Solve 6-3i/4+2i*(4+2i)/(4+2i)

$$\frac { 6 - 3 i } { 4 + 2 i } \cdot \frac { ( 4 + 2 i ) } { ( 4 + 2 i ) }$$

Evaluate

$$\frac{9}{10}-\frac{6}{5}i=0.9-1.2i$$

Solution Steps

Divide 4+2i by 4+2i to get 1.

$$\frac{6-3i}{4+2i}\times 1$$

Multiply both numerator and denominator of \frac{6-3i}{4+2i} by the complex conjugate of the denominator, 4-2i.

$$\frac{\left(6-3i\right)\left(4-2i\right)}{\left(4+2i\right)\left(4-2i\right)}\times 1$$

Do the multiplications in \frac{\left(6-3i\right)\left(4-2i\right)}{\left(4+2i\right)\left(4-2i\right)}.

$$\frac{18-24i}{20}\times 1$$

Divide 18-24i by 20 to get \frac{9}{10}-\frac{6}{5}i.

$$\left(\frac{9}{10}-\frac{6}{5}i\right)\times 1$$

Multiply \frac{9}{10}-\frac{6}{5}i and 1 to get \frac{9}{10}-\frac{6}{5}i.

$$\frac{9}{10}-\frac{6}{5}i$$

Real Part

$$\frac{9}{10} = 0.9$$

Solution Steps

Divide 4+2i by 4+2i to get 1.

$$Re(\frac{6-3i}{4+2i}\times 1)$$

Multiply both numerator and denominator of \frac{6-3i}{4+2i} by the complex conjugate of the denominator, 4-2i.

$$Re(\frac{\left(6-3i\right)\left(4-2i\right)}{\left(4+2i\right)\left(4-2i\right)}\times 1)$$

Do the multiplications in \frac{\left(6-3i\right)\left(4-2i\right)}{\left(4+2i\right)\left(4-2i\right)}.

$$Re(\frac{18-24i}{20}\times 1)$$

Divide 18-24i by 20 to get \frac{9}{10}-\frac{6}{5}i.

$$Re(\left(\frac{9}{10}-\frac{6}{5}i\right)\times 1)$$

Multiply \frac{9}{10}-\frac{6}{5}i and 1 to get \frac{9}{10}-\frac{6}{5}i.

$$Re(\frac{9}{10}-\frac{6}{5}i)$$

The real part of \frac{9}{10}-\frac{6}{5}i is \frac{9}{10}.

$$\frac{9}{10}$$