Solve 5i^9left(2-3iright)/2-i

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

Evaluate

$$4+7i$$

Solution Steps

Calculate i to the power of 9 and get i.

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

Multiply 5i and 2-3i to get 15+10i.

$$\frac{15+10i}{2-i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 2+i.

$$\frac{\left(15+10i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}$$

Do the multiplications in \frac{\left(15+10i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}.

$$\frac{20+35i}{5}$$

Divide 20+35i by 5 to get 4+7i.

$$4+7i$$

Real Part

$$4$$

Solution Steps

Calculate i to the power of 9 and get i.

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

Multiply 5i and 2-3i to get 15+10i.

$$Re(\frac{15+10i}{2-i})$$

Multiply both numerator and denominator of \frac{15+10i}{2-i} by the complex conjugate of the denominator, 2+i.

$$Re(\frac{\left(15+10i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)})$$

Do the multiplications in \frac{\left(15+10i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}.

$$Re(\frac{20+35i}{5})$$

Divide 20+35i by 5 to get 4+7i.

$$Re(4+7i)$$

The real part of 4+7i is 4.

$$4$$