# Imaginary part of z

#### idontknow

Find y such that $$\displaystyle Im(z)=0$$.

$$\displaystyle z=\dfrac{1}{1+yi}-\dfrac{1}{1+y^2 i }$$.

#### mathman

Forum Staff
y=0 and y=1 are solutions. With a little algebra, it can be shown these are the only solutions.

idontknow

#### skipjack

Forum Staff
Is $y$ required to be real?

Forum Staff

#### skipjack

Forum Staff
W|A finds a solution for which $y$ isn't real.

Forum Staff

#### skipjack

Forum Staff
-1 - 1.1474737891249609i approximately

idontknow

#### idontknow

$$\displaystyle Im(z)=[\dfrac{i}{i-y} - \dfrac{i}{i-y^2 }]_{i} ' =0 \;$$ ; $$\displaystyle \; (y-y^2 )(y^3 -1)=0$$.
$$\displaystyle y^2 =y$$ has real solutions $$\displaystyle y=0 \vee y=\pm 1$$.
$$\displaystyle y^3=1$$ has comlex solutions $$\displaystyle y=\dfrac{1}{2} \pm i\dfrac{\sqrt{3}}{2}$$.

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