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Question

Question: The minimum value of \(\left| {Z - 1 + 2i} \right| + \left| {4i - 3 - Z} \right|\) is a.\(\sqrt 5...

The minimum value of Z1+2i+4i3Z\left| {Z - 1 + 2i} \right| + \left| {4i - 3 - Z} \right| is
a.5\sqrt 5
b.5
c.2132\sqrt {13}
d.15\sqrt {15}

Explanation

Solution

At first let's rearrange the given expression and we can see that it is a sum of the distances of Z from two points P and Q and the sum is minimum in the line segment PQ and it is given by PQ=( real part of P - Q)2+(imaginary part of P - Q)2PQ = \sqrt {{{\left( {{\text{ real part of P - Q}}} \right)}^2} + {{\left( {{\text{imaginary part of P - Q}}} \right)}^2}}

Complete step-by-step answer:
Lets rearrange the given expression as Z(12i)+Z(3+4i)\left| {Z - \left( {1 - 2i} \right)} \right| + \left| {Z - \left( { - 3 + 4i} \right)} \right|
From this we can see that the given expression is the sum of the distances of Z from two points
Here the two points are P=12iP = 1 - 2i and Q=3+4iQ = - 3 + 4i
The sum is minimum when Z lies on the line segment PQ
Now to find PQ
PQ=( real part of P - Q)2+(imaginary part of P - Q)2\Rightarrow PQ = \sqrt {{{\left( {{\text{ real part of P - Q}}} \right)}^2} + {{\left( {{\text{imaginary part of P - Q}}} \right)}^2}}
PQ=(1+3)2+(24)2 PQ=(4)2+(6)2 PQ=16+36 PQ=52=413=213  \Rightarrow PQ = \sqrt {{{\left( {1 + 3} \right)}^2} + {{\left( { - 2 - 4} \right)}^2}} \\\ \Rightarrow PQ = \sqrt {{{\left( 4 \right)}^2} + {{\left( { - 6} \right)}^2}} \\\ \Rightarrow PQ = \sqrt {16 + 36} \\\ \Rightarrow PQ = \sqrt {52} = \sqrt {4*13} = 2\sqrt {13} \\\
Hence we get the minimum to be 2132\sqrt {13}
Therefore the correct option is c.

Note: The square root of i has both real and imaginary parts. The square root of a negative real number is purely imaginary, but the square root of a purely imaginary number has to have both real and imaginary parts
Any root of i has multiple unique solutions, and the N-th root has N unique solutions. For positive, real numbers, taking the square root (i.e., the second root) of that number gives you two possible solutions: a positive one and a negative one.
In an imaginary fraction, it actually matters whether the numerator or denominator has the “i” in it.