How to solve (3 - i)/(1 - 3i)

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (3 - i)/(1 - 3i) | Equatix

Question

$$\frac{3-i}{1-3i}$$

Evaluate

$\frac{3}{5}+\frac{4}{5}i=0.6+0.8i$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, $1+3i$.

$$\frac{\left(3-i\right)\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)}$$

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

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

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

$$\frac{\left(3-i\right)\left(1+3i\right)}{10}$$

Multiply complex numbers $3-i$ and $1+3i$ like you multiply binomials.

$$\frac{3\times 1+3\times \left(3i\right)-i-3i^{2}}{10}$$

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

$$\frac{3\times 1+3\times \left(3i\right)-i-3\left(-1\right)}{10}$$

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

$$\frac{3+9i-i+3}{10}$$

Combine the real and imaginary parts in $3+9i-i+3$.

$$\frac{3+3+\left(9-1\right)i}{10}$$

Do the additions in $3+3+\left(9-1\right)i$.

$$\frac{6+8i}{10}$$

Divide $6+8i$ by $10$ to get $\frac{3}{5}+\frac{4}{5}i$.

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

Show Solution

Real Part

$\frac{3}{5} = 0.6$

Solution Steps

Multiply both numerator and denominator of $\frac{3-i}{1-3i}$ by the complex conjugate of the denominator, $1+3i$.

$$Re(\frac{\left(3-i\right)\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)})$$

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

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

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

$$Re(\frac{\left(3-i\right)\left(1+3i\right)}{10})$$

Multiply complex numbers $3-i$ and $1+3i$ like you multiply binomials.

$$Re(\frac{3\times 1+3\times \left(3i\right)-i-3i^{2}}{10})$$

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

$$Re(\frac{3\times 1+3\times \left(3i\right)-i-3\left(-1\right)}{10})$$

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

$$Re(\frac{3+9i-i+3}{10})$$

Combine the real and imaginary parts in $3+9i-i+3$.

$$Re(\frac{3+3+\left(9-1\right)i}{10})$$

Do the additions in $3+3+\left(9-1\right)i$.

$$Re(\frac{6+8i}{10})$$

Divide $6+8i$ by $10$ to get $\frac{3}{5}+\frac{4}{5}i$.

$$Re(\frac{3}{5}+\frac{4}{5}i)$$

The real part of $\frac{3}{5}+\frac{4}{5}i$ is $\frac{3}{5}$.

$$\frac{3}{5}$$

Show Solution

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Real Part

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