IIT JEE , AIEEE Forum - Mathematics , Algebra

nithinjc1

the number of roots of z³+z(congugate)=0????????

feynmann

Simply note that ( a side change and then taking the absolute value )

|Z|= 1

so take z = exp ( i@)

put it in the eqn to get cos (3@ ) + cos @ = 0

sin ( 3@ ) = sin @

I think the above two eqns are simple enough to solve .

vnkt.swaroop

z conjugate=1/z.now solve it.

hsbhatt

venkat has the better approach:

$z^3 + \bar{z} = 0 \Rightarrow z^3 = - \bar{z} \Rightarrow |z|^3 = |-\bar{z}| = |z|$

This means either z = 0 or |z| = 1

$|z| = 1 \Rightarrow \bar{z} = \frac{1}{z}$

The equation simply becomes z^4 = -1 which has 4 complex roots

Thus total number of roots is 5