Complex numbers in rectangular form

[feral knoll]

Honestly have no idea where to start, any help would be amazing

[delicate pier]

a very useful fact to remember when dealing with complex numbers, is that any number multiplied by its conjugate is real

[limber vault]

The notation here is unfamiliar... is this:

$z_1 = 1 + 3i$

$z_2 = 1 - 2i$

$v=40\left(\cos{30^\circ}+i\sin{30^\circ}\right)$

Find $I=\frac{v}{z_1+z_2}$

??

[rocky coral]

Jay

[feral knoll]

ok thankyou!! i’ll have a go and then see if it’s correct 🙂

[feral knoll]

wait i read this completely wrong sorry i was out when i saw this

the question is:

[limber vault]

When it says $V=40\angle{30^\circ}$, does this mean that $|V|=40$ and $\arg{(V)}=30^\circ$?

[rocky coral]

Jay

[feral knoll]

It’s all good! thankyou for your help but i managed to figure it out!

[limber vault]

Great... but I am still interested in a notation I haven't seen before. Would you please explain? And maybe post the whole test?

[feral knoll]

yeah ofcourse, this is engineering mathmathis so this question is about imagery numbers ( sounds stupid i know)

So in this case J is just a value

but it’s “imaginary” so to speak

When you have j^2 it’s equal to -1

so say you have 20j^2 that would be -20

Rectangular form is the same as cartesian form, which means fully work out the answer, so say you had a 10cos30, you wouldn’t be able to leave it in that form, you would have to work it out using a calculator

When dealing with these equations we use a formula z=R x E^jtheta

for this example i will use the question z=4+j3

to find R we do 4^2+ 3^2= R^2 using the numbers from the given expression

so our R=5

[limber vault]

I know what complex numbers are 🙂

But I am used to $z = x+iy$ where $x,y\in\mathbb{R}$ and $i^2=-1$

Evidently $j$ is used in place of $i$ where you are studying.

[rocky coral]

Jay

[limber vault]

If, for $V = 40\angle{30^\circ}$, you have $z=Re^{j\theta}$,

you have $\theta = 30^\circ \equiv \frac{\pi}{6}$ radians...

do you have $R=40$?