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

Complex numbers
asked 2020-11-01
Find the principal argument and exponential form of \(\displaystyle{z}=\frac{{i}}{{{1}+{i}}}\) If z=x+iy is a complex number

Answers (1)


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]
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 \(\displaystyle{z}\frac{1}{\sqrt{{2}}}{e}^{{{i}{\left(\frac{\pi}{{4}}\right)}}}\)

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