1
Pavithra Ramamoorthy
·2009-03-24 21:18:41
no yaar.......... y do u say dis now??????? do u ve d sol wit u????????
1
ANKIT MAHATO
·2009-03-24 23:57:08
simple yaar
((1 + z)/z)6 = -1
now we can easily solve it ...
106
Asish Mahapatra
·2009-03-24 22:08:43
bhaiyya im getiing .. z = (-1±i)/2
1
Pavithra Ramamoorthy
·2009-03-24 22:02:43
YES BYAH... DIS S D WAY I DID .......
BUT I WEN WRON SOMEWHERE ELSE IN CALC.....
1
Pavithra Ramamoorthy
·2009-03-24 22:00:58
PLS TELL ME HOW TO SOLVE IT...................;-)
62
Lokesh Verma
·2009-03-24 22:00:11
does this help.. or am i blabbering!?
\left((Z+1)^3 \right)^2=(iZ^3)^2
21
tapanmast Vora
·2009-03-24 21:59:21
NO......
I say this coz :
the LOCUS u r suggesting is wrong :( jus take ne point on ur locus and c wether it satisfies :
(z+1)^6 + z^6 = 0
U shal hav ur answer
11
Mani Pal Singh
·2009-03-24 21:03:03
tapan wat r u trying to ask
please make it more clearer[1]
21
tapanmast Vora
·2009-03-24 21:17:24
Ram : | -1 | = 1 hota hai......... i hope u din make this mist
11
Mani Pal Singh
·2009-03-24 21:13:59
bhai roots chayiye ya fir locus
question will become different
if u need rootd then i think i have a method
21
tapanmast Vora
·2009-03-24 21:13:43
ya I did that!!!
Wats ur locus?? can u give ur eqtn in a proper form pl...
1
Pavithra Ramamoorthy
·2009-03-24 21:11:55
try solvin it by takin modulus on bth sides.....
21
tapanmast Vora
·2009-03-24 21:11:08
kaise yaar the power is "6" not 2
1
Pavithra Ramamoorthy
·2009-03-24 21:10:02
taking z= x+iy
locus s 2x2+2y2+2x+1=0????????????