Solve (4+i)-(3-5i)(-2+5i)

$$( 4 + i ) - ( 3 - 5 i ) ( - 2 + 5 i )$$

Evaluate

$$-15-24i$$

Solution Steps

Multiply complex numbers 3-5i and -2+5i like you multiply binomials.

$$4+i-\left(3\left(-2\right)+3\times \left(5i\right)-5i\left(-2\right)-5\times 5i^{2}\right)$$

By definition, i^{2} is -1.

$$4+i-\left(3\left(-2\right)+3\times \left(5i\right)-5i\left(-2\right)-5\times 5\left(-1\right)\right)$$

Do the multiplications in 3\left(-2\right)+3\times \left(5i\right)-5i\left(-2\right)-5\times 5\left(-1\right).

$$4+i-\left(-6+15i+10i+25\right)$$

Combine the real and imaginary parts in -6+15i+10i+25.

$$4+i-\left(-6+25+\left(15+10\right)i\right)$$

Do the additions in -6+25+\left(15+10\right)i.

$$4+i-\left(19+25i\right)$$

Subtract 19+25i from 4+i by subtracting corresponding real and imaginary parts.

$$4-19+\left(1-25\right)i$$

Subtract 19 from 4. Subtract 25 from 1.

$$-15-24i$$

Real Part

$$-15$$

Solution Steps

Multiply complex numbers 3-5i and -2+5i like you multiply binomials.

$$Re(4+i-\left(3\left(-2\right)+3\times \left(5i\right)-5i\left(-2\right)-5\times 5i^{2}\right))$$

By definition, i^{2} is -1.

$$Re(4+i-\left(3\left(-2\right)+3\times \left(5i\right)-5i\left(-2\right)-5\times 5\left(-1\right)\right))$$

Do the multiplications in 3\left(-2\right)+3\times \left(5i\right)-5i\left(-2\right)-5\times 5\left(-1\right).

$$Re(4+i-\left(-6+15i+10i+25\right))$$

Combine the real and imaginary parts in -6+15i+10i+25.

$$Re(4+i-\left(-6+25+\left(15+10\right)i\right))$$

Do the additions in -6+25+\left(15+10\right)i.

$$Re(4+i-\left(19+25i\right))$$

Subtract 19+25i from 4+i by subtracting corresponding real and imaginary parts.

$$Re(4-19+\left(1-25\right)i)$$

Subtract 19 from 4. Subtract 25 from 1.

$$Re(-15-24i)$$

The real part of -15-24i is -15.

$$-15$$