Solve 9-7i/3+i*(3-i)/(3-i)

$$\frac { 9 - 7 i } { 3 + i } \times \frac { ( 3 - i ) } { ( 3 - i ) }$$

Evaluate

$$2-3i$$

Solution Steps

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

$$\frac{9-7i}{3+i}\times 1$$

Multiply both numerator and denominator of \frac{9-7i}{3+i} by the complex conjugate of the denominator, 3-i.

$$\frac{\left(9-7i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}\times 1$$

Do the multiplications in \frac{\left(9-7i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}.

$$\frac{20-30i}{10}\times 1$$

Divide 20-30i by 10 to get 2-3i.

$$\left(2-3i\right)\times 1$$

Multiply 2-3i and 1 to get 2-3i.

$$2-3i$$

Real Part

$$2$$

Solution Steps

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

$$Re(\frac{9-7i}{3+i}\times 1)$$

Multiply both numerator and denominator of \frac{9-7i}{3+i} by the complex conjugate of the denominator, 3-i.

$$Re(\frac{\left(9-7i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}\times 1)$$

Do the multiplications in \frac{\left(9-7i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}.

$$Re(\frac{20-30i}{10}\times 1)$$

Divide 20-30i by 10 to get 2-3i.

$$Re(\left(2-3i\right)\times 1)$$

Multiply 2-3i and 1 to get 2-3i.

$$Re(2-3i)$$

The real part of 2-3i is 2.

$$2$$