The number of points z on the Argand plane wh | vedclass

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

Question

(B) Let $w = \frac{z-2}{z-4i}$. The given conditions are $\operatorname{Re}(w) = 0$ and $\operatorname{Im}(w) = 1$. This implies $w = 0 + 1i = i$. So,

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $w = \frac{z-2}{z-4i}$. The given conditions are $\operatorname{Re}(w) = 0$ and $\operatorname{Im}(w) = 1$. This implies $w = 0 + 1i = i$. So, $\frac{z-2}{z-4i} = i$. Multiplying both sides by $(z-4i)$, we get $z-2 = i(z-4i)$. $z-2 = iz - 4i^2$. Since $i^2 = -1$, we have $z-2 = iz + 4$. Rearranging the terms, $z - iz = 4 + 2$. $z(1-i) = 6$. $z = \frac{6}{1-i} = \frac{6(1+i)}{(1-i)(1+i)} = \frac{6(1+i)}{1^2 - i^2} = \frac{6(1+i)}{1+1} = \frac{6(1+i)}{2} = 3(1+i)$. Since there is a unique value for $z$, the number of points is $1$.

The number of points $z$ on the Argand plane which satisfy the conditions $\operatorname{Re}\left(\frac{z-2}{z-4i}\right)=0$ and $\operatorname{Im}\left(\frac{z-2}{z-4i}\right)=1$ simultaneously is

Similar Questions

The largest value of $r$ for which the region represented by the set $\{ \omega \in \mathbb{C} : |\omega - 4 - i| \le r \}$ is contained in the region represented by the set $\{ z \in \mathbb{C} : |z - 1| \le |z + i| \}$ is equal to

If $z = \sqrt{2} - i\sqrt{2}$ is rotated through an angle $45^{\circ}$ in the anti-clockwise direction about the origin,then the coordinates of its new position are

In the Argand plane, the distinct roots of $1+z+z^{3}+z^{4}=0$ ($z$ is a complex number) represent vertices of

If $\frac{z-1}{2z+1}$ is a purely imaginary number,then the locus of $z$ represents a circle. Find its radius.

What are the polar coordinates of the point $(-\sqrt{3}, 1)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module