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Complex Numbers question

2022 · 28 Jun · Shift 2 · Q48
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Complex Numbers question

2022 · 28 Jun · Shift 2 · Q48

JEE MainMathematicsComplex NumbersNumerical+4 / −1
Sum of squares of modulus of all the complex numbers z satisfying z‾=iz2+z2−z\overline z = i{z^2} + {z^2} - zz=iz2+z2−z is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. Let z=x+iy,z=x+iy,z=x+iy, where x,y∈Rx,y\in\mathbb Rx,y∈R. Then z‾=x−iy.\overline z=x-iy.z=x−iy. Also, z2=(x+iy)2=x2−y2+2ixy.z^2=(x+iy)^2=x^2-y^2+2ixy.z2=(x+iy)2=x2−y2+2ixy.

  2. The given equation is z‾=iz2+z2−z=(1+i)z2−z.\overline z=iz^2+z^2-z=(1+i)z^2-z.z=iz2+z2−z=(1+i)z2−z. So, x−iy=(1+i)(x2−y2+2ixy)−(x+iy).x-iy=(1+i)(x^2-y^2+2ixy)-(x+iy).x−iy=(1+i)(x2−y2+2ixy)−(x+iy).

  3. First compute (1+i)z2(1+i)z^2(1+i)z2. Let a=x2−y2,b=2xy,a=x^2-y^2,\qquad b=2xy,a=x2−y2,b=2xy, so that z2=a+ib.z^2=a+ib.z2=a+ib. Then (1+i)(a+ib)=(a−b)+i(a+b).(1+i)(a+ib)=(a-b)+i(a+b).(1+i)(a+ib)=(a−b)+i(a+b). Hence, (1+i)z2=(x2−y2−2xy)+i(x2−y2+2xy).(1+i)z^2=(x^2-y^2-2xy)+i(x^2-y^2+2xy).(1+i)z2=(x2−y2−2xy)+i(x2−y2+2xy).

Therefore,

\bigl(x^2-y^2-2xy-x\bigr)+i\bigl(x^2-y^2+2xy-y\bigr).$$ 4. Equating real and imaginary parts with $x-iy$ gives: $$x=x^2-y^2-2xy-x$$ and $$-y=x^2-y^2+2xy-y.$$ So, $$2x=x^2-y^2-2xy \quad ...(1)$$ and $$0=x^2-y^2+2xy. \quad ...(2)$$ 5. Add and subtract these equations. From (2), $$x^2-y^2=-2xy.$$ Substitute into (1): $$2x=(-2xy)-2xy=-4xy,$$ so $$x(1+2y)=0.$$ Now solve cases. ### Case 1: $x=0$ Then from (2), $$0^2-y^2+0=0 \implies y^2=0 \implies y=0.$$ Thus, $$z=0.$$ ### Case 2: $1+2y=0$ Then $$y=-\frac12.$$ Substitute into (2): $$x^2-\left(\frac14\right)+2x\left(-\frac12\right)=0$$ $$x^2-\frac14-x=0$$ $$x^2-x-\frac14=0.$$ Solving, $$x=\frac{1\pm\sqrt{1+1}}{2}=\frac{1\pm\sqrt2}{2}.$$ So the two solutions are $$z=\frac{1+\sqrt2}{2}-\frac{i}{2},\qquad z=\frac{1-\sqrt2}{2}-\frac{i}{2}.$$ 6. Now compute the sum of squares of moduli of all solutions. - For $z=0$, $$|z|^2=0.$$ - For $z=x-\frac{i}{2}$, $$|z|^2=x^2+\frac14.$$ Using the two values of $x$, let them be roots of $$x^2-x-\frac14=0.$$ Their sum is $1$ and product is $-\frac14$. Hence, $$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=1^2-2\left(-\frac14\right)=1+\frac12=\frac32.$$ Therefore, $$|z_1|^2+|z_2|^2=(x_1^2+x_2^2)+2\cdot\frac14=\frac32+\frac12=2.$$ Including $z=0$, total sum is $$0+2=2.$$ 7. Therefore, the required integer is $$\boxed{2}.$$
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