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

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

Evaluate

$$\frac{222}{25}-\frac{404}{25}i=8.88-16.16i$$

Solution Steps

Multiply 3 times 3-5i.

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

Do the multiplications in 3\times 3+3\times \left(-5i\right).

$$9-15i+\frac{3-5i}{4+3i}$$

Multiply both numerator and denominator of \frac{3-5i}{4+3i} by the complex conjugate of the denominator, 4-3i.

$$9-15i+\frac{\left(3-5i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}$$

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

$$9-15i+\frac{\left(3-5i\right)\left(4-3i\right)}{4^{2}-3^{2}i^{2}}$$

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

$$9-15i+\frac{\left(3-5i\right)\left(4-3i\right)}{25}$$

Multiply complex numbers 3-5i and 4-3i like you multiply binomials.

$$9-15i+\frac{3\times 4+3\times \left(-3i\right)-5i\times 4-5\left(-3\right)i^{2}}{25}$$

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

$$9-15i+\frac{3\times 4+3\times \left(-3i\right)-5i\times 4-5\left(-3\right)\left(-1\right)}{25}$$

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

$$9-15i+\frac{12-9i-20i-15}{25}$$

Combine the real and imaginary parts in 12-9i-20i-15.

$$9-15i+\frac{12-15+\left(-9-20\right)i}{25}$$

Do the additions in 12-15+\left(-9-20\right)i.

$$9-15i+\frac{-3-29i}{25}$$

Divide -3-29i by 25 to get -\frac{3}{25}-\frac{29}{25}i.

$$9-15i+\left(-\frac{3}{25}-\frac{29}{25}i\right)$$

Combine the real and imaginary parts.

$$9-\frac{3}{25}+\left(-15-\frac{29}{25}\right)i$$

Do the additions.

$$\frac{222}{25}-\frac{404}{25}i$$

Real Part

$$\frac{222}{25} = 8\frac{22}{25} = 8.88$$

Solution Steps

Multiply 3 times 3-5i.

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

Do the multiplications in 3\times 3+3\times \left(-5i\right).

$$Re(9-15i+\frac{3-5i}{4+3i})$$

Multiply both numerator and denominator of \frac{3-5i}{4+3i} by the complex conjugate of the denominator, 4-3i.

$$Re(9-15i+\frac{\left(3-5i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)})$$

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

$$Re(9-15i+\frac{\left(3-5i\right)\left(4-3i\right)}{4^{2}-3^{2}i^{2}})$$

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

$$Re(9-15i+\frac{\left(3-5i\right)\left(4-3i\right)}{25})$$

Multiply complex numbers 3-5i and 4-3i like you multiply binomials.

$$Re(9-15i+\frac{3\times 4+3\times \left(-3i\right)-5i\times 4-5\left(-3\right)i^{2}}{25})$$

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

$$Re(9-15i+\frac{3\times 4+3\times \left(-3i\right)-5i\times 4-5\left(-3\right)\left(-1\right)}{25})$$

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

$$Re(9-15i+\frac{12-9i-20i-15}{25})$$

Combine the real and imaginary parts in 12-9i-20i-15.

$$Re(9-15i+\frac{12-15+\left(-9-20\right)i}{25})$$

Do the additions in 12-15+\left(-9-20\right)i.

$$Re(9-15i+\frac{-3-29i}{25})$$

Divide -3-29i by 25 to get -\frac{3}{25}-\frac{29}{25}i.

$$Re(9-15i+\left(-\frac{3}{25}-\frac{29}{25}i\right))$$

Combine the real and imaginary parts in 9-15i+\left(-\frac{3}{25}-\frac{29}{25}i\right).

$$Re(9-\frac{3}{25}+\left(-15-\frac{29}{25}\right)i)$$

Do the additions in 9-\frac{3}{25}+\left(-15-\frac{29}{25}\right)i.

$$Re(\frac{222}{25}-\frac{404}{25}i)$$

The real part of \frac{222}{25}-\frac{404}{25}i is \frac{222}{25}.

$$\frac{222}{25}$$