The number of complex numbers z such that |z| | vedclass

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(B) Let $z = x + iy$. Given equation: $|z| + z - 3\bar{z} = 0$. Substituting $z = x + iy$ and $\bar{z} = x - iy$: $\sqrt{x^2 + y^2} + (x + iy) - 3(x -

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$1$

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(B) Let $z = x + iy$. Given equation: $|z| + z - 3\bar{z} = 0$. Substituting $z = x + iy$ and $\bar{z} = x - iy$: $\sqrt{x^2 + y^2} + (x + iy) - 3(x - iy) = 0$. $\sqrt{x^2 + y^2} + x + iy - 3x + 3iy = 0$. $\sqrt{x^2 + y^2} - 2x + 4iy = 0$. Equating the real and imaginary parts to zero: Imaginary part: $4y = 0 \implies y = 0$. Real part: $\sqrt{x^2 + 0^2} - 2x = 0 \implies |x| = 2x$. If $x \ge 0$,then $x = 2x \implies x = 0$. If $x Thus,the only solution is $x = 0$ and $y = 0$,which means $z = 0$. Therefore,there is only $1$ such complex number.

The number of complex numbers $z$ such that $|z| + z - 3\bar{z} = 0$ is equal to

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