If z = x + iy satisfies |z|-2=0 and |z-i|-|z+ | vedclass

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

Question

(C) Given $|z| - 2 = 0$,so $|z| = 2$. This implies $x^2 + y^2 = 4$. Given $|z - i| - |z + 5i| = 0$,so $|z - i| = |z + 5i|$. Substituting $z = x + iy$:

Vedclass · Accepted Answer

$x - 2y - 4 = 0$

Vedclass · Answer

(C) Given $|z| - 2 = 0$,so $|z| = 2$. This implies $x^2 + y^2 = 4$. Given $|z - i| - |z + 5i| = 0$,so $|z - i| = |z + 5i|$. Substituting $z = x + iy$: $|x + (y - 1)i| = |x + (y + 5)i|$ $x^2 + (y - 1)^2 = x^2 + (y + 5)^2$ $(y - 1)^2 = (y + 5)^2$ $y^2 - 2y + 1 = y^2 + 10y + 25$ $-12y = 24$ $y = -2$. Substituting $y = -2$ into $x^2 + y^2 = 4$: $x^2 + (-2)^2 = 4$ $x^2 + 4 = 4$ $x^2 = 0$,so $x = 0$. Thus,the point is $(0, -2)$. Checking the options for $(0, -2)$: $A: 0 + 2(-2) - 4 = -8 
eq 0$ $B: 0^2 + (-2) + 4 = 2 
eq 0$ $C: 0 - 2(-2) - 4 = 4 - 4 = 0$. Therefore,the correct option is $C$.

If $z = x + iy$ satisfies $|z|-2=0$ and $|z-i|-|z+5i|=0$,then

Similar Questions

If $x = 3 + i$,then $x^3 - 3x^2 - 8x + 15 = $

Convert the complex number $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$ into polar form.

$\frac{(1+i)^{2016}}{(1-i)^{2014}}$ is equal to

The complex numbers $\sin x + i \cos 2x$ and $\cos x - i \sin 2x$ (where $i = \sqrt{-1}$) are conjugate to each other for:

If $2 + 3i$ is one of the roots of the equation $2x^3 - 9x^2 + kx - 13 = 0,$ where $k \in R,$ then the real root of this equation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module