Solve 2-5i/1+8i*1-8i/1-8i=

$$\frac { 2 - 5 i } { 1 + 8 i } \cdot \frac { 1 - 8 i } { 1 - 8 i } =$$

Evaluate

$$-\frac{38}{65}-\frac{21}{65}i\approx -0.584615385-0.323076923i$$

Solution Steps

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

$$\frac{2-5i}{1+8i}\times 1$$

Multiply both numerator and denominator of \frac{2-5i}{1+8i} by the complex conjugate of the denominator, 1-8i.

$$\frac{\left(2-5i\right)\left(1-8i\right)}{\left(1+8i\right)\left(1-8i\right)}\times 1$$

Do the multiplications in \frac{\left(2-5i\right)\left(1-8i\right)}{\left(1+8i\right)\left(1-8i\right)}.

$$\frac{-38-21i}{65}\times 1$$

Divide -38-21i by 65 to get -\frac{38}{65}-\frac{21}{65}i.

$$\left(-\frac{38}{65}-\frac{21}{65}i\right)\times 1$$

Multiply -\frac{38}{65}-\frac{21}{65}i and 1 to get -\frac{38}{65}-\frac{21}{65}i.

$$-\frac{38}{65}-\frac{21}{65}i$$

Real Part

$$-\frac{38}{65} = -0.5846153846153846$$

Solution Steps

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

$$Re(\frac{2-5i}{1+8i}\times 1)$$

Multiply both numerator and denominator of \frac{2-5i}{1+8i} by the complex conjugate of the denominator, 1-8i.

$$Re(\frac{\left(2-5i\right)\left(1-8i\right)}{\left(1+8i\right)\left(1-8i\right)}\times 1)$$

Do the multiplications in \frac{\left(2-5i\right)\left(1-8i\right)}{\left(1+8i\right)\left(1-8i\right)}.

$$Re(\frac{-38-21i}{65}\times 1)$$

Divide -38-21i by 65 to get -\frac{38}{65}-\frac{21}{65}i.

$$Re(\left(-\frac{38}{65}-\frac{21}{65}i\right)\times 1)$$

Multiply -\frac{38}{65}-\frac{21}{65}i and 1 to get -\frac{38}{65}-\frac{21}{65}i.

$$Re(-\frac{38}{65}-\frac{21}{65}i)$$

The real part of -\frac{38}{65}-\frac{21}{65}i is -\frac{38}{65}.

$$-\frac{38}{65}$$