Solve 11i/1-i

$$\frac { 11 i } { 1 - i }$$

Evaluate

$$-\frac{11}{2}+\frac{11}{2}i=-5.5+5.5i$$

Solution Steps

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

$$\frac{11i\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}$$

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

$$\frac{11i\left(1+i\right)}{1^{2}-i^{2}}$$

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

$$\frac{11i\left(1+i\right)}{2}$$

Multiply 11i times 1+i.

$$\frac{11i\times 1+11i^{2}}{2}$$

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

$$\frac{11i\times 1+11\left(-1\right)}{2}$$

Do the multiplications in 11i\times 1+11\left(-1\right). Reorder the terms.

$$\frac{-11+11i}{2}$$

Divide -11+11i by 2 to get -\frac{11}{2}+\frac{11}{2}i.

$$-\frac{11}{2}+\frac{11}{2}i$$

Real Part

$$-\frac{11}{2} = -5\frac{1}{2} = -5.5$$

Solution Steps

Multiply both numerator and denominator of \frac{11i}{1-i} by the complex conjugate of the denominator, 1+i.

$$Re(\frac{11i\left(1+i\right)}{\left(1-i\right)\left(1+i\right)})$$

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

$$Re(\frac{11i\left(1+i\right)}{1^{2}-i^{2}})$$

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

$$Re(\frac{11i\left(1+i\right)}{2})$$

Multiply 11i times 1+i.

$$Re(\frac{11i\times 1+11i^{2}}{2})$$

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

$$Re(\frac{11i\times 1+11\left(-1\right)}{2})$$

Do the multiplications in 11i\times 1+11\left(-1\right). Reorder the terms.

$$Re(\frac{-11+11i}{2})$$

Divide -11+11i by 2 to get -\frac{11}{2}+\frac{11}{2}i.

$$Re(-\frac{11}{2}+\frac{11}{2}i)$$

The real part of -\frac{11}{2}+\frac{11}{2}i is -\frac{11}{2}.

$$-\frac{11}{2}$$