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

Last updated date: 17th Jul 2024
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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 $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$.

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.