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

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

Evaluate

$$2$$

Solution Steps

Divide 3-2i by 3-2i to get 1.

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

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

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

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

$$\frac{\frac{26}{13}}{1}$$

Divide 26 by 13 to get 2.

$$\frac{2}{1}$$

Anything divided by one gives itself.

$$2$$

Real Part

$$2$$

Solution Steps

Divide 3-2i by 3-2i to get 1.

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

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

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

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

$$Re(\frac{\frac{26}{13}}{1})$$

Divide 26 by 13 to get 2.

$$Re(\frac{2}{1})$$

Anything divided by one gives itself.

$$Re(2)$$

The real part of 2 is 2.

$$2$$