Solve (3-2i)^3+1/-1+i^19-9(2+i)/2-i-(2+i)(2-i)

$$( 3 - 2 i ) ^ { 3 } + \frac { 1 } { - 1 + i ^ { 19 } } - \frac { 9 ( 2 + i ) } { 2 - i } - ( 2 + i ) ( 2 - i )$$

Evaluate

$$-\frac{199}{10}-\frac{527}{10}i=-19.9-52.7i$$

Solution Steps

Calculate 3-2i to the power of 3 and get -9-46i.

$$-9-46i+\frac{1}{-1+i^{19}}-\frac{9\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right)$$

Calculate i to the power of 19 and get -i.

$$-9-46i+\frac{1}{-1-i}-\frac{9\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right)$$

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

$$-9-46i+\frac{1\left(-1+i\right)}{\left(-1-i\right)\left(-1+i\right)}-\frac{9\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right)$$

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

$$-9-46i+\frac{-1+i}{2}-\frac{9\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right)$$

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

$$-9-46i+\left(-\frac{1}{2}+\frac{1}{2}i\right)-\frac{9\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right)$$

Do the additions in -9-46i+\left(-\frac{1}{2}+\frac{1}{2}i\right).

$$-\frac{9\left(2+i\right)}{2-i}-\frac{19}{2}-\frac{91}{2}i-\left(2+i\right)\left(2-i\right)$$

Multiply 2+i and 2-i to get 5.

$$-\frac{9\left(2+i\right)}{2-i}-\frac{19}{2}-\frac{91}{2}i-5$$

Do the additions.

$$-\frac{9\left(2+i\right)}{2-i}-\frac{29}{2}-\frac{91}{2}i$$

Multiply 9 and 2+i to get 18+9i.

$$-\frac{18+9i}{2-i}-\frac{29}{2}-\frac{91}{2}i$$

Multiply both numerator and denominator of \frac{18+9i}{2-i} by the complex conjugate of the denominator, 2+i.

$$-\frac{\left(18+9i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}-\frac{29}{2}-\frac{91}{2}i$$

Do the multiplications in \frac{\left(18+9i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}.

$$-\frac{27+36i}{5}-\frac{29}{2}-\frac{91}{2}i$$

Divide 27+36i by 5 to get \frac{27}{5}+\frac{36}{5}i.

$$-\frac{27}{5}-\frac{36}{5}i-\frac{29}{2}-\frac{91}{2}i$$

Do the additions.

$$-\frac{199}{10}-\frac{527}{10}i$$

Real Part

$$-\frac{199}{10} = -19\frac{9}{10} = -19.9$$

Solution Steps

Calculate 3-2i to the power of 3 and get -9-46i.

$$Re(-9-46i+\frac{1}{-1+i^{19}}-\frac{9\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right))$$

Calculate i to the power of 19 and get -i.

$$Re(-9-46i+\frac{1}{-1-i}-\frac{9\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right))$$

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

$$Re(-9-46i+\frac{1\left(-1+i\right)}{\left(-1-i\right)\left(-1+i\right)}-\frac{9\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right))$$

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

$$Re(-9-46i+\frac{-1+i}{2}-\frac{9\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right))$$

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

$$Re(-9-46i+\left(-\frac{1}{2}+\frac{1}{2}i\right)-\frac{9\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right))$$

Do the additions in -9-46i+\left(-\frac{1}{2}+\frac{1}{2}i\right).

$$Re(-\frac{9\left(2+i\right)}{2-i}-\frac{19}{2}-\frac{91}{2}i-\left(2+i\right)\left(2-i\right))$$

Multiply 2+i and 2-i to get 5.

$$Re(-\frac{9\left(2+i\right)}{2-i}-\frac{19}{2}-\frac{91}{2}i-5)$$

Do the additions in -\frac{19}{2}-\frac{91}{2}i-5.

$$Re(-\frac{9\left(2+i\right)}{2-i}-\frac{29}{2}-\frac{91}{2}i)$$

Multiply 9 and 2+i to get 18+9i.

$$Re(-\frac{18+9i}{2-i}-\frac{29}{2}-\frac{91}{2}i)$$

Multiply both numerator and denominator of \frac{18+9i}{2-i} by the complex conjugate of the denominator, 2+i.

$$Re(-\frac{\left(18+9i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}-\frac{29}{2}-\frac{91}{2}i)$$

Do the multiplications in \frac{\left(18+9i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}.

$$Re(-\frac{27+36i}{5}-\frac{29}{2}-\frac{91}{2}i)$$

Divide 27+36i by 5 to get \frac{27}{5}+\frac{36}{5}i.

$$Re(-\frac{27}{5}-\frac{36}{5}i-\frac{29}{2}-\frac{91}{2}i)$$

Do the additions in -\frac{27}{5}-\frac{36}{5}i-\frac{29}{2}-\frac{91}{2}i.

$$Re(-\frac{199}{10}-\frac{527}{10}i)$$

The real part of -\frac{199}{10}-\frac{527}{10}i is -\frac{199}{10}.

$$-\frac{199}{10}$$