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Question: For \(x_{1},x_{2},y_{1},y_{2} \in R,\) if \(0 < x_{1} < x_{2},y_{1} = y_{2}\) and \(z_{1} = x_{1} +...

For x1,x2,y1,y2∈R,x_{1},x_{2},y_{1},y_{2} \in R, if 0<x1<x2,y1=y20 < x_{1} < x_{2},y_{1} = y_{2} and

z1=x1+iy1,z2=x2+iy2,z_{1} = x_{1} + iy_{1},z_{2} = x_{2} + iy_{2}, and z3=12(z1+z2),z_{3} = \frac{1}{2}(z_{1} + z_{2}), then z1,z2z_{1},z_{2} and z3z_{3} satisfy

A

∣z1∣=∣z2∣=∣z3∣|z_{1}| = |z_{2}| = |z_{3}|

B

∣z1∣<∣z2∣<∣z3∣|z_{1}| < |z_{2}| < |z_{3}|

C

∣z1∣>∣z2∣>∣z3∣|z_{1}| > |z_{2}| > |z_{3}|

D

∣z1∣<∣z3∣<∣z2∣|z_{1}| < |z_{3}| < |z_{2}|

Answer

∣z1∣<∣z3∣<∣z2∣|z_{1}| < |z_{3}| < |z_{2}|

Explanation

Solution

Sol. 0<x1<x2,y1=y20 < x_{1} < x_{2},y_{1} = y_{2} (Given)

∣z1∣=x12+y12,∣z2∣=x22+y22⇒∣z2∣>∣z1∣⇒∣z3∣=∣z1+z2∣2\left| z _ { 1 } \right| = \sqrt { x _ { 1 } ^ { 2 } + y _ { 1 } ^ { 2 } } , \left| z _ { 2 } \right| = \sqrt { x _ { 2 } ^ { 2 } + y _ { 2 } ^ { 2 } } \Rightarrow \left| z _ { 2 } \right| > \left| z _ { 1 } \right| \Rightarrow \left| z _ { 3 } \right| = \frac { \left| z _ { 1 } + z _ { 2 } \right| } { 2 } =(x1+x22)2+(y1+y22)2= \sqrt { \left( \frac { x _ { 1 } + x _ { 2 } } { 2 } \right) ^ { 2 } + \left( \frac { y _ { 1 } + y _ { 2 } } { 2 } \right) ^ { 2 } } (x1+x22)2+y12<∣z2∣>∣z1∣\sqrt{\left( \frac{x_{1} + x_{2}}{2} \right)^{2} + y_{1}^{2}} < |z_{2}| > |z_{1}|. Hence, ∣z1∣<∣z3∣<∣z2∣|z_{1}| < |z_{3}| < |z_{2}|