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forgot how to do easy proving in complex number :( (2 Viewers)

coeyz

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prove :

argz1 - argz2 = arg (z1/z2)

anyone please help~~?
thankss
 

coeyz

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arg (z-1 / z+1) = pie/2

find the locus of the point which represents z

and also this question thanks
 

Timothy.Siu

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arg (z-1 / z+1) = pie/2

find the locus of the point which represents z

and also this question thanks
its a semi circle origin at the centre and radius 1 above the x axis (so 1,0 and -1, 0 are circled)
 

GUSSSSSSSSSSSSS

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for the first:
divide z1 by z2 where z1=cos(n) + isin(n)
z2=cos(m) + isin(m)

realise the denominator
group into real nd imaginary
and then its just a compound angle formula job
(sorry bout all the shitty notation)


the second one end up as a semicircle

i cbb to explain at the moment sorry someone else will lol
 

coeyz

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thanks guys :)

for the first one,
ye i did the way u said,
but and then i get

z1/z2 = r1/r2 cis (日1-日2)

then i can only say arg(r1/r2) = 日1-日2
 

GUSSSSSSSSSSSSS

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thanks guys :)

for the first one,
ye i did the way u said,
but and then i get

z1/z2 = r1/r2 cis (日1-日2)

then i can only say arg(r1/r2) = 日1-日2
sorry i dun get ur notation

but up the begining where u defined r1 and r2 you gave their arg
then you can just replace that again down the bottom
if you get wat i mean??????
 

coeyz

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z1/z2 = r1 (cosN+isinN) (cosM-isinM)
r2 (cosM+isinM) (cosM-isinM)

= r1 cos (N-M) - isin(N-M)
r2

= r1 cis (N-M)
r2

then?
 

lolokay

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arg(z1/z2) = 日1-日2
arg(z1/z2) = arg(z1) - arg(z2)
 

GUSSSSSSSSSSSSS

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z1/z2 = r1 (cosN+isinN) (cosM-isinM)
r2 (cosM+isinM) (cosM-isinM)

= r1 cos (N-M) - isin(N-M)
r2

= r1 cis (N-M)
r2

then?
yea exactly as the guy above me said
r1=cosN+isinN
so arg(r1)=N
similarly for r2
so arg(r1/r2)=N-M=arg(r1)-arg(r2)
 

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