Solve s(2-3i)+(2)/(4-3i)*(2-5i)

$$s ( 2 - 3 i ) + ( 2 ) / ( 4 - 3 i ) * ( 2 - 5 i )$$

Evaluate

$$\left(2-3i\right)s+\left(\frac{46}{25}-\frac{28}{25}i\right)$$

Solution Steps

Multiply both numerator and denominator of \frac{2}{4-3i} by the complex conjugate of the denominator, 4+3i.

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

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

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

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

$$s\left(2-3i\right)+\frac{2\left(4+3i\right)}{25}\left(2-5i\right)$$

Multiply 2 times 4+3i.

$$s\left(2-3i\right)+\frac{2\times 4+2\times \left(3i\right)}{25}\left(2-5i\right)$$

Do the multiplications in 2\times 4+2\times \left(3i\right).

$$s\left(2-3i\right)+\frac{8+6i}{25}\left(2-5i\right)$$

Divide 8+6i by 25 to get \frac{8}{25}+\frac{6}{25}i.

$$s\left(2-3i\right)+\left(\frac{8}{25}+\frac{6}{25}i\right)\left(2-5i\right)$$

Multiply complex numbers \frac{8}{25}+\frac{6}{25}i and 2-5i like you multiply binomials.

$$s\left(2-3i\right)+\frac{8}{25}\times 2+\frac{8}{25}\times \left(-5i\right)+\frac{6}{25}i\times 2+\frac{6}{25}\left(-5\right)i^{2}$$

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

$$s\left(2-3i\right)+\frac{8}{25}\times 2+\frac{8}{25}\times \left(-5i\right)+\frac{6}{25}i\times 2+\frac{6}{25}\left(-5\right)\left(-1\right)$$

Do the multiplications in \frac{8}{25}\times 2+\frac{8}{25}\times \left(-5i\right)+\frac{6}{25}i\times 2+\frac{6}{25}\left(-5\right)\left(-1\right).

$$s\left(2-3i\right)+\frac{16}{25}-\frac{8}{5}i+\frac{12}{25}i+\frac{6}{5}$$

Combine the real and imaginary parts in \frac{16}{25}-\frac{8}{5}i+\frac{12}{25}i+\frac{6}{5}.

$$s\left(2-3i\right)+\frac{16}{25}+\frac{6}{5}+\left(-\frac{8}{5}+\frac{12}{25}\right)i$$

Do the additions in \frac{16}{25}+\frac{6}{5}+\left(-\frac{8}{5}+\frac{12}{25}\right)i.

$$s\left(2-3i\right)+\left(\frac{46}{25}-\frac{28}{25}i\right)$$

Expand

$$\left(2-3i\right)s+\left(\frac{46}{25}-\frac{28}{25}i\right)$$

Solution Steps

Multiply both numerator and denominator of \frac{2}{4-3i} by the complex conjugate of the denominator, 4+3i.

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

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

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

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

$$s\left(2-3i\right)+\frac{2\left(4+3i\right)}{25}\left(2-5i\right)$$

Multiply 2 times 4+3i.

$$s\left(2-3i\right)+\frac{2\times 4+2\times \left(3i\right)}{25}\left(2-5i\right)$$

Do the multiplications in 2\times 4+2\times \left(3i\right).

$$s\left(2-3i\right)+\frac{8+6i}{25}\left(2-5i\right)$$

Divide 8+6i by 25 to get \frac{8}{25}+\frac{6}{25}i.

$$s\left(2-3i\right)+\left(\frac{8}{25}+\frac{6}{25}i\right)\left(2-5i\right)$$

Multiply complex numbers \frac{8}{25}+\frac{6}{25}i and 2-5i like you multiply binomials.

$$s\left(2-3i\right)+\frac{8}{25}\times 2+\frac{8}{25}\times \left(-5i\right)+\frac{6}{25}i\times 2+\frac{6}{25}\left(-5\right)i^{2}$$

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

$$s\left(2-3i\right)+\frac{8}{25}\times 2+\frac{8}{25}\times \left(-5i\right)+\frac{6}{25}i\times 2+\frac{6}{25}\left(-5\right)\left(-1\right)$$

Do the multiplications in \frac{8}{25}\times 2+\frac{8}{25}\times \left(-5i\right)+\frac{6}{25}i\times 2+\frac{6}{25}\left(-5\right)\left(-1\right).

$$s\left(2-3i\right)+\frac{16}{25}-\frac{8}{5}i+\frac{12}{25}i+\frac{6}{5}$$

Combine the real and imaginary parts in \frac{16}{25}-\frac{8}{5}i+\frac{12}{25}i+\frac{6}{5}.

$$s\left(2-3i\right)+\frac{16}{25}+\frac{6}{5}+\left(-\frac{8}{5}+\frac{12}{25}\right)i$$

Do the additions in \frac{16}{25}+\frac{6}{5}+\left(-\frac{8}{5}+\frac{12}{25}\right)i.

$$s\left(2-3i\right)+\left(\frac{46}{25}-\frac{28}{25}i\right)$$