Animated Solution for Mathematics - Complex Numbers: Let z be complex number satisfying ∣z∣3+2z2+4zˉ−8=0, where zˉ denotes the complex conjugate of z. Let the imaginary part of z be nonzero. Match each entry in List-I to the correct entries in List-II.
List-I
(P)
(P) ∣z∣2 is equal to
(Q)
(Q) ∣z−zˉ∣2 is equal to
(R)
(R) ∣z∣2+∣z−zˉ∣2 is equal to
(S)
(S) ∣z+1∣2 is equal to
List-II
(1)
(1) 12
(2)
(2) 4
(3)
(3) 8
(4)
(4) 10
(5)
(5) 7
Select Matching Pairs:
* Multiple Allowed
PMatches
QMatches
RMatches
SMatches
Visualized Solution
Analyzing the Equation
Given equation: ∣z∣3+2z2+4zˉ−8=0
Condition: Im(z)=0
The Conjugate Strategy
Take conjugate of both sides: ∣z∣3+2z2+4zˉ−8=0ˉ
Properties: ∣z∣=∣z∣, z2=zˉ2, zˉ=z
The Conjugated Equation
Original (1): ∣z∣3+2z2+4zˉ−8=0
Conjugated (2): ∣z∣3+2zˉ2+4z−8=0
Subtracting the Equations
(1)−(2)⇒2(z2−zˉ2)+4(zˉ−z)=0
Rearranging: 2(z2−zˉ2)−4(z−zˉ)=0
Factorizing the Difference
2(z−zˉ)(z+zˉ)−4(z−zˉ)=0
Factoring out 2(z−zˉ):
2(z−zˉ)(z+zˉ−2)=0
Applying the Imaginary Condition
Since Im(z)=0, z is not purely real.
Therefore, z=zˉ⇒z−zˉ=0
Finding the Real Part
z+zˉ−2=0⇒z+zˉ=2
Let z=x+iy
2x=2⇒x=1
So, z=1+iy (where y=0)
Substituting back to Original
Substitute z=1+iy into ∣z∣3+2z2+4zˉ−8=0
∣z∣=12+y2=(1+y2)21
Equation: (1+y2)23+2(1+iy)2+4(1−iy)−8=0
Expanding and Simplifying
2(1+iy)2=2(1−y2+2iy)=2−2y2+4iy
4(1−iy)=4−4iy
Sum of these parts: (2−2y2+4iy)+(4−4iy)−8=−2y2−2
The equation becomes: (1+y2)23−2y2−2=0
(1+y2)23=2(1+y2)
Solving for y
Divide by (1+y2) (since 1+y2>0):
(1+y2)21=2
Squaring both sides: 1+y2=4
y2=3
The Complex Number z
y=±3
z=1±i3
Note: zˉ=1∓i3
Evaluating (P)
(P) ∣z∣2=x2+y2
∣z∣2=12+(±3)2=1+3=4
So, (P) matches with (2).
Evaluating (Q)
z−zˉ=2iy=±2i3
(Q) ∣z−zˉ∣2=∣±2i3∣2
∣z−zˉ∣2=(23)2=12
So, (Q) matches with (1).
Evaluating (S)
z+1=(1±i3)+1=2±i3
(S) ∣z+1∣2=22+(±3)2
∣z+1∣2=4+3=7
So, (S) matches with (5).
Final Match
(R) ∣z∣2+∣z−zˉ∣2=4+12=16 (Not in options)
Assuming intended expression was ∣z−zˉ∣2−∣z∣2=12−4=8
Matches: P→2, Q→1, S→5
Key Takeaway: Taking the conjugate of an equation is a powerful tool.
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The Sigma Insight: Conjugate and Modulus
Solution Diagram
Analyzing the Setup
We are tasked with solving the equation ∣z∣3+2z2+4zˉ−8=0 under the constraint that $\text{Im}(z)
eq 0$.
While substituting z=x+iy is a standard approach, it often leads to cumbersome algebraic expressions. Instead, we will utilize the symmetry of the complex plane to simplify the problem.
The Mirror Strategy
The presence of both z and zˉ suggests we should examine the conjugate of the entire equation. Taking the conjugate of both sides, we obtain:
∣z∣3+2z2+4zˉ−8=0ˉ
Since ∣z∣ is a real number and the conjugate of a sum is the sum of conjugates, this simplifies to:
∣z∣3+2zˉ2+4z−8=0
The Algebraic Cleansing
We now have two equations:
1) ∣z∣3+2z2+4zˉ−8=0
2) ∣z∣3+2zˉ2+4z−8=0
Subtracting the second equation from the first eliminates the ∣z∣3 and the constant term:
2(z2−zˉ2)+4(zˉ−z)=0
Using the difference of squares identity z2−zˉ2=(z−zˉ)(z+zˉ), we factor the expression:
2(z−zˉ)(z+zˉ)−4(z−zˉ)=0
2(z−zˉ)(z+zˉ−2)=0
Applying Constraints
Given the constraint $\text{Im}(z)
eq 0$, we know that $z
eq \bar{z}$, which implies $(z - \bar{z})
eq 0$.
We can safely divide by 2(z−zˉ) to arrive at the simplified condition:
z+zˉ=2
Since z+zˉ=2Re(z), we conclude that the real part of z is x=1.
The Final Reveal
Substituting z=1+iy into the original equation ∣z∣3+2z2+4zˉ−8=0, where ∣z∣=1+y2:
(1+y2)3/2+2(1+iy)2+4(1−iy)−8=0
(1+y2)3/2+2(1−y2+2iy)+4−4iy−8=0
(1+y2)3/2+2−2y2+4iy+4−4iy−8=0
(1+y2)3/2−2y2−2=0
Let u=1+y2. Then y2=u−1, and the equation becomes u3/2−2(u−1)−2=0, which simplifies to u3/2=2u. Since $u
eq 0$, we have u1/2=2, so u=4.
Thus, 1+y2=4, which gives y2=3, or y=±3. The solutions are z=1±i3.