Complex Numbers Question (1 Viewer)

[nrlwinner]

I've got a quick query here.

How would I find the modulus and argument of something like this

 

[bob fossil]

use trig forumlae, dont ask me how to do it tho lol

 

[iSplicer]

I've got a quick query here.

How would I find the modulus and argument of something like this

you need to force it into half angle form. i got mod z = 2cos(t/2), arg z = t/2.

 

[nrlwinner]

Thanks for the comment but I'm still unsure how to do this. For example find the modulus and argument of



using Trig rather than expanding the whole thing, which sometimes gets very messy.

 

[kaz1]

|z|²= (1 +cos(pi/3))² +sin²(pi/3)
1+2cos(pi/3) +cos²(pi/3) +sin²(pi/3)
2+1=3
|z|=rt3

tan (arg z)= sin(pi/3)/[1+cos(pi/3)]
=(rt3/2)/(3/2)
=1/rt3
tan (arg z)=1/rt3
argz=pi/6

 

[nrlwinner]

Sorry I'm not really good with Trig. Can you explain each of the formulas used for the step.

 

[study-freak]

EDIT: further simplified modulus as I noticed I could.

 

[kaz1]

Sorry I'm not really good with Trig. Can you explain each of the formulas used for the step.

It's good to look at it as a right angled triangle. For finding the modulus i used Pythagoras theorem where the real part is the horizontal side and the imaginary part is the vertical side.

For finding the argument I used the tan ratio where the Imaginary part is the opposite side and the real part is the adjacent side.

If you want to do it just using an algebraic way:

|z|²=Rez² +Imz²
tan argz =Imz/Rez

 

[GUSSSSSSSSSSSSS]

|z|²= (1 +cos(pi/3))² +sin²(pi/3)
1+2cos(pi/3) +cos²(pi/3) +sin²(pi/3)
2+1=3
|z|=rt3

tan (arg z)= sin(pi/3)/[1+cos(pi/3)]
=(rt3/2)/(3/2)
=1/rt3
tan (arg z)=1/rt3
argz=pi/6

Sorry I'm not really good with Trig. Can you explain each of the formulas used for the step.

the first line is just the standard method of finding the modulus (square root of the real part squared plus the imaginary part squared)
then he expands it
then uses cos²(pi/3) +sin²(pi/3) = 1 ...... and 2cos(pi/3) = 2*1/2 = 1
and therefore gets 1 + 1 + 1 = 3
therefore modulus is = sqrt3


for the argument he uses the standard method for finding the argument (tan inverse of imaginary part divided by the real part)
then he uses sin(pi/3 = sqrt3 / 2 ....... and cos(pi/3) = 1/2
and then simplifies to give his final answer

 

[nrlwinner]

Thanks study freak. I get the steps of what you're doing, but can you tell me the Trig identities you used.

 

[GUSSSSSSSSSSSSS]

Thanks study freak. I get the steps of what you're doing, but can you tell me the Trig identities you used.

cos²(@) +sin²(@) = 1


sin(2@) = 2sin(@)cos(@) (replace @ by @/2)

1 + cos(2@) = 2cos²(@) (replace @ by @/2)

 

[study-freak]

These are the ones I used.
lol GUESSSSSSSSSSSSS already posted it up.

 

[addikaye03]

use trig forumlae, dont ask me how to do it tho lol

That would only really be for simplifying ie.

[(1+cos@+isin@)/(1+cos@-isin@]^7

So cos2@=2cos^2@-1 ---> 1+cos2@=2cos^2@ ---> 1+cos@=2cos^2(@/2)

2sin@cos@=sin(2@) ---> sin(@)= 2sin(@/2)cos(@/2)

SO expression becomes

2cos^2(@/2)+i2cos(@/2)sin(@/2) on the numerator

Which can then be factorised further when you do the denominator.

then use de moivre

(nb. this Q can be done much more nicely geometrically, because its [(1+z)/(1+z*]^7, ie. equal moduli and equal angle between at point of origin]


EDIT: did'nt notice it asked for argument, it would work for that also haha

 

[The Nomad]

Umm...something a little different:

1 + Cos 2a + i Sin 2a = (1 + Cos 2a) + i Sin 2a = (1 + 2Cos^2 a - 1) + 2i Sin a Cos a = 2 Cos^2 a + 2i Sin a Cos a= 2Cos a (Cos a + i Sin a) = R Cis a, where R = modulus and a = argument.

So 1 + Cos 2a + i Sin 2a = 2 Cosa (Cos a + i Sin a), modulus = 2 Cos a, argument = a.

For your question, replace 2a with theta and it'll be fine.

 

[nrlwinner]

Also, is there any relationship between the modulus and argument of z and 1/z

 

[kaz1]

Also, is there any relationship between the modulus and argument of z and 1/z

Modulus are reciprocals and arguments are opposite in sign.

 

[ninetypercent]

Also, is there any relationship between the modulus and argument of z and 1/z

arg(1/z) = arg1 - argz = 0 - argz = -argz
the argument of 1/z is the negative of the argument of argz

modulus of 1/z
just the reciprocal of lzl

 

[nrlwinner]

Sorry for the questions guys. I need to fine tune this complex number stuff.

How would you find the relation between x and y if z=x+iy

 

[kaz1]

Sorry for the questions guys. I need to fine tune this complex number stuff.

How would you find the relation between x and y if z=x+iy

As far as I know x and y are independent of each other as x represents the real part of the complex number and y represents the imaginary part.

 

[ninetypercent]

Sorry for the questions guys. I need to fine tune this complex number stuff.

How would you find the relation between x and y if z=x+iy

no worries. ask more questions. its the best way to learn