If z is a complex number, then the number of common roots of the equation z¹⁹⁸⁵ + z¹⁰⁰ + 1 = 0 and z

Q: Why can't I just use polynomial long division to divide the big polynomial by the small one?

You absolutely could, but that would be a mathematical nightmare! The degree is 1985. The goal of this problem is to test your ability to identify roots and use the cyclic properties of $\omega$. Always look for the elegant path before brute-forcing. In JEE, elegance is usually the intended path.

Q: Wait, why did $z = -1$ fail? I thought it was a root of the cubic.

It is a root of the cubic, but for it to be a *common* root, it must satisfy *both* equations. When you plug $z = -1$ into $z^{1985} + z^{100} + 1$, you get $(-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1$. Since $1 \neq 0$, it fails the second test. It is a root of the cubic, but not a common root.

Q: How do I handle such huge powers like 1985 with $\omega$?

Remember that $\omega^3 = 1$. This means any power $\omega^n$ is equivalent to $\omega^{n \pmod 3}$. Just divide the exponent by 3 and look at the remainder. It turns a terrifying exponent into a simple 0, 1, or 2. For example, $1985 \div 3$ leaves a remainder of 2, so $\omega^{1985} = \omega^2$.

Question

If $z$ is a complex number, then the number of common roots of the equation $z^{1985} + z^{100} + 1 = 0$ and $z^3 + 2z^2 + 2z + 1 = 0$, is equal to :

Options:

(A) 1

(B) 2

(C) 0

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem presents a high-degree polynomial equation:
$$z^{1985} + z^{100} + 1 = 0$$

While this appears daunting, the solution lies in analyzing the roots of the auxiliary cubic equation:
$$z^3 + 2z^2 + 2z + 1 = 0$$

### The Art of Grouping

To solve the cubic equation, we rearrange the terms to reveal a hidden structure:
$$(z^3 + 1) + (2z^2 + 2z) = 0$$

Applying the sum of cubes identity, $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$, to the first group and factoring the second, we obtain:
$$(z+1)(z^2 - z + 1) + 2z(z+1) = 0$$

Factoring out the common term $(z+1)$, we get:
$$(z+1)[(z^2 - z + 1) + 2z] = 0$$
$$(z+1)(z^2 + z + 1) = 0$$

The roots of this equation are $z = -1$ and the complex cube roots of unity, denoted as $\omega$ and $\omega^2$, which satisfy $z^2 + z + 1 = 0$.

### The Testing Ground

We must now test these candidates against the original polynomial $P(z) = z^{1985} + z^{100} + 1$.

First, testing $z = -1$:
$$(-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1$$
Since $1 
eq 0$, $z = -1$ is **not** a common root.

### The Magic of Omega

Next, we evaluate the polynomial at $z = \omega$. Using the properties $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$, we simplify the powers:
$$\omega^{1985} = \omega^{3 	imes 661 + 2} = \omega^2$$
$$\omega^{100} = \omega^{3 	imes 33 + 1} = \omega$$

Substituting these back into the polynomial:
$$\omega^2 + \omega + 1 = 0$$
Thus, $\omega$ is a common root.

Finally, for $z = \omega^2$:
$$(\omega^2)^{1985} + (\omega^2)^{100} + 1 = \omega^{3970} + \omega^{200} + 1$$
Reducing the exponents modulo 3:
$$\omega^{3970} = \omega^1 = \omega$$
$$\omega^{200} = \omega^2$$
The expression becomes $\omega + \omega^2 + 1 = 0$, confirming that $\omega^2$ is also a common root.

### Final Conclusion

By testing the roots of the cubic equation, we have determined that $z = -1$ fails, while $z = \omega$ and $z = \omega^2$ satisfy the original equation.

Therefore, there are exactly **2** common roots.

Animated Solution for Mathematics - Complex Numbers: If $z$ is a complex number, then the number of common roots of the equation $z^{1985} + z^{100} + 1 = 0$ and $z^{3} + 2 z^{2} + 2 z + 1 = 0$ , is equal to :

Select Answer:

Visualized Solution

Problem Setup

Grouping the Cubic Equation

Applying Algebraic Identities

Factoring Out $(z + 1)$

Final Factorized Form

Visualizing on Argand Plane

Testing $z = - 1$

Property of $ω$

Testing $z = ω$

Testing $z = ω^{2}$

Final Conclusion

The Sigma Insight: Cube Roots and nth Roots of Unity

Analyzing the Setup

The Art of Grouping

The Testing Ground

The Magic of Omega

Final Conclusion

Similar Questions

Let a complex number $α, α \neq = 1$ , be a root of the equation $z^{p + q} - z^{p} - z^{q} + 1 = 0$ , where $p, q$ are distinct primes. Show that either $1 + α + α^{2} + \dots + α^{p - 1} = 0$ or $1 + α + α^{2} + \dots + α^{q - 1} = 0$ , but not both together.

If $z^{2} + z + 1 = 0$ , where $z$ is complex number, then the value of $(z + 1/ z)^{2} + (z^{2} + 1/ z^{2})^{2} + (z^{3} + 1/ z^{3})^{2} + \dots + (z^{6} + 1/ z^{6})^{2}$ is

Let $z_{0}$ be a root of the quadratic equation, $x^{2} + x + 1 = 0$ . If $z = 3 + 6 i z_{0}^{81} - 3 i z_{0}^{93}$ , then $ar g z$ is equal to:

Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9$ . Match the statements in List-I with those in List-II.

List-I

List-II

If $α, β \in C$ are the distinct roots, of the equation $x^{2} - x + 1 = 0$ , then $α^{101} + β^{107}$ is equal to :

The sum of $16 2^{t h}$ power of the roots of the equation $x^{3} - 2 x^{2} + 2 x -1 = 0$ is

Let $α = \frac{- 1 + i 3}{2}$ . If $a = (1 + α) \sum_{k = 0}^{100} α^{2 k}$ and $b = \sum_{k = 0}^{100} α^{3 k}$ , then $a$ and $b$ are the roots of the quadratic equation:

If $z^{2} + z + 1 = 0, z \in C$ , then $\sum_{n = 1}^{15} (z^{n} + (- 1)^{n} \frac{1}{z ^{n}})^{2}$ is equal to \_\_\_\_\_.

Let $z_{1}$ and $z_{2}$ be $n^{t h}$ roots of unity which subtend a right angle at the origin. Then $n$ must be of the form

Let $α = \frac{- 1 + i 3}{2}$ is a $= (1 + α) \sum_{k = 0}^{100} α^{2 k}$ and $b = \sum_{k = 0}^{100} α^{3 k}$ , then $a$ and $b$ are the roots of the quadratic equation :

Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z is a complex number, then the number of common roots of the equation z1985+z100+1=0 and z3+2z2+2z+1=0, is equal to :

Select Answer:

Visualized Solution

Problem Setup

Grouping the Cubic Equation

Applying Algebraic Identities

Factoring Out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (z+1)

Final Factorized Form

Visualizing on Argand Plane

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=−1

Property of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=ω

Testing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=ω2

Final Conclusion

The Sigma Insight: Cube Roots and nth Roots of Unity

Analyzing the Setup

The Art of Grouping

The Testing Ground

The Magic of Omega

Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z2+z+1=0, where z is complex number, then the value of (z+1/z)2+(z2+1/z2)2+(z3+1/z3)2+⋯+(z6+1/z6)2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z0​ be a root of the quadratic equation, x2+x+1=0. If z=3+6iz081​−3iz093​, then argz is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let zk​=cos(102kπ​)+isin(102kπ​);k=1,2,…,9. Match the statements in List-I with those in List-II.

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α,β∈C are the distinct roots, of the equation x2−x+1=0, then α101+β107 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of 162th power of the roots of the equation x3−2x2+2x–1=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α=2−1+i3​​. If a=(1+α)∑k=0100​α2k and b=∑k=0100​α3k, then a and b are the roots of the quadratic equation:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z2+z+1=0,z∈C, then ​∑n=115​(zn+(−1)nzn1​)2​ is equal to \_\_\_\_\_.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z1​ and z2​ be nth roots of unity which subtend a right angle at the origin. Then n must be of the form

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α=2−1+i3​​ is a =(1+α)∑k=0100​α2k and b=∑k=0100​α3k, then a and b are the roots of the quadratic equation :

Animated Solution for Mathematics - Complex Numbers: If $z$ is a complex number, then the number of common roots of the equation $z^{1985} + z^{100} + 1 = 0$ and $z^{3} + 2 z^{2} + 2 z + 1 = 0$ , is equal to :

Factoring Out $(z + 1)$

Testing $z = - 1$

Property of $ω$

Testing $z = ω$

Testing $z = ω^{2}$

If $z^{2} + z + 1 = 0$ , where $z$ is complex number, then the value of $(z + 1/ z)^{2} + (z^{2} + 1/ z^{2})^{2} + (z^{3} + 1/ z^{3})^{2} + \dots + (z^{6} + 1/ z^{6})^{2}$ is

Let $z_{0}$ be a root of the quadratic equation, $x^{2} + x + 1 = 0$ . If $z = 3 + 6 i z_{0}^{81} - 3 i z_{0}^{93}$ , then $ar g z$ is equal to:

Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9$ . Match the statements in List-I with those in List-II.

If $α, β \in C$ are the distinct roots, of the equation $x^{2} - x + 1 = 0$ , then $α^{101} + β^{107}$ is equal to :

The sum of $16 2^{t h}$ power of the roots of the equation $x^{3} - 2 x^{2} + 2 x -1 = 0$ is

Let $α = \frac{- 1 + i 3}{2}$ . If $a = (1 + α) \sum_{k = 0}^{100} α^{2 k}$ and $b = \sum_{k = 0}^{100} α^{3 k}$ , then $a$ and $b$ are the roots of the quadratic equation:

If $z^{2} + z + 1 = 0, z \in C$ , then $\sum_{n = 1}^{15} (z^{n} + (- 1)^{n} \frac{1}{z ^{n}})^{2}$ is equal to \_\_\_\_\_.

Let $z_{1}$ and $z_{2}$ be $n^{t h}$ roots of unity which subtend a right angle at the origin. Then $n$ must be of the form

Let $α = \frac{- 1 + i 3}{2}$ is a $= (1 + α) \sum_{k = 0}^{100} α^{2 k}$ and $b = \sum_{k = 0}^{100} α^{3 k}$ , then $a$ and $b$ are the roots of the quadratic equation :