Answer
Verified
376.8k+ views
Hint: In this type of question we have to use the concept of complex numbers. As the given complex number is in the fraction form first we have to rationalise it to convert it into the standard form \[z=x+iy\]. During this we also have to use the formulas \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]. We know that if \[z=x+iy\] then the modulus of \[z\] is denoted by \[\left| z \right|\] and defined as \[\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\]. Also the argument of \[z\] is denoted by \[\theta \] and defined as \[\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)\].
Complete step by step answer:
Now, we have to find the modulus and argument of the complex number: \[\dfrac{1+i}{1-i}\]
Let us consider, \[\dfrac{1+i}{1-i}\] by performing rationalisation we can write,
\[\Rightarrow \dfrac{1+i}{1-i}=\dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}\]
\[\Rightarrow \dfrac{1+i}{1-i}=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1-i \right)\left( 1+i \right)}\]
Now, as we know that, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\] we get,
\[\Rightarrow \dfrac{1+i}{1-i}=\dfrac{1+2i+{{i}^{2}}}{1-{{i}^{2}}}\]
Now as the value of \[{{i}^{2}}=-1\] we can write,
\[\begin{align}
& \Rightarrow \dfrac{1+i}{1-i}=\dfrac{1+2i-1}{1-\left( -1 \right)} \\
& \Rightarrow \dfrac{1+i}{1-i}=\dfrac{2i}{1+1} \\
& \Rightarrow \dfrac{1+i}{1-i}=\dfrac{2i}{2} \\
& \Rightarrow \dfrac{1+i}{1-i}=i \\
\end{align}\]
Thus we get, \[\dfrac{1+i}{1-i}\] is equal to \[i\] which we can write in standard form as \[0+i\].
As we know that, if \[z=x+iy\] then the modulus of \[z\] is defined as \[\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\] and the argument of \[z\] is defined as \[\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)\].
Here, on comparison we get, \[x=0\] and \[y=1\].
\[\Rightarrow \text{Modulus of }z\text{ = }\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
\[\begin{align}
& \Rightarrow \left| z \right|=\sqrt{{{0}^{2}}+{{1}^{2}}} \\
& \Rightarrow \left| z \right|=\sqrt{1} \\
& \Rightarrow \left| z \right|=1 \\
\end{align}\]
\[\Rightarrow \text{Argument of }z\text{ = }\theta \text{ = }{{\tan }^{-1}}\left( \dfrac{y}{x} \right)\]
\[\Rightarrow \theta \text{ = }{{\tan }^{-1}}\left( \dfrac{1}{0} \right)\]
\[\begin{align}
& \Rightarrow \tan \theta =\dfrac{1}{0} \\
& \Rightarrow \dfrac{\sin \theta }{\cos \theta }=\dfrac{1}{0} \\
& \Rightarrow \sin \theta =1\And \cos \theta =0 \\
\end{align}\]
Hence, the possible value of \[\theta \] is \[\dfrac{\pi }{2}\].
Thus the modulus of \[z\] is 1 and the argument of \[z\] is \[\dfrac{\pi }{2}\].
Note: In this type of question students have to note that the rationalisation of a given complex number is most important. In case of rationalisation students have to remember that they have to multiply the numerator and denominator by a complex conjugate of the denominator. Students have to note that the complex conjugate of a complex number \[z=x+iy\] is given by \[z=x-iy\]. Also students have to remember the value of \[{{i}^{2}}\] which is equal to -1.
Complete step by step answer:
Now, we have to find the modulus and argument of the complex number: \[\dfrac{1+i}{1-i}\]
Let us consider, \[\dfrac{1+i}{1-i}\] by performing rationalisation we can write,
\[\Rightarrow \dfrac{1+i}{1-i}=\dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}\]
\[\Rightarrow \dfrac{1+i}{1-i}=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1-i \right)\left( 1+i \right)}\]
Now, as we know that, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\] we get,
\[\Rightarrow \dfrac{1+i}{1-i}=\dfrac{1+2i+{{i}^{2}}}{1-{{i}^{2}}}\]
Now as the value of \[{{i}^{2}}=-1\] we can write,
\[\begin{align}
& \Rightarrow \dfrac{1+i}{1-i}=\dfrac{1+2i-1}{1-\left( -1 \right)} \\
& \Rightarrow \dfrac{1+i}{1-i}=\dfrac{2i}{1+1} \\
& \Rightarrow \dfrac{1+i}{1-i}=\dfrac{2i}{2} \\
& \Rightarrow \dfrac{1+i}{1-i}=i \\
\end{align}\]
Thus we get, \[\dfrac{1+i}{1-i}\] is equal to \[i\] which we can write in standard form as \[0+i\].
As we know that, if \[z=x+iy\] then the modulus of \[z\] is defined as \[\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\] and the argument of \[z\] is defined as \[\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)\].
Here, on comparison we get, \[x=0\] and \[y=1\].
\[\Rightarrow \text{Modulus of }z\text{ = }\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
\[\begin{align}
& \Rightarrow \left| z \right|=\sqrt{{{0}^{2}}+{{1}^{2}}} \\
& \Rightarrow \left| z \right|=\sqrt{1} \\
& \Rightarrow \left| z \right|=1 \\
\end{align}\]
\[\Rightarrow \text{Argument of }z\text{ = }\theta \text{ = }{{\tan }^{-1}}\left( \dfrac{y}{x} \right)\]
\[\Rightarrow \theta \text{ = }{{\tan }^{-1}}\left( \dfrac{1}{0} \right)\]
\[\begin{align}
& \Rightarrow \tan \theta =\dfrac{1}{0} \\
& \Rightarrow \dfrac{\sin \theta }{\cos \theta }=\dfrac{1}{0} \\
& \Rightarrow \sin \theta =1\And \cos \theta =0 \\
\end{align}\]
Hence, the possible value of \[\theta \] is \[\dfrac{\pi }{2}\].
Thus the modulus of \[z\] is 1 and the argument of \[z\] is \[\dfrac{\pi }{2}\].
Note: In this type of question students have to note that the rationalisation of a given complex number is most important. In case of rationalisation students have to remember that they have to multiply the numerator and denominator by a complex conjugate of the denominator. Students have to note that the complex conjugate of a complex number \[z=x+iy\] is given by \[z=x-iy\]. Also students have to remember the value of \[{{i}^{2}}\] which is equal to -1.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE