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

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Answer 1 · 2016-09-30T04:21:58

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

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, ...#