Let z_1 and z_2 be two distinct complex numbe | vedclass

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

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

Similar Questions

If $w = \frac{z}{z - \frac{1}{3}i}$ and $|w| = 1$,where $i = \sqrt{-1}$,then $z$ lies on

If $z = x + iy$,then the area of the triangle whose vertices are the points $z$,$iz$,and $z + iz$ is

$\alpha$ is the real root and $\beta, \gamma$ are the other roots of the equation $x^3-a^3=0$ $(a>0)$. Then the number of common points of the curves given by $|z-\beta|=\frac{\sqrt{3} a}{2}$ and $|z-\gamma|=\frac{\sqrt{3} a}{2}$ is

Let $z$ be a complex number such that the real part of $\frac{z-2i}{z+2i}$ is zero. Then,the maximum value of $|z-(6+8i)|$ is equal to:

Let $x_1, x_2, x_3, x_4$ be the roots of the equation $4x^4 + 8x^3 - 17x^2 - 12x + 9 = 0$. If $(4+x_1^2)(4+x_2^2)(4+x_3^2)(4+x_4^2) = \frac{125}{16}m$,then the value of $m$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module