Question

Evaluate the given expression in cartesian form $\left( a+ib \right)$.
$\dfrac{i}{1+i}$

Answer 1

Hint: To solve this question, we need to know that when a complex number is given in a denominator, then we have to rationalize the denominator by multiplying it with its conjugate complex number. If we have a complex number $\left( a+ib \right)$, then its conjugate is given by $\left( a-ib \right)$.

Complete step-by-step answer:
In this question, we have to simplify the given complex number to get its cartesian form, that is, we have to express $\dfrac{i}{1+i}$ in the form of $\left( a+ib \right)$. We know that we have been given $\dfrac{i}{1+i}$ to be simplified. We know that when we have to simplify terms like this, that have complex numbers in the denominator, then we have to rationalize the denominator by multiplying it with its conjugate. So, we will multiply both numerator and denominator with the conjugate of $\left( 1+i \right)$, which is $\left( 1-i \right)$. So, we get,
$\begin{aligned} & \dfrac{i}{1+i}=\left( \dfrac{i}{1+i} \right)\left( \dfrac{1-i}{1-i} \right) \\\ & \Rightarrow \dfrac{i}{1+i}=\dfrac{\left( i \right)\left( 1-i \right)}{\left( 1+i \right)\left( 1-i \right)} \\\ \end{aligned}$
On further simplifying, we get,
$\begin{aligned} & \dfrac{i}{1+i}=\dfrac{i-{{i}^{2}}}{1+i-i-{{i}^{2}}} \\\ & \Rightarrow \dfrac{i}{1+i}=\dfrac{i-\left( -1 \right)}{1-\left( -1 \right)} \\\ & \Rightarrow \dfrac{i}{1+i}=\dfrac{i+1}{1+1} \\\ & \Rightarrow \dfrac{i}{1+i}=\dfrac{1+i}{2} \\\ & \Rightarrow \dfrac{i}{1+i}=\dfrac{1}{2}+\dfrac{1}{2}i \\\ \end{aligned}$
Hence, after simplification, we get that $\dfrac{i}{1+i}$ can be further simplified and written in the cartesian form as $\left( \dfrac{1}{2}+\dfrac{1}{2}i \right)$ , where the real part and the imaginary part are both equal to $\dfrac{1}{2}$ .

Note: We have to remember that whenever we are given a complex number, having complex number in the denominator, that is in $\left( x+iy \right)$ form, then we have to multiply the numerator and denominator by $\left( x-iy \right)$ in order to rationalize the denominator. Also, we have to remember that ${{i}^{2}}=-1$ and ${{i}^{3}}=-i$. We might make a mistake by considering the value of $i=-1$ and the value of ${{i}^{3}}=1$ and end up with the wrong answers.