Sum of squares of modulus of all the complex numbers z satisfying / line 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 s(ii) From (i) and (ii) we get x=0 or y=-(1 / 2) When x=0 we get y=0 When y=-(1 / 2) we get x^2-x-(1 / 4)=0 => x=frac-1 ± sqrt(2)2 So there will be total 3 possible values of z, which are 0,(frac-1+sqrt(2)2)-(1 / 2) i and (frac-1-sqrt(2)2)-(1 / 2) i Sum of squares of modulus =0+(fracsqrt(2)-12)^2+(1 / 4)+(fracsqrt(2)+12)^2=+(1 / 4) \ =2

Sum of squares of modulus of all the complex numbers z ... | 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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