Question

If arg(z- w/z-w^2)=0 then prove that Re. (z)=-1/2( w and w^2 are real cube root of unity. )

Answer 1

Hey mate,

◆ Proof-
Let ω and ω^2 be cube roots of unity.

For given condition,
arg[(z-ω)/(z-ω^2)] = 0
We can say, the number z is a real number.

That is
Re(z-ω) = Re(z-ω^2) = 0
Re(z) - Re(ω) = 0
Re(z) = Re(ω)

We know that ω = -1/2 + √3i/2, i.e. Re(ω) = -1/2
Putting this value, we get-
Re(z) = -1/2


Hope this is useful...

Answer 2

Answer:

Step-by-step explanation:

concept:

If arg(z)=0 then Im(z)=0 where z=x+iy

Given:

$arg(\frac{z-\omega}{z-{\omega}^2})=0$

Then,

$Im(\frac{z-\omega}{z-{\omega}^2})=0$

Take z=x+iy,

$\omega=\frac{-1}{2}+i\frac{\sqrt{3}}{2}\:{\omega}^2=\frac{-1}{2}-i\frac{\sqrt{3}}{2}$

$\frac{z-\omega}{z-{\omega}^2}$

$=\frac{x+iy+\frac{1}{2}+i\frac{\sqrt{3}}{2}}{x+iy+\frac{1}{2}-i\frac{\sqrt{3}}{2}}$

$=\frac{(x+\frac{1}{2})+i(y+\frac{\sqrt{3}}{2})}{(x+\frac{1}{2})+i(y-\frac{\sqrt{3}}{2})}$

$Im(\frac{z-\omega}{z-{\omega}^2})=0$

implies

$\frac{(x+\frac{1}{2})(y+\frac{\sqrt{3}}{2})-(x+\frac{1}{2})(y-\frac{\sqrt{3}}{2})}{(x+\frac{1}{2})^2+(y+\frac{\sqrt{3}}{2})^2}=0$

$(x+\frac{1}{2})[(y+\frac{\sqrt{3}}{2})-(y-\frac{\sqrt{3}}{2})]=0$

$(x+\frac{1}{2})[(y+\frac{\sqrt{3}}{2})-y+\frac{\sqrt{3}}{2})]=0$

$(x+\frac{1}{2})\sqrt{3}=0$

$x+\frac{1}{2}=0$

$x=-\frac{1}{2}$

$Re(z)=-\frac{1}{2}$