Question

How do you write the complex number in trigonometric form `4i`?

Answer 1

4i=4 cis (pi/2).
For the general form,
4i = 4 cis ((2k+1/2)pi)`, k = 0, +-1, +-2, +-3, ...`

Explanation:

Any complex number in rectangular cartesian form is

z = (x, y) = (real part) x + (imaginary part ) iy), where x and y are real.

(x, y) in polar form is

`r(cos theta, sin theta)`,

`r(cos theta+isin theta)` and, in brief,

=r cis `theta`

The conversion is from

`r=sqrt(x^2+y^2)>=0`,

`cos theta = x/sqrt(x^2+y^2) and sin theta = y/sqrt(x^2+y^2)`

Here, x = 0, y = 4, and so,

`r = sqrt(4^2+0)=4` ( the principal square root ),

`cos theta = 0 and sin theta = 4/4 = 1`.

The value of theta =`pi/2 in Q_1`.

The general value is `2kpi+pi/2 in Q_1, k = 0, +-1, +-2, +-3, ...`

All values point to the same direction..

So, seemingly, the general form might be viewed as irrelevant. Yet,

for rotation problems, with `theta = theta` (time ) = ct, the same point

is reached in a cycle of period `2pi`. And so, it cannot be ignored,

Answer: 4i=4 cis (pi/2).

For the general form,

4i = 4 cis ((2k+1/2)pi)`, k = 0, +-1, +-2, +-3, ...`