Solve -1+2i/3-i*3+i/3+i=

$$\frac { - 1 + 2 i } { 3 - i } \cdot \frac { 3 + i } { 3 + i } =$$

Evaluate

$$-\frac{1}{2}+\frac{1}{2}i=-0.5+0.5i$$

Solution Steps

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

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

Multiply both numerator and denominator of \frac{-1+2i}{3-i} by the complex conjugate of the denominator, 3+i.

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

Do the multiplications in \frac{\left(-1+2i\right)\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}.

$$\frac{-5+5i}{10}\times 1$$

Divide -5+5i by 10 to get -\frac{1}{2}+\frac{1}{2}i.

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

Multiply -\frac{1}{2}+\frac{1}{2}i and 1 to get -\frac{1}{2}+\frac{1}{2}i.

$$-\frac{1}{2}+\frac{1}{2}i$$

Real Part

$$-\frac{1}{2} = -0.5$$

Solution Steps

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

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

Multiply both numerator and denominator of \frac{-1+2i}{3-i} by the complex conjugate of the denominator, 3+i.

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

Do the multiplications in \frac{\left(-1+2i\right)\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}.

$$Re(\frac{-5+5i}{10}\times 1)$$

Divide -5+5i by 10 to get -\frac{1}{2}+\frac{1}{2}i.

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

Multiply -\frac{1}{2}+\frac{1}{2}i and 1 to get -\frac{1}{2}+\frac{1}{2}i.

$$Re(-\frac{1}{2}+\frac{1}{2}i)$$

The real part of -\frac{1}{2}+\frac{1}{2}i is -\frac{1}{2}.

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