How to solve (2 - 5i)/(1 - 2i)

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 (2 - 5i)/(1 - 2i) | Equatix

Question

$$\frac{2-5i}{1-2i}$$

Evaluate

$\frac{12}{5}-\frac{1}{5}i=2.4-0.2i$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, $1+2i$.

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

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

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

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

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

Multiply complex numbers $2-5i$ and $1+2i$ like you multiply binomials.

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

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

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

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

$$\frac{2+4i-5i+10}{5}$$

Combine the real and imaginary parts in $2+4i-5i+10$.

$$\frac{2+10+\left(4-5\right)i}{5}$$

Do the additions in $2+10+\left(4-5\right)i$.

$$\frac{12-i}{5}$$

Divide $12-i$ by $5$ to get $\frac{12}{5}-\frac{1}{5}i$.

$$\frac{12}{5}-\frac{1}{5}i$$

Show Solution

Real Part

$\frac{12}{5} = 2\frac{2}{5} = 2.4$

Solution Steps

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

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

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

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

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

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

Multiply complex numbers $2-5i$ and $1+2i$ like you multiply binomials.

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

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

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

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

$$Re(\frac{2+4i-5i+10}{5})$$

Combine the real and imaginary parts in $2+4i-5i+10$.

$$Re(\frac{2+10+\left(4-5\right)i}{5})$$

Do the additions in $2+10+\left(4-5\right)i$.

$$Re(\frac{12-i}{5})$$

Divide $12-i$ by $5$ to get $\frac{12}{5}-\frac{1}{5}i$.

$$Re(\frac{12}{5}-\frac{1}{5}i)$$

The real part of $\frac{12}{5}-\frac{1}{5}i$ is $\frac{12}{5}$.

$$\frac{12}{5}$$

Show Solution

Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

Statistics

Finite Math

Linear Algebra

Chemistry

Example

Question

Evaluate

Solution Steps

Real Part

Solution Steps