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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 -

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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

Similar Questions

If ${z_1}$ and ${z_2}$ are two complex numbers,then $|{z_1} - {z_2}|$ is

If $Z_1=2+i$ and $Z_2=3-4i$,and $\frac{\overline{Z_1}}{\overline{Z_2}}=a+bi$,then the value of $-7a+b$ is (where $i=\sqrt{-1}$ and $a, b \in \mathbb{R}$)

Find the modulus and argument of the complex number: $\frac{1}{1+i}$

If $x+iy = \frac{a+ib}{a-ib}$,prove that $x^{2}+y^{2}=1$.

Find the conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module