Solve =5+2i/frac{3{4}+2/3i}

$$= \frac { 5 + 2 i } { \frac { 3 } { 4 } + \frac { 2 } { 3 } i }$$

Evaluate

$$\frac{732}{145}-\frac{264}{145}i\approx 5.048275862-1.820689655i$$

Solution Steps

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

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

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

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

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

$$\frac{\left(5+2i\right)\left(\frac{3}{4}-\frac{2}{3}i\right)}{\frac{145}{144}}$$

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

$$\frac{5\times \frac{3}{4}+5\times \left(-\frac{2}{3}i\right)+2i\times \frac{3}{4}+2\left(-\frac{2}{3}\right)i^{2}}{\frac{145}{144}}$$

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

$$\frac{5\times \frac{3}{4}+5\times \left(-\frac{2}{3}i\right)+2i\times \frac{3}{4}+2\left(-\frac{2}{3}\right)\left(-1\right)}{\frac{145}{144}}$$

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

$$\frac{\frac{15}{4}-\frac{10}{3}i+\frac{3}{2}i+\frac{4}{3}}{\frac{145}{144}}$$

Combine the real and imaginary parts in \frac{15}{4}-\frac{10}{3}i+\frac{3}{2}i+\frac{4}{3}.

$$\frac{\frac{15}{4}+\frac{4}{3}+\left(-\frac{10}{3}+\frac{3}{2}\right)i}{\frac{145}{144}}$$

Do the additions in \frac{15}{4}+\frac{4}{3}+\left(-\frac{10}{3}+\frac{3}{2}\right)i.

$$\frac{\frac{61}{12}-\frac{11}{6}i}{\frac{145}{144}}$$

Divide \frac{61}{12}-\frac{11}{6}i by \frac{145}{144} to get \frac{732}{145}-\frac{264}{145}i.

$$\frac{732}{145}-\frac{264}{145}i$$

Real Part

$$\frac{732}{145} = 5\frac{7}{145} = 5.048275862068966$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(5+2i\right)\left(\frac{3}{4}-\frac{2}{3}i\right)}{\frac{145}{144}})$$

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

$$Re(\frac{5\times \frac{3}{4}+5\times \left(-\frac{2}{3}i\right)+2i\times \frac{3}{4}+2\left(-\frac{2}{3}\right)i^{2}}{\frac{145}{144}})$$

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

$$Re(\frac{5\times \frac{3}{4}+5\times \left(-\frac{2}{3}i\right)+2i\times \frac{3}{4}+2\left(-\frac{2}{3}\right)\left(-1\right)}{\frac{145}{144}})$$

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

$$Re(\frac{\frac{15}{4}-\frac{10}{3}i+\frac{3}{2}i+\frac{4}{3}}{\frac{145}{144}})$$

Combine the real and imaginary parts in \frac{15}{4}-\frac{10}{3}i+\frac{3}{2}i+\frac{4}{3}.

$$Re(\frac{\frac{15}{4}+\frac{4}{3}+\left(-\frac{10}{3}+\frac{3}{2}\right)i}{\frac{145}{144}})$$

Do the additions in \frac{15}{4}+\frac{4}{3}+\left(-\frac{10}{3}+\frac{3}{2}\right)i.

$$Re(\frac{\frac{61}{12}-\frac{11}{6}i}{\frac{145}{144}})$$

Divide \frac{61}{12}-\frac{11}{6}i by \frac{145}{144} to get \frac{732}{145}-\frac{264}{145}i.

$$Re(\frac{732}{145}-\frac{264}{145}i)$$

The real part of \frac{732}{145}-\frac{264}{145}i is \frac{732}{145}.

$$\frac{732}{145}$$