How Do You Convert Complex Numbers Between Cartesian and Polar Forms?

[vorcil]

Take,

P = 4e^{-j\frac{\pi}{3}}

Q = 4-3j

R = 2e^{j\frac{\pi}{2}}

S = 5

note: I'm using j to be a complex number, it's equivalent to i in mathematics

-

A: express p q r s in both cartesian (a+ib) and polar (re^itheta) forms

B: sketch p q r and s on the complex plane

--------------------------------------------------------------
I'm not too certain as to how P and R work,
I know from the taylor series that r(cos(theta)+isin(theta)) = re^itheta

P is already in polar form, P = 4e^{-j\frac{\pi}{3}}
so the Cartesian form,
4(cos(pi/3)+j(sin(pi/3)))

(a+jb) = 4 cos\frac{\pi}{3} = a and 4 sin \frac{\pi}{3} = b

so it is 2+j(3.464) or 2+j\sqrt{12}
NOTE I AM NOT SURE, BECAUSE THE ORIGINAL EQUATION SHOWS -J
SO SHOULD I PUT, 2-j\sqrt{12} ??
----------------------------------------------------------------
Q = 4-3j
Q is already in cartesian form,

|z| q = 5, = root 4^2 + -(3^2)
\theta = cos^{-1} \frac{4}{5} = 0.643 rad

so the polar form is , 5cis(0.643) = 5(cos(0.643) + jsin(0.643)) = 5e^(j0.643)

----------------------------------------------------------------
R = 2e^{j\frac{pi}{2}}
R is already in polar form,

2 = |z|

2 cos(pi/2) = 0,
2 sin(pi/2) = 2,
this means if i wear to imagine the vector R, it would be going straight up,

so in cartesian form, the equivalent equation of R = 2e^{j\frac{pi}{2}} is j2

-------------------------------------------------------------------------------------------------
S=5
I'm assuming S is already in cartesian form, since it is 5+j0

|z| =5
the angle it makes with the x axis, is 0,
so i think the equation for S in polar form is, 5e^{j * 0 }

 

[vorcil]

Determine numberical answers for each of the following:
either in cartesian form or in polar form with the angle in degrees

------------------------------------------------------------------------

1) (PQ)^1/2
2) (R/P)^1/3

-------------------------------------------------------------------------------------

 

[vorcil]

i hate macs,

I calculate pq to be,

20e^{-j(\frac{\pi}{3} - 0.643)}

how do i square root it?

\sqrt{20e^{-j(\frac{\pi}{3} - 0.643)}}

 

[ammontgo]

=>P is already in Exponential form. To convert to Cartesian:

Euler's Identity:
[PLAIN]https://dl.dropbox.com/u/4645835/MATH/EulersID.gif

so:
[PLAIN]https://dl.dropbox.com/u/4645835/MATH/Pcart.gif

=>Q is already in Cartesian; so convert to Polar:
[PLAIN]https://dl.dropbox.com/u/4645835/MATH/Qpol.gif

=>Convert Q to Exponential:
[PLAIN]https://dl.dropbox.com/u/4645835/MATH/Qexp.gif


I think these are right and I hope they help.

 

[HallsofIvy]

Take,

P = 4e^{-j\frac{\pi}{3}}

Q = 4-3j

R = 2e^{j\frac{\pi}{2}}

S = 5

note: I'm using j to be a complex number, it's equivalent to i in mathematics

-

A: express p q r s in both cartesian (a+ib) and polar (re^itheta) forms

B: sketch p q r and s on the complex plane

--------------------------------------------------------------
I'm not too certain as to how P and R work,
I know from the taylor series that r(cos(theta)+isin(theta)) = re^itheta

P is already in polar form, P = 4e^{-j\frac{\pi}{3}}
so the Cartesian form,
4(cos(pi/3)+j(sin(pi/3)))

No, \theta= -\frac{\pi/3}, not \frac{\pi}{3}.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> (a+jb) = 4 cos\frac{\pi}{3} = a and 4 sin \frac{\pi}{3} = b<br /> <br /> so it is 2+j(3.464) or 2+j\sqrt{12}<br /> NOTE I AM NOT SURE, BECAUSE THE ORIGINAL EQUATION SHOWS -J<br /> SO SHOULD I PUT, 2-j\sqrt{12} ??<br /> ----------------------------------------------------------------<br /> Q = 4-3j<br /> Q is already in cartesian form, <br /> <br /> |z| q = 5, = root 4^2 + -(3^2) <br /> \theta = cos^{-1} \frac{4}{5} = 0.643 rad<br /> <br /> so the polar form is , 5cis(0.643) = 5(cos(0.643) + jsin(0.643)) = 5e^(j0.643)<br /> <br /> ----------------------------------------------------------------<br /> R = 2e^{j\frac{pi}{2}}<br /> R is already in polar form,<br /> <br /> 2 = |z| <br /> <br /> 2 cos(pi/2) = 0,<br /> 2 sin(pi/2) = 2,<br /> this means if i wear to imagine the vector R, it would be going straight up,<br /> <br /> so in cartesian form, the equivalent equation of R = 2e^{j\frac{pi}{2}} is j2<br /> <br /> -------------------------------------------------------------------------------------------------<br /> S=5<br /> I'm assuming S is already in cartesian form, since it is 5+j0<br /> <br /> |z| =5 <br /> the angle it makes with the x axis, is 0,<br /> so i think the equation for S in polar form is, 5e^{j * 0 } </div> </div> </blockquote>

 

[ammontgo]

It is possible for an expression to fit into more than one category or into none of the categories. I'm not sure how this works anymore..as I did this a long time ago. But I think S satisfies exponential and cartesian.