Solve (2-i)2+4i/1+i

$$( 2 - i ) \frac { 2 + 4 i } { 1 + i }$$

Evaluate

$$7-i$$

Solution Steps

Multiply both numerator and denominator of \frac{2+4i}{1+i} by the complex conjugate of the denominator, 1-i.

$$\left(2-i\right)\times \frac{\left(2+4i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(2-i\right)\times \frac{\left(2+4i\right)\left(1-i\right)}{1^{2}-i^{2}}$$

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

$$\left(2-i\right)\times \frac{\left(2+4i\right)\left(1-i\right)}{2}$$

Multiply complex numbers 2+4i and 1-i like you multiply binomials.

$$\left(2-i\right)\times \frac{2\times 1+2\left(-i\right)+4i\times 1+4\left(-1\right)i^{2}}{2}$$

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

$$\left(2-i\right)\times \frac{2\times 1+2\left(-i\right)+4i\times 1+4\left(-1\right)\left(-1\right)}{2}$$

Do the multiplications in 2\times 1+2\left(-i\right)+4i\times 1+4\left(-1\right)\left(-1\right).

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

Combine the real and imaginary parts in 2-2i+4i+4.

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

Do the additions in 2+4+\left(-2+4\right)i.

$$\left(2-i\right)\times \frac{6+2i}{2}$$

Divide 6+2i by 2 to get 3+i.

$$\left(2-i\right)\left(3+i\right)$$

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

$$2\times 3+2i-i\times 3-i^{2}$$

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

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

Do the multiplications.

$$6+2i-3i+1$$

Combine the real and imaginary parts.

$$6+1+\left(2-3\right)i$$

Do the additions.

$$7-i$$

Real Part

$$7$$

Solution Steps

Multiply both numerator and denominator of \frac{2+4i}{1+i} by the complex conjugate of the denominator, 1-i.

$$Re(\left(2-i\right)\times \frac{\left(2+4i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)})$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\left(2-i\right)\times \frac{\left(2+4i\right)\left(1-i\right)}{1^{2}-i^{2}})$$

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

$$Re(\left(2-i\right)\times \frac{\left(2+4i\right)\left(1-i\right)}{2})$$

Multiply complex numbers 2+4i and 1-i like you multiply binomials.

$$Re(\left(2-i\right)\times \frac{2\times 1+2\left(-i\right)+4i\times 1+4\left(-1\right)i^{2}}{2})$$

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

$$Re(\left(2-i\right)\times \frac{2\times 1+2\left(-i\right)+4i\times 1+4\left(-1\right)\left(-1\right)}{2})$$

Do the multiplications in 2\times 1+2\left(-i\right)+4i\times 1+4\left(-1\right)\left(-1\right).

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

Combine the real and imaginary parts in 2-2i+4i+4.

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

Do the additions in 2+4+\left(-2+4\right)i.

$$Re(\left(2-i\right)\times \frac{6+2i}{2})$$

Divide 6+2i by 2 to get 3+i.

$$Re(\left(2-i\right)\left(3+i\right))$$

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

$$Re(2\times 3+2i-i\times 3-i^{2})$$

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

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

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

$$Re(6+2i-3i+1)$$

Combine the real and imaginary parts in 6+2i-3i+1.

$$Re(6+1+\left(2-3\right)i)$$

Do the additions in 6+1+\left(2-3\right)i.

$$Re(7-i)$$

The real part of 7-i is 7.

$$7$$