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

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

Evaluate

$$-\frac{9}{10}-\frac{3}{10}i=-0.9-0.3i$$

Solution Steps

Calculate i to the power of 2 and get -1.

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

Do the additions in -1+3i+1.

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

Calculate i to the power of 2 and get -1.

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

Do the additions in -1+i+1.

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

Multiply 3i and i to get -3.

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

Calculate 1 to the power of 2 and get 1.

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

Do the additions in 1-i+2.

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

Multiply both numerator and denominator by the complex conjugate of the denominator, 3+i.

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

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

$$\frac{-9-3i}{10}$$

Divide -9-3i by 10 to get -\frac{9}{10}-\frac{3}{10}i.

$$-\frac{9}{10}-\frac{3}{10}i$$

Real Part

$$-\frac{9}{10} = -0.9$$

Solution Steps

Calculate i to the power of 2 and get -1.

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

Do the additions in -1+3i+1.

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

Calculate i to the power of 2 and get -1.

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

Do the additions in -1+i+1.

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

Multiply 3i and i to get -3.

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

Calculate 1 to the power of 2 and get 1.

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

Do the additions in 1-i+2.

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

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

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

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

$$Re(\frac{-9-3i}{10})$$

Divide -9-3i by 10 to get -\frac{9}{10}-\frac{3}{10}i.

$$Re(-\frac{9}{10}-\frac{3}{10}i)$$

The real part of -\frac{9}{10}-\frac{3}{10}i is -\frac{9}{10}.

$$-\frac{9}{10}$$