Solve -2+5i/3-4i*1+7/3+4i

$$\frac { - 2 + 5 i } { 3 - 4 i } \cdot \frac { 1 + 7 } { 3 + 4 i }$$

Evaluate

$$-\frac{16}{25}+\frac{8}{5}i=-0.64+1.6i$$

Solution Steps

Multiply both numerator and denominator of \frac{-2+5i}{3-4i} by the complex conjugate of the denominator, 3+4i.

$$\frac{\left(-2+5i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}\times \frac{1+7}{3+4i}$$

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

$$\frac{\left(-2+5i\right)\left(3+4i\right)}{3^{2}-4^{2}i^{2}}\times \frac{1+7}{3+4i}$$

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

$$\frac{\left(-2+5i\right)\left(3+4i\right)}{25}\times \frac{1+7}{3+4i}$$

Multiply complex numbers -2+5i and 3+4i like you multiply binomials.

$$\frac{-2\times 3-2\times \left(4i\right)+5i\times 3+5\times 4i^{2}}{25}\times \frac{1+7}{3+4i}$$

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

$$\frac{-2\times 3-2\times \left(4i\right)+5i\times 3+5\times 4\left(-1\right)}{25}\times \frac{1+7}{3+4i}$$

Do the multiplications in -2\times 3-2\times \left(4i\right)+5i\times 3+5\times 4\left(-1\right).

$$\frac{-6-8i+15i-20}{25}\times \frac{1+7}{3+4i}$$

Combine the real and imaginary parts in -6-8i+15i-20.

$$\frac{-6-20+\left(-8+15\right)i}{25}\times \frac{1+7}{3+4i}$$

Do the additions in -6-20+\left(-8+15\right)i.

$$\frac{-26+7i}{25}\times \frac{1+7}{3+4i}$$

Divide -26+7i by 25 to get -\frac{26}{25}+\frac{7}{25}i.

$$\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{1+7}{3+4i}$$

Add 1 and 7 to get 8.

$$\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{8}{3+4i}$$

Multiply both numerator and denominator of \frac{8}{3+4i} by the complex conjugate of the denominator, 3-4i.

$$\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{8\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}$$

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

$$\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{8\left(3-4i\right)}{3^{2}-4^{2}i^{2}}$$

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

$$\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{8\left(3-4i\right)}{25}$$

Multiply 8 times 3-4i.

$$\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{8\times 3+8\times \left(-4i\right)}{25}$$

Do the multiplications in 8\times 3+8\times \left(-4i\right).

$$\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{24-32i}{25}$$

Divide 24-32i by 25 to get \frac{24}{25}-\frac{32}{25}i.

$$\left(-\frac{26}{25}+\frac{7}{25}i\right)\left(\frac{24}{25}-\frac{32}{25}i\right)$$

Multiply complex numbers -\frac{26}{25}+\frac{7}{25}i and \frac{24}{25}-\frac{32}{25}i like you multiply binomials.

$$-\frac{26}{25}\times \frac{24}{25}-\frac{26}{25}\times \left(-\frac{32}{25}i\right)+\frac{7}{25}i\times \frac{24}{25}+\frac{7}{25}\left(-\frac{32}{25}\right)i^{2}$$

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

$$-\frac{26}{25}\times \frac{24}{25}-\frac{26}{25}\times \left(-\frac{32}{25}i\right)+\frac{7}{25}i\times \frac{24}{25}+\frac{7}{25}\left(-\frac{32}{25}\right)\left(-1\right)$$

Do the multiplications.

$$-\frac{624}{625}+\frac{832}{625}i+\frac{168}{625}i+\frac{224}{625}$$

Combine the real and imaginary parts.

$$-\frac{624}{625}+\frac{224}{625}+\left(\frac{832}{625}+\frac{168}{625}\right)i$$

Do the additions.

$$-\frac{16}{25}+\frac{8}{5}i$$

Real Part

$$-\frac{16}{25} = -0.64$$

Solution Steps

Multiply both numerator and denominator of \frac{-2+5i}{3-4i} by the complex conjugate of the denominator, 3+4i.

$$Re(\frac{\left(-2+5i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}\times \frac{1+7}{3+4i})$$

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

$$Re(\frac{\left(-2+5i\right)\left(3+4i\right)}{3^{2}-4^{2}i^{2}}\times \frac{1+7}{3+4i})$$

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

$$Re(\frac{\left(-2+5i\right)\left(3+4i\right)}{25}\times \frac{1+7}{3+4i})$$

Multiply complex numbers -2+5i and 3+4i like you multiply binomials.

$$Re(\frac{-2\times 3-2\times \left(4i\right)+5i\times 3+5\times 4i^{2}}{25}\times \frac{1+7}{3+4i})$$

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

$$Re(\frac{-2\times 3-2\times \left(4i\right)+5i\times 3+5\times 4\left(-1\right)}{25}\times \frac{1+7}{3+4i})$$

Do the multiplications in -2\times 3-2\times \left(4i\right)+5i\times 3+5\times 4\left(-1\right).

$$Re(\frac{-6-8i+15i-20}{25}\times \frac{1+7}{3+4i})$$

Combine the real and imaginary parts in -6-8i+15i-20.

$$Re(\frac{-6-20+\left(-8+15\right)i}{25}\times \frac{1+7}{3+4i})$$

Do the additions in -6-20+\left(-8+15\right)i.

$$Re(\frac{-26+7i}{25}\times \frac{1+7}{3+4i})$$

Divide -26+7i by 25 to get -\frac{26}{25}+\frac{7}{25}i.

$$Re(\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{1+7}{3+4i})$$

Add 1 and 7 to get 8.

$$Re(\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{8}{3+4i})$$

Multiply both numerator and denominator of \frac{8}{3+4i} by the complex conjugate of the denominator, 3-4i.

$$Re(\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{8\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)})$$

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

$$Re(\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{8\left(3-4i\right)}{3^{2}-4^{2}i^{2}})$$

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

$$Re(\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{8\left(3-4i\right)}{25})$$

Multiply 8 times 3-4i.

$$Re(\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{8\times 3+8\times \left(-4i\right)}{25})$$

Do the multiplications in 8\times 3+8\times \left(-4i\right).

$$Re(\left(-\frac{26}{25}+\frac{7}{25}i\right)\times \frac{24-32i}{25})$$

Divide 24-32i by 25 to get \frac{24}{25}-\frac{32}{25}i.

$$Re(\left(-\frac{26}{25}+\frac{7}{25}i\right)\left(\frac{24}{25}-\frac{32}{25}i\right))$$

Multiply complex numbers -\frac{26}{25}+\frac{7}{25}i and \frac{24}{25}-\frac{32}{25}i like you multiply binomials.

$$Re(-\frac{26}{25}\times \frac{24}{25}-\frac{26}{25}\times \left(-\frac{32}{25}i\right)+\frac{7}{25}i\times \frac{24}{25}+\frac{7}{25}\left(-\frac{32}{25}\right)i^{2})$$

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

$$Re(-\frac{26}{25}\times \frac{24}{25}-\frac{26}{25}\times \left(-\frac{32}{25}i\right)+\frac{7}{25}i\times \frac{24}{25}+\frac{7}{25}\left(-\frac{32}{25}\right)\left(-1\right))$$

Do the multiplications in -\frac{26}{25}\times \frac{24}{25}-\frac{26}{25}\times \left(-\frac{32}{25}i\right)+\frac{7}{25}i\times \frac{24}{25}+\frac{7}{25}\left(-\frac{32}{25}\right)\left(-1\right).

$$Re(-\frac{624}{625}+\frac{832}{625}i+\frac{168}{625}i+\frac{224}{625})$$

Combine the real and imaginary parts in -\frac{624}{625}+\frac{832}{625}i+\frac{168}{625}i+\frac{224}{625}.

$$Re(-\frac{624}{625}+\frac{224}{625}+\left(\frac{832}{625}+\frac{168}{625}\right)i)$$

Do the additions in -\frac{624}{625}+\frac{224}{625}+\left(\frac{832}{625}+\frac{168}{625}\right)i.

$$Re(-\frac{16}{25}+\frac{8}{5}i)$$

The real part of -\frac{16}{25}+\frac{8}{5}i is -\frac{16}{25}.

$$-\frac{16}{25}$$