Question

Convert the given complex number in polar form : $i$.

Answer 1

In Polar form, complex numbers are represented as the combination of $r$ and $\theta$, where $\theta$ is called the argument of complex number and $r$ is called the modulus of complex number. The complex number $z = x + iy$ in polar form is $z = r(\cos \theta + i\sin \theta )$, where $r = |z| = \sqrt {{x^2} + {y^2}}$ and $\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ for $x \geqslant 0$ and $\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) + {180^ \circ }$ for $x < 0$. Using these formulas, we can convert the given complex number in polar form.

Complete step by step answer:
We need to convert $i$ in polar form.Writing $i$ in the form of $z = x + iy$, we have
$z = i = 0 + i(1)$.
Comparing it with $z = x + iy$, we get
$\Rightarrow x = 0 - - - - - - (1)$
$\Rightarrow y = 1 - - - - - - (2)$

Let the polar form of $z = i = 0 + i(1)$ be $z = r(\cos \theta + i\sin \theta ) - - - - - - (*)$,
Where $r = |z| = \sqrt {{x^2} + {y^2}}$
And $\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ as $x = 0$
First of all, finding $r$using (1) and (2)
$r = \sqrt {{x^2} + {y^2}} = \sqrt {{0^2} + {1^2}}$
As ${a^2} = a \times a$
$\sqrt {0 + 1} = \sqrt 1 = 1$
$\therefore r = 1 - - - - - - (3)$

Finding $\theta$ using (1) and (2).
$\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{0}} \right)$
We know $\dfrac{1}{0} = \infty$. Hence, using this above, we have
$\Rightarrow \theta = {\tan ^{ - 1}}\left( \infty \right)$
As we know, $\tan \left( {\dfrac{\pi }{2}} \right) = \infty \Rightarrow {\tan ^{ - 1}}\left( \infty \right) = \dfrac{\pi }{2}$, we get
$\Rightarrow \theta = {\tan ^{ - 1}}\left( \infty \right) = \dfrac{\pi }{2}$
Hence, we get $\theta = \dfrac{\pi }{2} - - - - - - (4)$
Using equations (2),(3) and (4). We can obtained the polar form as,
$\therefore z = 1\left( {\cos \left( {\dfrac{\pi }{2}} \right) + i\sin \left( {\dfrac{\pi }{2}} \right)} \right)$

Therefore, polar form of $z = i$ is $z = 1\left( {\cos \left( {\dfrac{\pi }{2}} \right) + i\sin \left( {\dfrac{\pi }{2}} \right)} \right)$.

Note: We need to be very thorough with the trigonometric as well as inverse trigonometric formulas. Usually, we ignore the conditions that are to be considered while applying a particular formula like in this question, when we have to find $\theta$, we have to consider the two conditions, whether $x < 0$ or $x > 0$. When we need to convert it in polar form, we need to remember the formulas for finding the argument and the modulus values. While comparing, we should keep in mind the negative and positive signs and then while finding the argument we need to consider the quadrants according to the signs.