Complex numbers

oxford · 2009-02-18 21:42:44

Find the solutions of the equation

in polar form.

JaneFairfax · 2009-02-19 00:38:02

oxford · 2009-02-19 15:39:01

another one:
If the real part of

is zero, find the locus of points representing z in the complex plane.
Any ideas on what to do first?

mathsyperson · 2009-02-19 22:41:51

I'd think of it in modulus-argument form.

|a/b| = |a|/|b|, which makes finding the modulus of your number fairly easy to find.
arg(a/b) = arg(a) - arg(b). Due to symmetry, there's a nice way of expressing one argument in terms of the other one, and that should help you find an expression for arg(a/b).

Look at any restrictions you have on the modulus and on the argument, and figure out what the locus has to be.

Kurre · 2009-02-20 01:48:19

let w=z-1. then:
(z+1)/(z-1)=(w+2)/w=1+2/w=1+2w*/|w|. If this is imaginary, 2Re(w)/|w|=-1. But Re(w)/|w| is equal to Cos(Arg(w)).

JaneFairfax · 2009-02-20 11:26:41

I could have sworn there was a smiley in Kurres post originally.

Kurre · 2009-02-20 11:33:45

Yeah, I misread and thought (z+1)/(z-1) should be real, then I failed to edit and accidently posted a new post instead, so I deleted my first post, therefore no "edit"

JaneFairfax · 2009-02-20 11:49:16

No, I am referring to the fact that your post looked like this originally:

let w=z-1. then:
(z+1)/(z-1)=(w+2)/w=1+2/w=1+2w*/|w|. If this is imaginary, 2Re(w)/|w|=-1. But Re(w)/|w| is equal to Cos(Arg(w)).

Kurre · 2009-02-20 19:59:47

aha ok
thats weird. Website update?

JaneFairfax · 2009-02-20 22:56:43

Maybe Mathsy quietly edited and fixed your post.

mathsyperson · 2009-02-21 00:54:06

Super sleuthing.

I didn't touch the actual post though, I just ticked the box that stops text getting turned into smilies. There's a list of options just below Image Upload.

oxford · 2009-02-21 11:42:54

Thanks guys and girls.
I need on another.

|z-4-i|+|z+4+i| > 1

mathsyperson · 2009-02-21 12:28:38

That looks to be true everywhere.

Something encased in a modulus is always non-negative, so a solution to |z+4+i| > 1 would be a solution to your question. Similarly, any solution to |z-4-i| > 1 would work.

Getting more specific again, any solution to |Re (z+4+i)| = |Re (z) + 4| > 1 will be a solution to |z+4+i| > 1.

So, any value of z that does not satisfy -5 ≤ Re(z) ≤ -3 will work.

But using a similar process on |z-4-i| > 1, you get that any z that does not satisfy 3 ≤ Re(z) ≤ 5 will work.

Clearly any z satisfies one or both of these conditions, and so any z will solve the original inequality.

JaneFairfax · 2009-02-21 14:40:05

mathsyperson wrote:

That looks to be true everywhere.

mathsyperson · 2009-02-21 23:36:21

Yeah, the triangle inequality was such a better way of doing it. I came here to add that to my post but you'd got there already.

Math Is Fun Forum

#1 2009-02-18 21:42:44

Complex numbers

#2 2009-02-19 00:38:02

Re: Complex numbers

#3 2009-02-19 15:39:01

Re: Complex numbers

#4 2009-02-19 22:41:51

Re: Complex numbers

#5 2009-02-20 01:48:19

Re: Complex numbers

#6 2009-02-20 11:26:41

Re: Complex numbers

#7 2009-02-20 11:33:45

Re: Complex numbers

#8 2009-02-20 11:49:16

Re: Complex numbers

#9 2009-02-20 19:59:47

Re: Complex numbers

#10 2009-02-20 22:56:43

Re: Complex numbers

#11 2009-02-21 00:54:06

Re: Complex numbers

#12 2009-02-21 11:42:54

Re: Complex numbers

#13 2009-02-21 12:28:38

Re: Complex numbers

#14 2009-02-21 14:40:05

Re: Complex numbers

#15 2009-02-21 23:36:21

Re: Complex numbers

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