Solve Write (3 + 2i)/(4 - 3i) . in the form a + bi . | Equatix - Math Solver

Fractions & Decimals

Ratios & Proportions

Basic Number Theory

Integers & Absolute Values

Factors & Multiples

Prime Factorization

Basic Equations & Inequalities

Coordinate Plane & Graphing

Linear Equations & Inequalities

Quadratic Equations

Polynomials & Factoring

Functions & Graphs

Systems of Equations

Trigonometric Functions

Trigonometric Identities & Equations

Right Triangle Trigonometry

Graphing Trigonometric Functions

Limits & Continuity

Derivatives & Applications

Integrals & Applications

Series & Sequences

Multivariable Calculus

Functions & Graphs

Polynomial & Rational Functions

Exponential & Logarithmic Functions

Sequences & Series

Descriptive Statistics

Normal Distribution

Hypothesis Testing

Regression Analysis

Linear Programming

Sets & Counting

Eigenvalues & Eigenvectors

Linear Transformations

Atomic Structure

Thermochemistry

Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac { 3 + 2 i } { 4 - 3 i }$$

Evaluate

$\frac{6}{25}+\frac{17}{25}i=0.24+0.68i$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, $4+3i$.

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

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

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

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

$$\frac{\left(3+2i\right)\left(4+3i\right)}{25}$$

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

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

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

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

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

$$\frac{12+9i+8i-6}{25}$$

Combine the real and imaginary parts in $12+9i+8i-6$.

$$\frac{12-6+\left(9+8\right)i}{25}$$

Do the additions in $12-6+\left(9+8\right)i$.

$$\frac{6+17i}{25}$$

Divide $6+17i$ by $25$ to get $\frac{6}{25}+\frac{17}{25}i$.

$$\frac{6}{25}+\frac{17}{25}i$$

Real Part

$\frac{6}{25} = 0.24$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(3+2i\right)\left(4+3i\right)}{25})$$

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

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

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

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

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

$$Re(\frac{12+9i+8i-6}{25})$$

Combine the real and imaginary parts in $12+9i+8i-6$.

$$Re(\frac{12-6+\left(9+8\right)i}{25})$$

Do the additions in $12-6+\left(9+8\right)i$.

$$Re(\frac{6+17i}{25})$$

Divide $6+17i$ by $25$ to get $\frac{6}{25}+\frac{17}{25}i$.

$$Re(\frac{6}{25}+\frac{17}{25}i)$$

The real part of $\frac{6}{25}+\frac{17}{25}i$ is $\frac{6}{25}$.

$$\frac{6}{25}$$