Animated Solution for Mathematics - Complex Numbers: For z∈C if the minimum value of (∣z−32∣+∣z−p2i∣) is 52, then a value of p is
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Visualized Solution
Visualizing the Argand Plane
Consider the complex number z in the Argand plane.
The given expression is ∣z−32∣+∣z−p2i∣.
Identifying Fixed Points
Let z1=32 and z2=p2i.
The expression becomes ∣z−z1∣+∣z−z2∣.
Geometric Meaning
∣z−z1∣ is the distance between z and z1.
∣z−z2∣ is the distance between z and z2.
We need to minimize this sum.
The Triangle Inequality
By the Triangle Inequality, ∣z−z1∣+∣z−z2∣≥∣z1−z2∣.
Condition for Minimum
The minimum value occurs when z lies exactly on the line segment joining z1 and z2.
Thus, Minimum Value =∣z1−z2∣.
Equating to the Given Minimum
We are given that the minimum value is 52.
Therefore, ∣z1−z2∣=52.
Distance Formula Setup
Substitute z1=32 and z2=p2i into the distance formula:
∣32−p2i∣=52.
Applying the Magnitude Formula
The magnitude of a+bi is a2+b2.
So, (32)2+(−p2)2=52.
Squaring Both Sides
Squaring both sides to remove the square root:
(32)2+(−p2)2=(52)2.
Expanding the Squares
Calculate the squares:
18+2p2=50.
Isolating p2
Subtract 18 from both sides:
2p2=50−18⟹2p2=32.
Solving for p
Divide by 2:
p2=16.
Taking the square root gives p=±4.
Final Answer Selection
Checking the given options: 3, 27, 4, 29.
The valid value is p=4.
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The Sigma Insight: Geometrical Applications of Complex Numbers
Solution Diagram
Analyzing the Setup
Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are exploring the elegant geometry hidden within the complex plane.
Imagine the Argand plane as your canvas. We are given the expression ∣z−32∣+∣z−p2i∣ and told that its minimum value is 52.
Translating Algebra into Geometry
Look closely at the terms inside the modulus. In the complex plane, ∣z−z0∣ represents the distance between the variable point z and a fixed point z0.
By defining z1=32 and z2=p2i, our expression transforms into the sum of the distances from z to these two fixed points. We are essentially asking: "If I have two fixed stakes in the ground at z1 and z2, where should I stand so that the total distance to both is minimized?"
The Power of the Triangle Inequality
This is where the magic happens. The Triangle Inequality tells us that for any point z, the sum of the distances to two fixed points z1 and z2 is always greater than or equal to the direct distance between z1 and z2.
Mathematically, this is expressed as:
∣z−z1∣+∣z−z2∣≥∣z1−z2∣
The minimum occurs when z lies directly on the line segment connecting z1 and z2. In this state of perfect alignment, the sum of the distances is exactly equal to the distance between the two fixed points.
The Engine Room
Calculation
Now that we have identified the geometric condition, the rest is a beautiful, straightforward calculation. We are given that the minimum value is 52.
Therefore, we set the distance between our fixed points equal to this value:
∣z1−z2∣=52
Substituting our values, we get ∣32−p2i∣=52. To find the magnitude of this complex number, we use the formula a2+b2, where a=32 and b=−p2.
This gives us:
(32)2+(−p2)2=52
Squaring both sides to eliminate the radical, we obtain:
(32)2+(−p2)2=(52)2
Expanding these squares, we find:
18+2p2=50
Subtracting 18 from both sides yields 2p2=32, which simplifies to p2=16.
Thus, p=±4. Since 4 is one of our options, we have arrived at our destination.