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

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

Evaluate

$$-\frac{2}{3}+\frac{2}{3}i\approx -0.666666667+0.666666667i$$

Solution Steps

Multiply both numerator and denominator of \frac{1+i}{3i} by imaginary unit i.

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

Divide -1+i by -3 to get \frac{1}{3}-\frac{1}{3}i.

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

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

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

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

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

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

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

Add \frac{1}{3}-\frac{1}{3}i and -1+i to get -\frac{2}{3}+\frac{2}{3}i.

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

Real Part

$$-\frac{2}{3} = -0.6666666666666666$$

Solution Steps

Multiply both numerator and denominator of \frac{1+i}{3i} by imaginary unit i.

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

Divide -1+i by -3 to get \frac{1}{3}-\frac{1}{3}i.

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

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

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

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

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

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

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

Add \frac{1}{3}-\frac{1}{3}i and -1+i to get -\frac{2}{3}+\frac{2}{3}i.

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

The real part of -\frac{2}{3}+\frac{2}{3}i is -\frac{2}{3}.

$$-\frac{2}{3}$$