Solve 1-i/1+i+3i

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

Evaluate

$$2i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

$$\frac{1-i-i-1}{2}+3i$$

Combine the real and imaginary parts in 1-i-i-1.

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

Do the additions in 1-1+\left(-1-1\right)i.

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

Divide -2i by 2 to get -i.

$$-i+3i$$

Add -i and 3i to get 2i.

$$2i$$

Real Part

$$0$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

Combine the real and imaginary parts in 1-i-i-1.

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

Do the additions in 1-1+\left(-1-1\right)i.

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

Divide -2i by 2 to get -i.

$$Re(-i+3i)$$

Add -i and 3i to get 2i.

$$Re(2i)$$

The real part of 2i is 0.

$$0$$