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

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

Evaluate

$$8-5i$$

Solution Steps

Multiply 2-3i and 1+i to get 5-i.

$$5-i-\left(1+2i\right)^{2}$$

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

$$5-i-\left(-3+4i\right)$$

Subtract -3+4i from 5-i to get 8-5i.

$$8-5i$$

Real Part

$$8$$

Solution Steps

Multiply 2-3i and 1+i to get 5-i.

$$Re(5-i-\left(1+2i\right)^{2})$$

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

$$Re(5-i-\left(-3+4i\right))$$

Subtract -3+4i from 5-i to get 8-5i.

$$Re(8-5i)$$

The real part of 8-5i is 8.

$$8$$