Question

Evaluate the quotient, and write the result in the form $a+bi$.

$\frac{2-3i}{1-2i}$

Short answer

The quotient of complex numbers is $\frac{8}{5}+\frac{1}{5}i$.

Step-by-step solution

Step 1. Concept of complex number.

A complex number is of the form $a+bi$. Where a and b are some real numbers and i is the imaginary number.

The number a is known as the real part of the complex number and b is known as imaginary part of the complex number.

Step 2. Concept of quotient of complex numbers.

To divide two complex numbers, multiply the numerator and denominator by the conjugate of the denominator to rationalize it.

The complex conjugate of the complex number$a+bi$ is represented by a horizontal bar over it, that is, the complex conjugate of$a+bi$ is denoted by $\overline{a+bi}$.

The complex conjugate of any complex number can be obtained by changing the sign of the imaginary part of the complex number. The complex conjugate of$a+bi$ is $a-bi$.

The product of a complex number and its complex conjugate is given by

$\left(a+bi\right)\left(a-bi\right)=a^2+b^2$

The imaginary number i is defined as the square root of negative one. That is, $i^2=-1$.

Step 3. Find the quotient.

To find the quotient of the given complex numbers, multiply the numerator and denominator by the complex conjugate of denominator, that is,$\overline{1-2i}=1+2i$ and simplify further as shown below

$$\begin{gathered} \frac{2-3i}{1-2i}=\frac{2-3i}{1-2i}\cdot \frac{1+2i}{1+2i} \\ =\frac{\left(2-3i\right)\left(1+2i\right)}{\left(1-2i\right)\left(1+2i\right)} \\ =\frac{2\cdot 1+2\cdot \left(2i\right)-3i\cdot \left(1\right)-3i\cdot \left(2i\right)}{{\left(1\right)}^2+{\left(2\right)}^2} \\ =\frac{2+4i-3i-6i^2}{1+4} \\ =\frac{2+i-6\left(-1\right)}{5} \\ =\frac{2+i+6}{5} \\ =\frac{8}{5}+\frac{1}{5}i \end{gathered}$$

Thus, the quotient of the complex numbers is $\frac{8}{5}+\frac{1}{5}i$.