
Find the locus of the complex number z = x + iy, satisfying relations $\arg (z-1)=\dfrac{\pi }{4}\text{ and }\left| z-2-3i \right|=2$. Illustrate the locus on the Argand plane.
Answer
602.7k+ views
Hint: Mathematics includes the study of topics which are related to quantity, structure, space and change. Mathematics deals with a variety of numbers ranging from natural numbers to irrational numbers. In this problem we are given a complex number and its argument. With this information we can formulate relations between the variables x and y. So, we have two variables and two conditions to evaluate this system.
Complete step-by-step answer:
We are required to find the locus of complex numbers. Let the complex number be z = x + iy.
Now, by using the first information: $\arg (z-1)=\dfrac{\pi }{4}$.
As we know that argument of a complex number is expressed as: $\arg (z)=\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$, where y belongs to imaginary part and x belongs to real part. Also, addition and subtraction can be done only with corresponding terms such that real can operate with real part and imaginary with imaginary part.
$\begin{align}
& \arg (z-1)=\dfrac{\pi }{4} \\
& \because z-1=x+iy-1 \\
& z-1=x-1+iy \\
\end{align}$
Now, by using argument definition, ${{\tan }^{-1}}\left( \dfrac{y}{x-1} \right)=\dfrac{\pi }{4}$.
Now, applying tan to both sides we get,
$\begin{align}
& \tan \left( {{\tan }^{-1}}\left( \dfrac{y}{x-1} \right) \right)=\tan \dfrac{\pi }{4} \\
& \dfrac{y}{x-1}=1 \\
& \therefore y=x-1 \\
\end{align}$
Hence, the locus for this part will be a straight line.
Now, by using the second information: $\left| z-2-3i \right|=2$.
Also, for the modulus of a complex number z we have, $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
Therefore first obtaining the proper complex number from information,
$\begin{align}
& z-2-3i=x+iy-2-3i \\
& =(x-2)+i(y-3) \\
\end{align}$
Now, applying the modulus expansion we get,
$\begin{align}
& \left| z-2-3i \right|=\sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}} \\
& \sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}}=2 \\
\end{align}$
Squaring both sides, we get,
\[{{\left( x-2 \right)}^{2}}+{{(y-3)}^{2}}=4\]
Hence, the locus for this part is a circle with center (2,3) and radius 4.
Now, to illustrate both the locus on argand plane:
Note: The key step in solving this problem is the knowledge of complex numbers particularly in modulus and argument topics. On correct use of formulas, we can establish a relationship between variables x and y with the given information. After obtaining all the equations, we can easily draw them in an argand plane.
Complete step-by-step answer:
We are required to find the locus of complex numbers. Let the complex number be z = x + iy.
Now, by using the first information: $\arg (z-1)=\dfrac{\pi }{4}$.
As we know that argument of a complex number is expressed as: $\arg (z)=\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$, where y belongs to imaginary part and x belongs to real part. Also, addition and subtraction can be done only with corresponding terms such that real can operate with real part and imaginary with imaginary part.
$\begin{align}
& \arg (z-1)=\dfrac{\pi }{4} \\
& \because z-1=x+iy-1 \\
& z-1=x-1+iy \\
\end{align}$
Now, by using argument definition, ${{\tan }^{-1}}\left( \dfrac{y}{x-1} \right)=\dfrac{\pi }{4}$.
Now, applying tan to both sides we get,
$\begin{align}
& \tan \left( {{\tan }^{-1}}\left( \dfrac{y}{x-1} \right) \right)=\tan \dfrac{\pi }{4} \\
& \dfrac{y}{x-1}=1 \\
& \therefore y=x-1 \\
\end{align}$
Hence, the locus for this part will be a straight line.
Now, by using the second information: $\left| z-2-3i \right|=2$.
Also, for the modulus of a complex number z we have, $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
Therefore first obtaining the proper complex number from information,
$\begin{align}
& z-2-3i=x+iy-2-3i \\
& =(x-2)+i(y-3) \\
\end{align}$
Now, applying the modulus expansion we get,
$\begin{align}
& \left| z-2-3i \right|=\sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}} \\
& \sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}}=2 \\
\end{align}$
Squaring both sides, we get,
\[{{\left( x-2 \right)}^{2}}+{{(y-3)}^{2}}=4\]
Hence, the locus for this part is a circle with center (2,3) and radius 4.
Now, to illustrate both the locus on argand plane:
Note: The key step in solving this problem is the knowledge of complex numbers particularly in modulus and argument topics. On correct use of formulas, we can establish a relationship between variables x and y with the given information. After obtaining all the equations, we can easily draw them in an argand plane.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

What happens to glucose which enters nephron along class 10 biology CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Write a dialogue with at least ten utterances between class 10 english CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

A circle is inscribed in an equilateral triangle and class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Which of the following does not have a fundamental class 10 physics CBSE

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

