Sum of squares of modulus of all the complex numbers z satisfying z = iz^2 + z^2 - z is equal to ___________.

Let z=x+i y So 2 x=(1+i)(x^2-y^2+2 x y i) 2 x=x^2-y^2-2 x y ...(i) and x^2-y^2+2 x y=0(ii) From (i) and (ii) we get x=0 or y=-12 When x=0 we get y=0 When y=-12 we get x^2-x-14=0 x=-1 22 So there will be total 3 possible values of z, which are 0,(-1+22)-12 i and (-1-22)-12 i Sum of squares of modulus a

Sum of squares of modulus of all the complex numbers z satis: null (Mathematics)

Let

z=x+i y

2 x=(1+i)\left(x^{2}-y^{2}+2 x y i\right)

\Rightarrow 2 x=x^{2}-y^{2}-2 x y\quad

...(i) and

x^{2}-y^{2}+2 x y=0\quad\dots(ii)

From (i) and (ii) we get

x=0 \text { or } y=-\frac{1}{2}

When

x=0

we get

y=0

When

y=-\frac{1}{2}

we get

x^{2}-x-\frac{1}{4}=0

\Rightarrow \quad x=\frac{-1 \pm \sqrt{2}}{2}

So there will be total 3 possible values of

z

, which are

0,\left(\frac{-1+\sqrt{2}}{2}\right)-\frac{1}{2} i

and

\left(\frac{-1-\sqrt{2}}{2}\right)-\frac{1}{2} i

Sum of squares of modulus

\begin{aligned} &=0+\left(\frac{\sqrt{2}-1}{2}\right)^{2}+\frac{1}{4}+\left(\frac{\sqrt{2}+1}{2}\right)^{2}=+\frac{1}{4} \\\\ &=2 \end{aligned}

Explanation

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