Solve frac{(20+20texttt{i})*-40texttt{i}}{20-20texttt{i}}

$$\frac{ (20+20 \texttt{i} ) \times -40 \texttt{i} }{ 20-20 \texttt{i} }$$

Evaluate

$$40$$

Solution Steps

Multiply 20+20i times -40i.

$$\frac{20\times \left(-40i\right)+20\left(-40\right)i^{2}}{20-20i}$$

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

$$\frac{20\times \left(-40i\right)+20\left(-40\right)\left(-1\right)}{20-20i}$$

Do the multiplications in 20\times \left(-40i\right)+20\left(-40\right)\left(-1\right). Reorder the terms.

$$\frac{800-800i}{20-20i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 20+20i.

$$\frac{\left(800-800i\right)\left(20+20i\right)}{\left(20-20i\right)\left(20+20i\right)}$$

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

$$\frac{\left(800-800i\right)\left(20+20i\right)}{20^{2}-20^{2}i^{2}}$$

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

$$\frac{\left(800-800i\right)\left(20+20i\right)}{800}$$

Multiply complex numbers 800-800i and 20+20i like you multiply binomials.

$$\frac{800\times 20+800\times \left(20i\right)-800i\times 20-800\times 20i^{2}}{800}$$

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

$$\frac{800\times 20+800\times \left(20i\right)-800i\times 20-800\times 20\left(-1\right)}{800}$$

Do the multiplications in 800\times 20+800\times \left(20i\right)-800i\times 20-800\times 20\left(-1\right).

$$\frac{16000+16000i-16000i+16000}{800}$$

Combine the real and imaginary parts in 16000+16000i-16000i+16000.

$$\frac{16000+16000+\left(16000-16000\right)i}{800}$$

Do the additions in 16000+16000+\left(16000-16000\right)i.

$$\frac{32000}{800}$$

Divide 32000 by 800 to get 40.

$$40$$

Real Part

$$40$$

Solution Steps

Multiply 20+20i times -40i.

$$Re(\frac{20\times \left(-40i\right)+20\left(-40\right)i^{2}}{20-20i})$$

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

$$Re(\frac{20\times \left(-40i\right)+20\left(-40\right)\left(-1\right)}{20-20i})$$

Do the multiplications in 20\times \left(-40i\right)+20\left(-40\right)\left(-1\right). Reorder the terms.

$$Re(\frac{800-800i}{20-20i})$$

Multiply both numerator and denominator of \frac{800-800i}{20-20i} by the complex conjugate of the denominator, 20+20i.

$$Re(\frac{\left(800-800i\right)\left(20+20i\right)}{\left(20-20i\right)\left(20+20i\right)})$$

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

$$Re(\frac{\left(800-800i\right)\left(20+20i\right)}{20^{2}-20^{2}i^{2}})$$

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

$$Re(\frac{\left(800-800i\right)\left(20+20i\right)}{800})$$

Multiply complex numbers 800-800i and 20+20i like you multiply binomials.

$$Re(\frac{800\times 20+800\times \left(20i\right)-800i\times 20-800\times 20i^{2}}{800})$$

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

$$Re(\frac{800\times 20+800\times \left(20i\right)-800i\times 20-800\times 20\left(-1\right)}{800})$$

Do the multiplications in 800\times 20+800\times \left(20i\right)-800i\times 20-800\times 20\left(-1\right).

$$Re(\frac{16000+16000i-16000i+16000}{800})$$

Combine the real and imaginary parts in 16000+16000i-16000i+16000.

$$Re(\frac{16000+16000+\left(16000-16000\right)i}{800})$$

Do the additions in 16000+16000+\left(16000-16000\right)i.

$$Re(\frac{32000}{800})$$

Divide 32000 by 800 to get 40.

$$Re(40)$$

The real part of 40 is 40.

$$40$$