# Find the modulus and argument of the complex number $\dfrac{1+2i}{1-3i}$.

Last updated date: 23rd Mar 2023

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Answer

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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.

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