Solve (3/5-4/5i)(3/5+4/5i)=

$$( \frac { 3 } { 5 } - \frac { 4 } { 5 } i ) ( \frac { 3 } { 5 } + \frac { 4 } { 5 } i ) =$$

Evaluate

$$1$$

Solution Steps

Multiply complex numbers \frac{3}{5}-\frac{4}{5}i and \frac{3}{5}+\frac{4}{5}i like you multiply binomials.

$$\frac{3}{5}\times \frac{3}{5}+\frac{3}{5}\times \left(\frac{4}{5}i\right)-\frac{4}{5}i\times \frac{3}{5}-\frac{4}{5}\times \frac{4}{5}i^{2}$$

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

$$\frac{3}{5}\times \frac{3}{5}+\frac{3}{5}\times \left(\frac{4}{5}i\right)-\frac{4}{5}i\times \frac{3}{5}-\frac{4}{5}\times \frac{4}{5}\left(-1\right)$$

Do the multiplications.

$$\frac{9}{25}+\frac{12}{25}i-\frac{12}{25}i+\frac{16}{25}$$

Combine the real and imaginary parts.

$$\frac{9}{25}+\frac{16}{25}+\left(\frac{12}{25}-\frac{12}{25}\right)i$$

Do the additions.

$$1$$

Real Part

$$1$$

Solution Steps

Multiply complex numbers \frac{3}{5}-\frac{4}{5}i and \frac{3}{5}+\frac{4}{5}i like you multiply binomials.

$$Re(\frac{3}{5}\times \frac{3}{5}+\frac{3}{5}\times \left(\frac{4}{5}i\right)-\frac{4}{5}i\times \frac{3}{5}-\frac{4}{5}\times \frac{4}{5}i^{2})$$

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

$$Re(\frac{3}{5}\times \frac{3}{5}+\frac{3}{5}\times \left(\frac{4}{5}i\right)-\frac{4}{5}i\times \frac{3}{5}-\frac{4}{5}\times \frac{4}{5}\left(-1\right))$$

Do the multiplications in \frac{3}{5}\times \frac{3}{5}+\frac{3}{5}\times \left(\frac{4}{5}i\right)-\frac{4}{5}i\times \frac{3}{5}-\frac{4}{5}\times \frac{4}{5}\left(-1\right).

$$Re(\frac{9}{25}+\frac{12}{25}i-\frac{12}{25}i+\frac{16}{25})$$

Combine the real and imaginary parts in \frac{9}{25}+\frac{12}{25}i-\frac{12}{25}i+\frac{16}{25}.

$$Re(\frac{9}{25}+\frac{16}{25}+\left(\frac{12}{25}-\frac{12}{25}\right)i)$$

Do the additions in \frac{9}{25}+\frac{16}{25}+\left(\frac{12}{25}-\frac{12}{25}\right)i.

$$Re(1)$$

The real part of 1 is 1.

$$1$$