Solve 2/1-i+2i=

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

Evaluate

$$1+3i$$

Solution Steps

Multiply both numerator and denominator of \frac{2}{1-i} by the complex conjugate of the denominator, 1+i.

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

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

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

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

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

Multiply 2 times 1+i.

$$\frac{2\times 1+2i}{2}+2i$$

Do the multiplications in 2\times 1+2i.

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

Divide 2+2i by 2 to get 1+i.

$$1+i+2i$$

Combine the real and imaginary parts in numbers 1+i and 2i.

$$1+\left(1+2\right)i$$

Add 1 to 2.

$$1+3i$$

Real Part

$$1$$

Solution Steps

Multiply both numerator and denominator of \frac{2}{1-i} by the complex conjugate of the denominator, 1+i.

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

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

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

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

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

Multiply 2 times 1+i.

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

Do the multiplications in 2\times 1+2i.

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

Divide 2+2i by 2 to get 1+i.

$$Re(1+i+2i)$$

Combine the real and imaginary parts in numbers 1+i and 2i.

$$Re(1+\left(1+2\right)i)$$

Add 1 to 2.

$$Re(1+3i)$$

The real part of 1+3i is 1.

$$1$$