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]

\(\displaystyle=\frac{{{i}{\left({1}-{i}\right)}}}{{{\left({1}+{i}\right)}{\left({1}-{i}\right)}}}=\frac{{{i}-{i}^{{2}}}}{{{1}^{{2}}-{i}^{{2}}}}={\left({i}+{1}\right)}{\left({1}+{1}\right)}=\frac{{{i}+{1}}}{{2}}=\frac{{1}}{{2}}+\frac{{1}}{{2}}{i}\)

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 \(\displaystyle\theta={{\tan}^{

\(\displaystyle{z}=\frac{{i}}{{{1}+{i}}}=\frac{{i}}{{{1}+{i}}}\cdot\frac{{{1}-{i}}}{{{1}-{i}}}\) [multiply and divide by the conjugate 1−i of 1+i]

\(\displaystyle=\frac{{{i}{\left({1}-{i}\right)}}}{{{\left({1}+{i}\right)}{\left({1}-{i}\right)}}}=\frac{{{i}-{i}^{{2}}}}{{{1}^{{2}}-{i}^{{2}}}}={\left({i}+{1}\right)}{\left({1}+{1}\right)}=\frac{{{i}+{1}}}{{2}}=\frac{{1}}{{2}}+\frac{{1}}{{2}}{i}\)

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 \(\displaystyle\theta={{\tan}^{