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

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

Evaluate

$$\frac{9}{5}i=1.8i$$

Solution Steps

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

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

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

$$\left(1+i\right)\left(\frac{6-2i}{10}-\frac{1+i}{3-i}\right)+i\left(i+2\right)$$

Divide 6-2i by 10 to get \frac{3}{5}-\frac{1}{5}i.

$$\left(1+i\right)\left(\frac{3}{5}-\frac{1}{5}i-\frac{1+i}{3-i}\right)+i\left(i+2\right)$$

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

$$\left(1+i\right)\left(\frac{3}{5}-\frac{1}{5}i-\frac{\left(1+i\right)\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}\right)+i\left(i+2\right)$$

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

$$\left(1+i\right)\left(\frac{3}{5}-\frac{1}{5}i-\frac{2+4i}{10}\right)+i\left(i+2\right)$$

Divide 2+4i by 10 to get \frac{1}{5}+\frac{2}{5}i.

$$\left(1+i\right)\left(\frac{3}{5}-\frac{1}{5}i+\left(-\frac{1}{5}-\frac{2}{5}i\right)\right)+i\left(i+2\right)$$

Add \frac{3}{5}-\frac{1}{5}i and -\frac{1}{5}-\frac{2}{5}i to get \frac{2}{5}-\frac{3}{5}i.

$$\left(1+i\right)\left(\frac{2}{5}-\frac{3}{5}i\right)+i\left(i+2\right)$$

Multiply 1+i and \frac{2}{5}-\frac{3}{5}i to get 1-\frac{1}{5}i.

$$1-\frac{1}{5}i+i\left(i+2\right)$$

Use the distributive property to multiply i by i+2.

$$1-\frac{1}{5}i+\left(-1+2i\right)$$

Do the additions.

$$\frac{9}{5}i$$

Real Part

$$0$$

Solution Steps

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

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

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

$$Re(\left(1+i\right)\left(\frac{6-2i}{10}-\frac{1+i}{3-i}\right)+i\left(i+2\right))$$

Divide 6-2i by 10 to get \frac{3}{5}-\frac{1}{5}i.

$$Re(\left(1+i\right)\left(\frac{3}{5}-\frac{1}{5}i-\frac{1+i}{3-i}\right)+i\left(i+2\right))$$

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

$$Re(\left(1+i\right)\left(\frac{3}{5}-\frac{1}{5}i-\frac{\left(1+i\right)\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}\right)+i\left(i+2\right))$$

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

$$Re(\left(1+i\right)\left(\frac{3}{5}-\frac{1}{5}i-\frac{2+4i}{10}\right)+i\left(i+2\right))$$

Divide 2+4i by 10 to get \frac{1}{5}+\frac{2}{5}i.

$$Re(\left(1+i\right)\left(\frac{3}{5}-\frac{1}{5}i+\left(-\frac{1}{5}-\frac{2}{5}i\right)\right)+i\left(i+2\right))$$

Add \frac{3}{5}-\frac{1}{5}i and -\frac{1}{5}-\frac{2}{5}i to get \frac{2}{5}-\frac{3}{5}i.

$$Re(\left(1+i\right)\left(\frac{2}{5}-\frac{3}{5}i\right)+i\left(i+2\right))$$

Multiply 1+i and \frac{2}{5}-\frac{3}{5}i to get 1-\frac{1}{5}i.

$$Re(1-\frac{1}{5}i+i\left(i+2\right))$$

Use the distributive property to multiply i by i+2.

$$Re(1-\frac{1}{5}i+\left(-1+2i\right))$$

Do the additions in 1-\frac{1}{5}i-1+2i.

$$Re(\frac{9}{5}i)$$

The real part of \frac{9}{5}i is 0.

$$0$$