Question

I need to write 8i in polar form. how to find angle when 8/0 not defined ?

Answer 1

`8i = 8{cos(pi/2)+isin(pi/2)}`

Explanation:

We seek a polar form of:

`omega = 8i`

When measuring the trigonometric angle, `theta`, (which we call the argument) in an anticlockwise direction from the positive `Ox`-axis, so we call see from geometry that `omega` lies on the positive `Oy`-axis, so the angle we seek for `omega` is `pi/2`. We can readily confirm this by writing the polar form:

`z = 8{cos(pi/2)+isin(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:

`{ (a = rcos theta), (b = rsin theta) :} => a^2 + b^2 = r^2`

so that:

`r = sqrt(a^2 + b^2)`

And we calculate `theta` by using elementary trigonometry:

`tan theta = (rsin theta)/(r cos theta) => tan theta = b/a`

so that:

`theta = arctan(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).

So in the above example, with `omega = 8i`:

Then we have:

`| omega | = sqrt(0^2+8^2) = 8`

And:

`tan theta = 8/0 = oo => "arg" \ omega = pi/2`

Thus:

`8i = 8{cos(pi/2)+isin(pi/2)}`