Animated Solution for Mathematics - Complex Numbers: Let z be a complex number such that ∣z+2iz−i∣=1 and ∣z∣=25. Then the value of ∣z+3i∣ is:
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Visualized Solution
Initial Setup in Argand Plane
Given: z+2iz−i=1 and ∣z∣=25
Objective: Find the value of ∣z+3i∣
Let z=x+iy be the complex number.
Locus of ∣z−i∣=∣z+2i∣
Rewrite as: ∣z−i∣=∣z+2i∣
Geometrically, z is equidistant from points (0,1) and (0,−2).
This represents the perpendicular bisector of the segment joining these points.
Solving for the Imaginary Part y
Substitute z=x+iy: x2+(y−1)2=x2+(y+2)2
Expand: y2−2y+1=y2+4y+4
Simplify: −6y=3⇒y=−21
Modulus Condition ∣z∣=25
Given: ∣z∣=25
Squaring both sides: ∣z∣2=425
In Cartesian form: x2+y2=425
This represents a circle centered at the origin with radius 25.
Calculating the Real Part x2
Substitute y=−21 into the circle equation:
x2+(−21)2=425
x2+41=425
x2=424=6
Defining the Target ∣z+3i∣
We need to find ∣z+3i∣.
Geometrically, this is the distance between z(x,y) and the point (0,−3).
Using distance formula: ∣z+3i∣=x2+(y+3)2
Substituting x2 and y
Substitute x2=6 and y=−21:
∣z+3i∣=6+(−21+3)2
Simplify inside the bracket: −21+3=25
Final Calculation
∣z+3i∣=6+(25)2
∣z+3i∣=6+425
∣z+3i∣=424+25=449=27
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The Sigma Insight: Geometrical Applications of Complex Numbers
Solution Diagram
Analyzing the Setup
Imagine you are standing on the Argand plane, a vast, two-dimensional landscape where every point is a complex number z=x+iy. We are embarking on a journey to find the hidden location of a point z that obeys two strict, beautiful laws.
Our first law is given by the equation:
z+2iz−i=1
By cross-multiplying, we transform this into ∣z−i∣=∣z−(−2i)∣. This is a profound geometric statement; it tells us that the distance from z to the point i (which is (0,1)) is exactly equal to the distance from z to the point −2i (which is (0,−2)).
The Line of Symmetry
The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them. The midpoint between (0,1) and (0,−2) is (0,−0.5), and since the segment is vertical, the perpendicular bisector must be a horizontal line.
Algebraically, if we set z=x+iy, the equation becomes:
x2+(y−1)2=x2+(y+2)2
Watch as the x2 terms vanish, leaving us with y2−2y+1=y2+4y+4. Simplifying this, we find −6y=3, or:
y=−21
This horizontal line is the home of our complex number z.
The Circle of Constraints
Now, we introduce our second law: ∣z∣=25. This is the equation of a circle centered at the origin with a radius of 25.
In Cartesian terms, this is:
x2+y2=425
We now have two constraints: our point z must lie on the line y=−21 and on the circle x2+y2=425. By substituting y=−21 into the circle equation, we get:
x2+(−21)2=425
This simplifies to x2+41=425, which leads us to x2=6.
The Final Destination
The question asks for the value of ∣z+3i∣. Geometrically, this is the distance between our point z(x,y) and the point (0,−3).
Using the distance formula, we have:
∣z+3i∣=x2+(y+3)2
We already know x2=6 and y=−21. Substituting these values, we get:
∣z+3i∣=6+(−21+3)2
Simplifying the term inside the parenthesis, we have −21+3=25. Squaring this gives 425.