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Find the solutions of the equation
in polar form.Offline
another one:
If the real part of
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.
Why did the vector cross the road?
It wanted to be normal.
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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)).
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I could have sworn there was a smiley in Kurres post originally.
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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"
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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)).
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aha ok
thats weird. Website update?
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Maybe Mathsy quietly edited and fixed your post.
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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.
Why did the vector cross the road?
It wanted to be normal.
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Thanks guys and girls.
I need on another.
|z-4-i|+|z+4+i| > 1
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.
Why did the vector cross the road?
It wanted to be normal.
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That looks to be true everywhere.
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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.
Why did the vector cross the road?
It wanted to be normal.
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