Let z_1 and z_2 be two distinct complex numbe | vedclass

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Question

(A) Given $z = (1-t)z_1 + tz_2$,where $0 < t < 1$. This implies that $z$ lies on the line segment joining $z_1$ and $z_2$.$1$. Since $z$ lies on the s

Vedclass · Accepted Answer

$(A), (C), (D)$

Vedclass · Answer

(A) Given $z = (1-t)z_1 + tz_2$,where $0 $1$. Since $z$ lies on the segment $AB$,the sum of distances $|z-z_1| + |z-z_2|$ is equal to the total distance $|z_1-z_2|$. Thus,$(A)$ is true.$2$. The vector $z-z_1$ is in the same direction as $z_2-z_1$ because $z-z_1 = t(z_2-z_1)$ and $t > 0$. Therefore,$\operatorname{Arg}(z-z_1) = \operatorname{Arg}(z_2-z_1)$. Thus,$(D)$ is true.$3$. The condition $\frac{z-z_1}{z_2-z_1} = t$ (where $t$ is real) implies that the ratio is purely real. This is equivalent to $\frac{z-z_1}{z_2-z_1} = \overline{\left(\frac{z-z_1}{z_2-z_1}ight)} = \frac{\bar{z}-\bar{z}_1}{\bar{z}_2-\bar{z}_1}$. Cross-multiplying gives $(z-z_1)(\bar{z}_2-\bar{z}_1) - (\bar{z}-\bar{z}_1)(z_2-z_1) = 0$,which is the determinant form $\left|\begin{array}{cc} z-z_1 & \bar{z}-\bar{z}_1 \ z_2-z_1 & \bar{z}_2-\bar{z}_1 \end{array}ight| = 0$. Thus,$(C)$ is true.$4$. $\operatorname{Arg}(z-z_1)$ and $\operatorname{Arg}(z-z_2)$ represent the angles of vectors $P-A$ and $P-B$. Since $P$ is between $A$ and $B$,these vectors point in opposite directions,so $\operatorname{Arg}(z-z_1) 
eq \operatorname{Arg}(z-z_2)$. Thus,$(B)$ is false.Therefore,the correct options are $(A), (C), (D)$.

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