Let alpha and beta be the sum and the product | vedclass

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

Question

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $z = x + iy$. Then $\bar{z} = x - iy$ and $|z| = \sqrt{x^2 + y^2}$.Given equation: $(x - iy)^2 + \sqrt{x^2 + y^2} = 0$.$(x^2 - y^2 - 2ixy) + \sqrt{x^2 + y^2} = 0$.Equating the imaginary part to zero: $-2xy = 0 \implies x = 0$ or $y = 0$.Case $1$: If $x = 0$,then $-y^2 + |y| = 0 \implies |y|^2 = |y|$. Since $z 
eq 0$,$|y| = 1$,so $y = 1$ or $y = -1$. Thus,$z_1 = i$ and $z_2 = -i$.Case $2$: If $y = 0$,then $x^2 + |x| = 0 \implies |x|^2 + |x| = 0$. Since $|x| \geq 0$,this implies $|x| = 0$,so $x = 0$,which gives $z = 0$ (not a non-zero solution).The non-zero solutions are $z_1 = i$ and $z_2 = -i$.Sum $\alpha = i + (-i) = 0$.Product $\beta = i 	imes (-i) = -i^2 = 1$.Therefore,$4(\alpha^2 + \beta^2) = 4(0^2 + 1^2) = 4(1) = 4$.

Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^2+|z|=0$,where $z \in \mathbb{C}$. Then $4(\alpha^2+\beta^2)$ is equal to:

Similar Questions

If $\frac{2z_1}{3z_2}$ is a purely imaginary number,then $\left| \frac{z_1 - z_2}{z_1 + z_2} \right| =$

If $e^{it} = \cos t + i \sin t$ and $e^{-it} = \cos t - i \sin t$, then $\cosh(x + iy) - \cosh(x - iy) =$

Let $z \in \mathbb{C}$ be such that $\frac{z^2+3i}{z-2+i}=2+3i$. Then the sum of all possible values of $z^2$ is

If $2i$ is a root of $f(z) = z^4 + z^3 + 2z^2 + 4z - 8 = 0$,then which among the following cannot be a root of $f(z) = 0$?

The modulus-amplitude form of $\frac{(1-i)^3(2-i)}{(2+i)(1+i)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module