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

Sep 30, 2016

4i=4 cis (pi/2).
For the general form,
4i = 4 cis ((2k+1/2)pi)$, k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

#### 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

$r \left(\cos \theta , \sin \theta\right)$,

$r \left(\cos \theta + i \sin \theta\right)$ and, in brief,

=r cis $\theta$

The conversion is from

$r = \sqrt{{x}^{2} + {y}^{2}} \ge 0$,

$\cos \theta = \frac{x}{\sqrt{{x}^{2} + {y}^{2}}} \mathmr{and} \sin \theta = \frac{y}{\sqrt{{x}^{2} + {y}^{2}}}$

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

$r = \sqrt{{4}^{2} + 0} = 4$ ( the principal square root ),

$\cos \theta = 0 \mathmr{and} \sin \theta = \frac{4}{4} = 1$.

The value of theta =$\frac{\pi}{2} \in {Q}_{1}$.

The general value is $2 k \pi + \frac{\pi}{2} \in {Q}_{1} , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

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 $2 \pi$. And so, it cannot be ignored,

4i = 4 cis ((2k+1/2)pi)$, k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$