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

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

Evaluate

$$-\frac{79}{5}-\frac{237}{5}i=-15.8-47.4i$$

Solution Steps

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

$$-9-46i+\frac{1-i}{-1+i^{19}}-\frac{3\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right)$$

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

$$-9-46i+\frac{1-i}{-1-i}-\frac{3\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right)$$

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

$$-9-46i+\frac{\left(1-i\right)\left(-1+i\right)}{\left(-1-i\right)\left(-1+i\right)}-\frac{3\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right)$$

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

$$-9-46i+\frac{2i}{2}-\frac{3\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right)$$

Divide 2i by 2 to get i.

$$-9-46i+i-\frac{3\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right)$$

Do the additions in -9-46i+i.

$$-\frac{3\left(2+i\right)}{2-i}-9-45i-\left(2+i\right)\left(2-i\right)$$

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

$$-\frac{3\left(2+i\right)}{2-i}-9-45i-5$$

Do the additions.

$$-\frac{3\left(2+i\right)}{2-i}-14-45i$$

Multiply 3 and 2+i to get 6+3i.

$$-\frac{6+3i}{2-i}-14-45i$$

Multiply both numerator and denominator of \frac{6+3i}{2-i} by the complex conjugate of the denominator, 2+i.

$$-\frac{\left(6+3i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}-14-45i$$

Do the multiplications in \frac{\left(6+3i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}.

$$-\frac{9+12i}{5}-14-45i$$

Divide 9+12i by 5 to get \frac{9}{5}+\frac{12}{5}i.

$$-\frac{9}{5}-\frac{12}{5}i-14-45i$$

Do the additions.

$$-\frac{79}{5}-\frac{237}{5}i$$

Real Part

$$-\frac{79}{5} = -15\frac{4}{5} = -15.8$$

Solution Steps

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

$$Re(-9-46i+\frac{1-i}{-1+i^{19}}-\frac{3\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right))$$

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

$$Re(-9-46i+\frac{1-i}{-1-i}-\frac{3\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right))$$

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

$$Re(-9-46i+\frac{\left(1-i\right)\left(-1+i\right)}{\left(-1-i\right)\left(-1+i\right)}-\frac{3\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right))$$

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

$$Re(-9-46i+\frac{2i}{2}-\frac{3\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right))$$

Divide 2i by 2 to get i.

$$Re(-9-46i+i-\frac{3\left(2+i\right)}{2-i}-\left(2+i\right)\left(2-i\right))$$

Do the additions in -9-46i+i.

$$Re(-\frac{3\left(2+i\right)}{2-i}-9-45i-\left(2+i\right)\left(2-i\right))$$

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

$$Re(-\frac{3\left(2+i\right)}{2-i}-9-45i-5)$$

Do the additions in -9-45i-5.

$$Re(-\frac{3\left(2+i\right)}{2-i}-14-45i)$$

Multiply 3 and 2+i to get 6+3i.

$$Re(-\frac{6+3i}{2-i}-14-45i)$$

Multiply both numerator and denominator of \frac{6+3i}{2-i} by the complex conjugate of the denominator, 2+i.

$$Re(-\frac{\left(6+3i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}-14-45i)$$

Do the multiplications in \frac{\left(6+3i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}.

$$Re(-\frac{9+12i}{5}-14-45i)$$

Divide 9+12i by 5 to get \frac{9}{5}+\frac{12}{5}i.

$$Re(-\frac{9}{5}-\frac{12}{5}i-14-45i)$$

Do the additions in -\frac{9}{5}-\frac{12}{5}i-14-45i.

$$Re(-\frac{79}{5}-\frac{237}{5}i)$$

The real part of -\frac{79}{5}-\frac{237}{5}i is -\frac{79}{5}.

$$-\frac{79}{5}$$