
Find the modulus and argument of the complex number $\dfrac{1+2i}{1-3i}$.
Answer
631.8k+ views
Hint: To solve this question, we should be aware about the basic properties of complex numbers. For a complex number, say a+bi, we have,
Modulus of this complex number = $\sqrt{{{a}^{2}}+{{b}^{2}}}$
Argument = ${{\tan }^{-1}}\dfrac{b}{a}$
Further, to solve this problem, we will make use of rationalization along with the above two properties to solve. Also, keep in mind that ${{i}^{2}}=-1$.
Complete step-by-step answer:
We have, the complex number,
=$\dfrac{1+2i}{1-3i}$
To bring it in the form, a+bi, we do rationalization, thus, we have,
=$\dfrac{1+2i}{1-3i}\times \dfrac{1+3i}{1+3i}$
Thus, multiplying, we have,
=$\dfrac{(1+2i)(1+3i)}{(1-3i)(1+3i)}$
=$\dfrac{1+3i+2i+6{{i}^{2}}}{({{1}^{2}}-{{(3i)}^{2}})}$
Now, we use, ${{i}^{2}}=-1$ to solve further,
=$\dfrac{1+5i-6}{1+9}$
=$\dfrac{5i-5}{10}$
=$\dfrac{5(i-1)}{10}$
=$\dfrac{i-1}{2}$
=-0.5+0.5i -- (A)
Now, we have got the expression in the form a+bi, thus, we can find modulus and argument using the following properties-
Modulus of this complex number = $\sqrt{{{a}^{2}}+{{b}^{2}}}$ -- (1)
Argument = ${{\tan }^{-1}}\dfrac{b}{a}$ -- (2)
Thus, we have,
Modulus of -0.5+0.5i =$\sqrt{{{(-0.5)}^{2}}+{{0.5}^{2}}}=\sqrt{0.25+0.25}=\sqrt{0.5}$=0.707 (approximately)
Argument of -0.5+0.5i = ${{\tan }^{-1}}\left( \dfrac{0.5}{-0.5} \right)={{\tan }^{-1}}(-1)=\dfrac{3\pi }{4}$
It is important to note that the argument would not be $\dfrac{-\pi }{4}$since the real part (concerning cos$\theta $) is negative and imaginary part (concerning sin$\theta $) is negative, thus the angle should belong to the second quadrant. Thus, for angles in the second quadrant, only $\tan \left( \dfrac{3\pi }{4} \right)=-1$.
Hence, the modulus and argument of the complex number $\dfrac{1+2i}{1-3i}$ are 0.707 and $\dfrac{3\pi }{4}$respectively.
Note: While solving problems related to complex numbers concerning modulus and argument of the complex number, it is important to keep in mind the basic properties of complex numbers like addition, subtraction, multiplication and division of the complex number. While, addition and subtraction are quite intuitive since they are similar to the properties of real numbers, one should be careful while performing multiplication and division. While multiplication, one must keep in mind the fact that ${{i}^{2}}=-1$ and for division, in most cases, one needs to perform rationalization. Further, it is also suggested to know about the basic properties of inverse trigonometric functions since they are used while finding the argument of the complex number.
Modulus of this complex number = $\sqrt{{{a}^{2}}+{{b}^{2}}}$
Argument = ${{\tan }^{-1}}\dfrac{b}{a}$
Further, to solve this problem, we will make use of rationalization along with the above two properties to solve. Also, keep in mind that ${{i}^{2}}=-1$.
Complete step-by-step answer:
We have, the complex number,
=$\dfrac{1+2i}{1-3i}$
To bring it in the form, a+bi, we do rationalization, thus, we have,
=$\dfrac{1+2i}{1-3i}\times \dfrac{1+3i}{1+3i}$
Thus, multiplying, we have,
=$\dfrac{(1+2i)(1+3i)}{(1-3i)(1+3i)}$
=$\dfrac{1+3i+2i+6{{i}^{2}}}{({{1}^{2}}-{{(3i)}^{2}})}$
Now, we use, ${{i}^{2}}=-1$ to solve further,
=$\dfrac{1+5i-6}{1+9}$
=$\dfrac{5i-5}{10}$
=$\dfrac{5(i-1)}{10}$
=$\dfrac{i-1}{2}$
=-0.5+0.5i -- (A)
Now, we have got the expression in the form a+bi, thus, we can find modulus and argument using the following properties-
Modulus of this complex number = $\sqrt{{{a}^{2}}+{{b}^{2}}}$ -- (1)
Argument = ${{\tan }^{-1}}\dfrac{b}{a}$ -- (2)
Thus, we have,
Modulus of -0.5+0.5i =$\sqrt{{{(-0.5)}^{2}}+{{0.5}^{2}}}=\sqrt{0.25+0.25}=\sqrt{0.5}$=0.707 (approximately)
Argument of -0.5+0.5i = ${{\tan }^{-1}}\left( \dfrac{0.5}{-0.5} \right)={{\tan }^{-1}}(-1)=\dfrac{3\pi }{4}$
It is important to note that the argument would not be $\dfrac{-\pi }{4}$since the real part (concerning cos$\theta $) is negative and imaginary part (concerning sin$\theta $) is negative, thus the angle should belong to the second quadrant. Thus, for angles in the second quadrant, only $\tan \left( \dfrac{3\pi }{4} \right)=-1$.
Hence, the modulus and argument of the complex number $\dfrac{1+2i}{1-3i}$ are 0.707 and $\dfrac{3\pi }{4}$respectively.
Note: While solving problems related to complex numbers concerning modulus and argument of the complex number, it is important to keep in mind the basic properties of complex numbers like addition, subtraction, multiplication and division of the complex number. While, addition and subtraction are quite intuitive since they are similar to the properties of real numbers, one should be careful while performing multiplication and division. While multiplication, one must keep in mind the fact that ${{i}^{2}}=-1$ and for division, in most cases, one needs to perform rationalization. Further, it is also suggested to know about the basic properties of inverse trigonometric functions since they are used while finding the argument of the complex number.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

The largest wind power cluster is located in the state class 11 social science CBSE

Explain zero factorial class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Which among the following are examples of coming together class 11 social science CBSE

Can anyone list 10 advantages and disadvantages of friction

