Sum of squares of modulus of all the complex numbers z satisfying z+ z=(1+i)z^2 is equal t…

Question

Sum of squares of modulus of all the complex numbers $z$ satisfying $z+\bar z=(1+i)z^2$ is equal to \_\_\_\_\_

Answer: 2

Accepted Answer

Let $z=x+iy$, so $z+\bar z=2x$. Then $2x=(1+i)(x^2-y^2+2xyi)$:
$$2x = \big[(x^2-y^2)-2xy\big]+i\big[2xy+(x^2-y^2)\big].$$

Real: $2x=(x^2-y^2)-2xy$. Imaginary (must vanish, since LHS is real): $x^2-y^2+2xy=0$.

Subtracting: $4xy=-2x \implies x=0$ or $y=-\dfrac12$.

$x=0$: then $-y^2=0 \implies y=0$, giving $z=0$.

$y=-\dfrac12$: $x^2-\dfrac14-x=0 \implies x=\dfrac{1\pm\sqrt2}2$.

So the solutions are $z=0,\ \dfrac{1+\sqrt2}2-\dfrac i2,\ \dfrac{1-\sqrt2}2-\dfrac i2$.

$$\sum|z|^2 = 0+\left[\left(\frac{1+\sqrt2}2\right)^2+\frac14\right]+\left[\left(\frac{1-\sqrt2}2\right)^2+\frac14\right].$$

$$\left(\frac{1+\sqrt2}2\right)^2+\left(\frac{1-\sqrt2}2\right)^2=\frac{3+2\sqrt2}4+\frac{3-2\sqrt2}4=\frac64=\frac32.$$

$$\sum|z|^2=\frac32+\frac12=2.$$

Solutions · 1