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

Marvin Mccormick 2020-11-01 Answered
Find the principal argument and exponential form of $z=\frac{i}{1+i}$ If z=x+iy is a complex number
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Rewrite the above complex number in the usual form z=x+iy as follows.
$z=\frac{i}{1+i}=\frac{i}{1+i}\cdot \frac{1-i}{1-i}$ [multiply and divide by the conjugate 1−i of 1+i]
$=\frac{i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}=\frac{i-{i}^{2}}{{1}^{2}-{i}^{2}}=\left(i+1\right)\left(1+1\right)=\frac{i+1}{2}=\frac{1}{2}+\frac{1}{2}i$
Compare the complex number $z=\frac{1}{2}+\frac{1}{2}i$ with z=x+iy and obtain $x=\frac{1}{2}$ and $y=\frac{1}{2}.$
The argument of a complex number z=x+iy is given by $\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$

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

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

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

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

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

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

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

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