divide 6 by 2 to get 3.

$$Re(3)$$

The real part of 3 is 3.

$$3$$