Solve (1+2i)^2*i^47/(3-2i)-(2+i)

$$\frac { ( 1 + 2 i ) ^ { 2 } \cdot i ^ { 47 } } { ( 3 - 2 i ) - ( 2 + i ) }$$

Evaluate

$$-\frac{1}{2}+\frac{3}{2}i=-0.5+1.5i$$

Solution Steps

Calculate 1+2i to the power of 2 and get -3+4i.

$$\frac{\left(-3+4i\right)i^{47}}{3-2i-\left(2+i\right)}$$

Calculate i to the power of 47 and get -i.

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

Multiply -3+4i and -i to get 4+3i.

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

Subtract 2+i from 3-2i to get 1-3i.

$$\frac{4+3i}{1-3i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 1+3i.

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

Do the multiplications in \frac{\left(4+3i\right)\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)}.

$$\frac{-5+15i}{10}$$

Divide -5+15i by 10 to get -\frac{1}{2}+\frac{3}{2}i.

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

Real Part

$$-\frac{1}{2} = -0.5$$

Solution Steps

Calculate 1+2i to the power of 2 and get -3+4i.

$$Re(\frac{\left(-3+4i\right)i^{47}}{3-2i-\left(2+i\right)})$$

Calculate i to the power of 47 and get -i.

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

Multiply -3+4i and -i to get 4+3i.

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

Subtract 2+i from 3-2i to get 1-3i.

$$Re(\frac{4+3i}{1-3i})$$

Multiply both numerator and denominator of \frac{4+3i}{1-3i} by the complex conjugate of the denominator, 1+3i.

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

Do the multiplications in \frac{\left(4+3i\right)\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)}.

$$Re(\frac{-5+15i}{10})$$

Divide -5+15i by 10 to get -\frac{1}{2}+\frac{3}{2}i.

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

The real part of -\frac{1}{2}+\frac{3}{2}i is -\frac{1}{2}.

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