Solve 8/8-8i

$$\frac{ 8 }{ 8-8i }$$

Evaluate

$$\frac{1}{2}+\frac{1}{2}i=0.5+0.5i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 8+8i.

$$\frac{8\left(8+8i\right)}{\left(8-8i\right)\left(8+8i\right)}$$

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

$$\frac{8\left(8+8i\right)}{8^{2}-8^{2}i^{2}}$$

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

$$\frac{8\left(8+8i\right)}{128}$$

Multiply 8 times 8+8i.

$$\frac{8\times 8+8\times \left(8i\right)}{128}$$

Do the multiplications in 8\times 8+8\times \left(8i\right).

$$\frac{64+64i}{128}$$

Divide 64+64i by 128 to get \frac{1}{2}+\frac{1}{2}i.

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

Real Part

$$\frac{1}{2} = 0.5$$

Solution Steps

Multiply both numerator and denominator of \frac{8}{8-8i} by the complex conjugate of the denominator, 8+8i.

$$Re(\frac{8\left(8+8i\right)}{\left(8-8i\right)\left(8+8i\right)})$$

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

$$Re(\frac{8\left(8+8i\right)}{8^{2}-8^{2}i^{2}})$$

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

$$Re(\frac{8\left(8+8i\right)}{128})$$

Multiply 8 times 8+8i.

$$Re(\frac{8\times 8+8\times \left(8i\right)}{128})$$

Do the multiplications in 8\times 8+8\times \left(8i\right).

$$Re(\frac{64+64i}{128})$$

Divide 64+64i by 128 to get \frac{1}{2}+\frac{1}{2}i.

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

The real part of \frac{1}{2}+\frac{1}{2}i is \frac{1}{2}.

$$\frac{1}{2}$$