Find the principal argument and exponential form of z=i/(1+i) If z=x+iy is a complex number

Rewrite the above complex number in the usual form z=x+iy as follows.
${\displaystyle z=\frac{i}{1+i}=\frac{i}{1+i}\cdot \frac{1-i}{1-i}}$ [multiply and divide by the conjugate 1−i of 1+i]
${\displaystyle =\frac{i(1-i)}{(1+i)(1-i)}=\frac{i-i^2}{1^2-i^2}=(i+1)(1+1)=\frac{i+1}{2}=\frac{1}{2}+\frac{1}{2}i}$
Compare the complex number ${\displaystyle z=\frac{1}{2}+\frac{1}{2}i}$ with z=x+iy and obtain ${\displaystyle x=\frac{1}{2}}$ and ${\displaystyle y=\frac{1}{2}.}$
The argument of a complex number z=x+iy is given by $\theta ={\tan }^{-1}(\frac{y}{x})$

Here, for $z=\frac{1}{2}+\frac{1}{2i}$

${\displaystyle \theta -{\tan }^{-1}\left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)={\tan }^{-1}\left(1\right)=\frac{\pi }{4}}$

Since both ${\displaystyle x=\frac{1}{2}}$ and ${\displaystyle y=\frac{1}{2}}$ are positive, the complex number ${\displaystyle z=\frac{1}{2}+\frac{1}{2}i}$ lie in the 1st quadrant.

Hence, its principal argument is the same as ${\displaystyle \theta =\frac{\pi }{4}}$

The exponential form of a complex number z=x+iy with principal argument theta is given by ${\displaystyle z=r{e}^{i\theta }}$ where ${\displaystyle r=\sqrt{x^2+y^2}}$

For ${\displaystyle z=\frac{1}{2}+\frac{1}{2}i}$ the value ${\displaystyle r=\sqrt{x^2+y^2}=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}}$

Hence, the exponential form of ${\displaystyle z=\frac{1}{2}+\frac{1}{2}i}$ is ${\displaystyle z=\frac{1}{2}+\frac{1}{2}i}$ is ${\displaystyle z=\frac{1}{\sqrt{2}}{e}^{i\left(\frac{\pi }{4}\right)}}$

Thus, the principal argument of ${\displaystyle z=\frac{i}{1+i}}$ is ${\displaystyle 0=\frac{\pi }{4}}$ and the exponential form is ${\displaystyle z\frac{1}{\sqrt{2}}{e}^{i\left(\frac{\pi }{4}\right)}}$