Home
Class 12
MATHS
The solutions of the equation z^4 + 4i z...

The solutions of the equation `z^4 + 4i z^3 - 6z^2 - 4iz -i =0 ` represent vertices of a convex polygon in the complex plane. The area of the polygon is :

A

`2^(1//2)`

B

`2^(3//2)`

C

`2^(5//2)`

D

`2^(5//4)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the equation \( z^4 + 4iz^3 - 6z^2 - 4iz - i = 0 \) and find the area of the polygon represented by its roots in the complex plane, we can follow these steps: ### Step 1: Rewrite the equation We start with the given polynomial: \[ z^4 + 4iz^3 - 6z^2 - 4iz - i = 0 \] ### Step 2: Identify the transformation We notice that the polynomial can be transformed by substituting \( z = w + i \) to simplify the expression. This substitution will help us eliminate the imaginary unit \( i \) from the constant term. ### Step 3: Substitute and simplify Substituting \( z = w + i \): \[ (w + i)^4 + 4i(w + i)^3 - 6(w + i)^2 - 4i(w + i) - i = 0 \] Expanding this expression will yield a polynomial in \( w \). ### Step 4: Expand the polynomial Using the binomial theorem, we expand each term: 1. \( (w + i)^4 = w^4 + 4w^3i + 6w^2(-1) + 4w(-i) + 1 \) 2. \( 4i(w + i)^3 = 4i(w^3 + 3w^2i - 3w - i) \) 3. \( -6(w + i)^2 = -6(w^2 + 2wi - 1) \) 4. \( -4i(w + i) = -4iw - 4i^2 = -4iw + 4 \) Combining all these expansions will lead to a polynomial in \( w \). ### Step 5: Solve for roots Once we have the polynomial in \( w \), we can solve for its roots. The roots \( w_1, w_2, w_3, w_4 \) can be found using numerical methods or factoring if possible. ### Step 6: Convert back to \( z \) Convert the roots back to \( z \) using \( z = w + i \). ### Step 7: Identify the vertices of the polygon The roots \( z_1, z_2, z_3, z_4 \) will represent the vertices of a convex polygon in the complex plane. ### Step 8: Calculate the area of the polygon To find the area of the polygon formed by the vertices \( z_1, z_2, z_3, z_4 \), we can use the formula for the area of a polygon given its vertices: \[ \text{Area} = \frac{1}{2} \left| \sum_{j=1}^{n} (x_j y_{j+1} - x_{j+1} y_j) \right| \] where \( (x_{n+1}, y_{n+1}) = (x_1, y_1) \). ### Step 9: Final calculation After calculating the area using the coordinates of the vertices, we will arrive at the final answer.
Promotional Banner

Topper's Solved these Questions

  • COMPLEX NUMBERS

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise EXERCISE-3:COMPREHENSION TYPE PROBLEMS|8 Videos
  • COMPLEX NUMBERS

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise EXERCISE-4:MATCHING TYPE PROBLEMS|2 Videos
  • COMPLEX NUMBERS

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise EXERCISE-5 : SUBJECTIVE TYPE PROBLEMS|8 Videos
  • CIRCLE

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise Exercise - 5 : Subjective Type Problems|13 Videos
  • COMPOUND ANGLES

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise Exercise-5 : Subjective Type Problems|31 Videos

Similar Questions

Explore conceptually related problems

The roots of the equation z^(4) + az^(3) + (12 + 9i)z^(2) + bz = 0 (where a and b are complex numbers) are the vertices of a square. Then The area of the square is

The equation zbarz+(4-3i)z+(4+3i)barz+5=0 represents a circle of radius

The equation 3y+4z=0 represents a

The set of values of k for which the equation zbarz+(-3+4i)barz-(3+4i)z+k=0 represents a circle, is

If z = x + yi, omega = (2-iz)/(2z-i) and |omega|=1 , find the locus of z in the complex plane

If P(z_(1)),Q(z_(2)),R(z_(3)) " and " S(z_(4)) are four complex numbers representing the vertices of a rhombus taken in order on the complex plane, which one of the following is held good?

If z_(1),z_(2),z_(3)………….z_(n) are in G.P with first term as unity such that z_(1)+z_(2)+z_(3)+…+z_(n)=0 . Now if z_(1),z_(2),z_(3)……..z_(n) represents the vertices of n -polygon, then the distance between incentre and circumcentre of the polygon is

How many solutions the system of equations ||z + 4 |-|z-3i|| =5 and |z| = 4 has ?

Show that if iz^3+z^2-z+i=0 , then |z|=1

Show that if iz^3+z^2-z+i=0 , then |z|=1