Solve 1/(1-i)*5=

$$\frac { 1 } { ( 1 - i ) \cdot 5 } =$$

Evaluate

$$\frac{1}{10}+\frac{1}{10}i=0.1+0.1i$$

Solution Steps

Multiply 1-i times 5.

$$\frac{1}{1\times 5-i\times 5}$$

Do the multiplications in 1\times 5-i\times 5.

$$\frac{1}{5-5i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 5+5i.

$$\frac{1\left(5+5i\right)}{\left(5-5i\right)\left(5+5i\right)}$$

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

$$\frac{1\left(5+5i\right)}{5^{2}-5^{2}i^{2}}$$

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

$$\frac{1\left(5+5i\right)}{50}$$

Multiply 1 and 5+5i to get 5+5i.

$$\frac{5+5i}{50}$$

Divide 5+5i by 50 to get \frac{1}{10}+\frac{1}{10}i.

$$\frac{1}{10}+\frac{1}{10}i$$

Real Part

$$\frac{1}{10} = 0.1$$

Solution Steps

Multiply 1-i times 5.

$$Re(\frac{1}{1\times 5-i\times 5})$$

Do the multiplications in 1\times 5-i\times 5.

$$Re(\frac{1}{5-5i})$$

Multiply both numerator and denominator of \frac{1}{5-5i} by the complex conjugate of the denominator, 5+5i.

$$Re(\frac{1\left(5+5i\right)}{\left(5-5i\right)\left(5+5i\right)})$$

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

$$Re(\frac{1\left(5+5i\right)}{5^{2}-5^{2}i^{2}})$$

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

$$Re(\frac{1\left(5+5i\right)}{50})$$

Multiply 1 and 5+5i to get 5+5i.

$$Re(\frac{5+5i}{50})$$

Divide 5+5i by 50 to get \frac{1}{10}+\frac{1}{10}i.

$$Re(\frac{1}{10}+\frac{1}{10}i)$$

The real part of \frac{1}{10}+\frac{1}{10}i is \frac{1}{10}.

$$\frac{1}{10}$$