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

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(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,

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$1$

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(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

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