Find all Complex no. Z which satisfy the following expression
(1)Z(bar) = i(Z2) without using Z=x+iy
(2) question
Z2+|Z|=0
622nd question..
z^2=|z|
so z^2 is a real number..
hence z is real... or purely imaginary...
hence |z|=-z^2
but then if z is real, only solution is zero..
if z is imaginary, then z=±i are the only solution..
621st question,
see the thing here is that taking modulus, we can say that |z|=1
now use reiθ to finish it ;)
1708(2) Z2+|Z|=0
Z2 = -|Z|
taking modulus on both side we get
|Z|2 = |Z|
or |Z|2-|Z|=0
|Z|(|Z|-1) = 0
so Z=0 and |Z|=1 means Z=+i or -i (from question )
Nishant sir is it right or not.
66Q1) Since
\overline{z}=iz^2
Taking mod both sides we get
|z|=|z|^2
(since |z|=|\overline{z}|)
So we get |z|=0 or |z|=1.
If |z|=0\quad\Rightarrow\ \boxed{z=0}
If |z|=1, then multiplying the original equation with z on both sides we get
|z|^2=iz^3\quad\Rightarrow \ iz^3=1\quad\Rightarrow \ z^3 = -i
Transform as follows:
z^3+i=0\quad\Rightarrow \ z^3-i^3=0\quad\Rightarrow \ (z-i)(z-i\omega)(z-i\omega^2)=0
where \omega is a cube root of unity. From where we get
z=i, \ i\omega,\ i\omega^2