Question

I need to write \[\;8i\] in polar form. how to find angle when \[\dfrac{8}{0}\] is not defined?

Answer 1

Hint:
\[\omega \,\]lies in the positive y-axis, so the angle of \[\omega \,\] is $\dfrac{\pi }{2}$.
We are going to find the polar form of the given complex number, then since we have a condition which cannot be defined, we are going to prove that the found polar form is true since $\tan \theta $ is periodic.

Complete step by step solution:
When measuring the trigonometric angle\[
  z = 8\{ cos(\dfrac{\pi }{2}) + isin(\dfrac{\pi }{2})\} \;\; \\
   = 8\{ (0) + i(1)\} \;\; \\
   = 8i \\
\], (which we call the argument) in an anticlockwise direction from the positive x-axis, so we call see from geometry that ω lies on the positive y-axis, so the angle we seek for \[\omega \] is \[\dfrac{\pi }{2}\]. We can readily confirm this by writing the polar form:
\[
  z = 8\{ cos(\dfrac{\pi }{2}) + isin(\dfrac{\pi }{2})\} \;\; \\
   = 8\{ (0) + i(1)\} \;\; \\
   = 8i \\
\]
Now, Consider the general case. We seek \[r\] and \[\theta \]
Such that:
\[r\{ cos\theta + isin\theta \} = a + ib\]

We calculate r by equating real and imaginary parts then:
$
  \left \{
  a = rcos\theta \\
  b = rsin\theta \\
\right\} \\
   \Rightarrow a2 + b2 = r2 \\
$
From that we can get,
\[r = \sqrt {{a^2} + {b^2}} \]

And we calculate \[\theta \] by using elementary trigonometry:
\[tan\theta = \dfrac{{rsin\theta }}{{rcos\theta }} \Rightarrow tan\theta = \dfrac{b}{a}\] \[tan\theta = \dfrac{{rsin\theta }}{{rcos\theta }} \Rightarrow tan\theta = \dfrac{b}{a}\]
From the above, we get
\[\theta = arctan(\dfrac{b}{a})\]
Remembering that tangent is periodic, we cannot assume that we seek the principal value, so take care to gain the appropriate angle (and sign) by consideration of the quadrant of the required point (typically by a very quick argand diagram sketch).
Then, we have
\[|\omega | = \sqrt {{0^2} + {8^2}} = 8\]
Then, we have been given a condition, which we will equate to it, we get
\[tan\theta = \dfrac{8}{0} = \infty \Rightarrow arg\;\omega = \dfrac{\pi }{2}\]
Therefore, we get
\[8i = 8\{ cos(\dfrac{\pi }{2}) + isin(\dfrac{\pi }{2})\} \;\;\]

Note:
We should be careful with angles when we are finding the polar form as we should know the values of $\cos \theta $and $\sin \theta $. So that the angles do not get mixed, as they both have the same values but at different angles.