Hey pple ,I'm currently at university as a mature aged student and am struggling with complex numbers .
Here is one z=3/2+3/2isqrt3
how do i find the Arg (z ) of it ...
thanks
$z = \dfrac 3 2 + i \dfrac{3 \sqrt{3}}{2}$
just use the arctan function on the ratio of the imaginary to the real part paying attention to what quadrant it will fall into.
here both components are positive so we're in the first quadrant
$\theta = \arctan\left(\dfrac{\dfrac{3\sqrt{3}}{2}}{\dfrac 3 2}\right) = \arctan\left(\sqrt{3}\right) = \dfrac \pi 3$
The formula that Romsek and Plato have quoted are useful.
At this early stage though I would suggest you draw an Argand diagram of the complex number, construct a right angled triangle in the appropriate quadrant, and calculated the acute angle made with the x-axis (using basic trig).
Then remember that Arg(z) is measured anticlockwise (from pos x-axis) for positive angles up to (and including) pi, and measured anticlockwise (from pos -axis) for negative angles to -pi.
So:
..if your acute angle is A and your complex number is in Quadrant 1, then Arg(z) =A
..if your acute angle is A and your complex number is in Quadrant 2, then Arg(z) = pi - A
..if your acute angle is A and your complex number is in Quadrant 3, then Arg(z) = -(pi- A)
..if your acute angle is A and your complex number is in Quadrant 4, then Arg(z) = - A
Make sure you understand WHY.
Not sure what you mean? But do you know the value of
tan(pi/3)
tan(pi/6)
and
tan(pi/4)?
Or are you wondering where the pi/3 came from in romsek's answer?
I'm not rally sure of those tan values tan(pi/30 ) ,tan(pi/6) and tan(pi/4) ...plus where pi/3 came from ...I know on a unit circle sine is the x plane and cos is the y plane ,I 'm not so sure about tan ....sorry pretty rusty on these concepts
Prerequisite knowledge for this topic includes knowing these values. Have a look at Trigonometric Ratios of Special Angles: 0, 30, 45, 60, 90 (solutions, examples, videos, worksheets, activities)
It makes the rest so much easier if you know, for example that tan 60 deg = tan (pi/3) = sqrt 3. Romsek's answer comes from knowing that fact.