1708thanks Nishant Sir and hsbhatt sir
nice solution using Triangle Inequality.
\hspace{-16}\bold{(Q)::}\; $If $\mathbf{p(z)=|2z-1-i|+|3z-2-2i|+|4z-3-3i|}$\\\\ Then find Min. of $\mathbf{p(z)}$ and also find corrosponding Complex no.$\mathbf{z}$
-
UP 0 DOWN 0 0 3
3 Answers
62From geometry, after a bit of manipulation, we can say that the point lines on the line R(z)=Im(z)
For that we just have to take a point which is the projection along the line Re(z)=Im(z) and see that it has a lower value of P(z) using triangle inequality
After that it does not seem to be too difficult! does it?
341We can write the expression as
2|z - a| + 3 \left|z - \frac{4a}{3} \right| + 4\left|z - \frac{3a}{2} \right|
where
a = \frac{1+i}{2}
Notice that a , \frac{4a}{3} ,\frac{3a}{2} are collinear.
Now, you have to get clear that the candidates for minimizing the function will lie on this line. The reason, as Nishant sir pointed out, is that for any point not on the line, you can consider the foot of the perpendicular from that point to the line and this point will be closer to the points.
Now
LHS \ge 2\left(|z-a| + \left|z - \frac{4a}{3} \right|\right) + \left|z - \frac{4a}{3} \right| + 4\left|z - \frac{3a}{2} \right|
\ge 2\left| \frac{a}{3} \right| + \left|z - \frac{4a}{3} \right| + 4\left|z - \frac{3a}{2} \right|
(Since |x| + |y| ≥|x+y|)
The equality
|z - a| + \left|z - \frac{4a}{3} \right| \ge \left| \frac{a}{3} \right|
holds for those points lying between z=a and z=4a/3 (end points included). You can see that the expression gets minimised when z = 4a/3.
You can similarly experiment with any pair among these points and you will obtain the min at z=4a/3
which is 4 \left| \frac{a}{3} \right| = \frac{2\sqrt2}{3}
1708