Question

How do you divide \[\dfrac{1-2i}{1-3i}\] in trigonometric form?

Answer 1

Hint: This question is from the topic of pre-calculus. In this question, we are going to first understand how to find the complex conjugate of a complex number. After that, we will understand how to rationalize any complex number. After that, we will solve the given term \[\dfrac{1-2i}{1-3i}\]. We will first rationalize the term \[\dfrac{1-2i}{1-3i}\] by multiplying the complex conjugate of denominator of the term \[\dfrac{1-2i}{1-3i}\] to both numerator and denominator. After solving the further question, we will get our answer.

Complete step by step answer:
Let us solve this question.
In this question, we will divide the term \[\dfrac{1-2i}{1-3i}\] in trigonometric form. Or, we can say we will solve the term \[\dfrac{1-2i}{1-3i}\].
Let us first know about complex conjugate.
Suppose, we have a complex number \[a-ib\], then the complex conjugate of \[a-ib\] will be \[a+ib\].
Now, let us know about rationalization.
Whenever we have to rationalize a complex number, we will multiply the complex conjugate of that complex number to the numerator and denominator.
The given term which we have to solve is
\[\dfrac{1-2i}{1-3i}\]
We can write the above term after rationalizing as
\[\dfrac{1-2i}{1-3i}=\dfrac{1-2i}{1-3i}\times \dfrac{1+3i}{1+3i}\]
Using the formula \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\], we can write the above equation as
\[\Rightarrow \dfrac{1-2i}{1-3i}=\dfrac{1-2i}{{{1}^{2}}-{{\left( 3i \right)}^{2}}}\times \dfrac{1+3i}{1}\]
Now, multiplying the above terms of numerator, we can write
\[\Rightarrow \dfrac{1-2i}{1-3i}=\dfrac{1+3i-2i-6{{i}^{2}}}{{{1}^{2}}-{{\left( 3i \right)}^{2}}}\]
The above equation can also be written as
\[\Rightarrow \dfrac{1-2i}{1-3i}=\dfrac{1+i-6{{i}^{2}}}{1-9{{\left( i \right)}^{2}}}\]
Using the formula \[{{\left( i \right)}^{2}}=-1\], we can write
\[\Rightarrow \dfrac{1-2i}{1-3i}=\dfrac{1+i-6\left( -1 \right)}{1-9\left( -1 \right)}\]
\[\Rightarrow \dfrac{1-2i}{1-3i}=\dfrac{1+i+6}{1+9}\]
The above equation can also be written as
\[\Rightarrow \dfrac{1-2i}{1-3i}=\dfrac{7+i}{10}\]
\[\Rightarrow \dfrac{1-2i}{1-3i}=\dfrac{7}{10}+\dfrac{i}{10}\]
Now, we have solved the term \[\dfrac{1-2i}{1-3i}\] and our answer is \[\dfrac{7}{10}+\dfrac{i}{10}\].

Note:
We should have a better knowledge on the topic of pre-calculus.
For solving this type of question easily, we should know how to do rationalization and find the complex conjugate of complex numbers. For example, the rationalization of \[\dfrac{1}{a+ib}\] will be \[\dfrac{1}{a+ib}\times \dfrac{a-ib}{a-ib}\].
We should remember the following formulas for solving this type of question easily:
\[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]
\[{{\left( i \right)}^{2}}=-1\]