106Let Z = z-1eiθ + eiθz-1
let z=x+iy
then z-1eiθ = ((x-1)+iy)e-iθ
and eiθz-1 = eiθ(x-1+iy)x2+y2
im(Z) = 0
=>y(cosθ) -(x-1)sinθ + (x-1)sinθ + ycosθx2+y2 = 0
dunno whether this will help
If the imaginary part of expression z-1eiθ +eiθz-1is zero,then find locus of z
i tried by taking z-1eiθ =t ..another complex no.. but in vain..
plz suggest a better method
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9 Answers
1061if any one has deep study material on complex nos can u plz email it to me karanmehtaforu@yahoo.com
it must have locus,arguments and tough questions
24where does this lead us asish...
its a nice work[1]..but i dont think it helps ..[2]
1357Let W=(z-1)/eiθ
Let Z'=W+1/W=W+W/|W|2
Now arg(Z')=0
So from graph we have |W|=1
or |(z-1)/eiθ|=1 or |z-1|=1
so locus is (x-1)2+y2=1
See if I m missing something
1357I missed one thing
we have either |w|=1 or W is real
I found it for |w|=1, you do it for next part
1357W+1/W=x+iy+(x-iy)/(x2+y2)
Im part is zero implies y(1-1/(x2+y2))=0
or y(x2+y2-1)=0
this might clear your doubt
1357Now (z-1)/eiθ= real
implies arg(z-1)=argeiθ
or tan-1[y/(x-1)]=θ
or y=(x-1)tanθ
So I m getting the locus as
y=(x-1)tanθ
or (x-1)2+y2=1
1i have a good method
z - z=2i(imginary part)
z - z=0
(z-1eiθ +eiθz-1) - (z-1ei[-θ] +ei[-θ]z-1) = 0
(z-1eiθ - z-1ei[-θ]) =(eiθz-1 -ei[-θ]z-1)
cancelling off and solving u get
(x-1)2+y2=1
sorry if there is any typo mistake