Solve (-2+3i)^2*i^5+13-26i/3+2i-(1-i)*(1+i)

$$( - 2 + 3 i ) ^ { 2 } \cdot i ^ { 5 } + \frac { 13 - 26 i } { 3 + 2 i } - ( 1 - i ) \cdot ( 1 + i )$$

Evaluate

$$9-13i$$

Solution Steps

Calculate -2+3i to the power of 2 and get -5-12i.

$$\left(-5-12i\right)i^{5}+\frac{13-26i}{3+2i}-\left(1-i\right)\left(1+i\right)$$

Calculate i to the power of 5 and get i.

$$\left(-5-12i\right)i+\frac{13-26i}{3+2i}-\left(1-i\right)\left(1+i\right)$$

Multiply -5-12i and i to get 12-5i.

$$12-5i+\frac{13-26i}{3+2i}-\left(1-i\right)\left(1+i\right)$$

Multiply both numerator and denominator of \frac{13-26i}{3+2i} by the complex conjugate of the denominator, 3-2i.

$$12-5i+\frac{\left(13-26i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)}-\left(1-i\right)\left(1+i\right)$$

Do the multiplications in \frac{\left(13-26i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)}.

$$12-5i+\frac{-13-104i}{13}-\left(1-i\right)\left(1+i\right)$$

Divide -13-104i by 13 to get -1-8i.

$$12-5i+\left(-1-8i\right)-\left(1-i\right)\left(1+i\right)$$

Add 12-5i and -1-8i to get 11-13i.

$$11-13i-\left(1-i\right)\left(1+i\right)$$

Multiply 1-i and 1+i to get 2.

$$11-13i-2$$

Subtract 2 from 11-13i to get 9-13i.

$$9-13i$$

Real Part

$$9$$

Solution Steps

Calculate -2+3i to the power of 2 and get -5-12i.

$$Re(\left(-5-12i\right)i^{5}+\frac{13-26i}{3+2i}-\left(1-i\right)\left(1+i\right))$$

Calculate i to the power of 5 and get i.

$$Re(\left(-5-12i\right)i+\frac{13-26i}{3+2i}-\left(1-i\right)\left(1+i\right))$$

Multiply -5-12i and i to get 12-5i.

$$Re(12-5i+\frac{13-26i}{3+2i}-\left(1-i\right)\left(1+i\right))$$

Multiply both numerator and denominator of \frac{13-26i}{3+2i} by the complex conjugate of the denominator, 3-2i.

$$Re(12-5i+\frac{\left(13-26i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)}-\left(1-i\right)\left(1+i\right))$$

Do the multiplications in \frac{\left(13-26i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)}.

$$Re(12-5i+\frac{-13-104i}{13}-\left(1-i\right)\left(1+i\right))$$

Divide -13-104i by 13 to get -1-8i.

$$Re(12-5i+\left(-1-8i\right)-\left(1-i\right)\left(1+i\right))$$

Add 12-5i and -1-8i to get 11-13i.

$$Re(11-13i-\left(1-i\right)\left(1+i\right))$$

Multiply 1-i and 1+i to get 2.

$$Re(11-13i-2)$$

Subtract 2 from 11-13i to get 9-13i.

$$Re(9-13i)$$

The real part of 9-13i is 9.

$$9$$