Let $a, b$ be nonzero real numbers. If $p(x)=x^n+c_{n-1} x^{n-1}+\cdots+c_0$, where $c_{n-1}, \ldots...

To determine which statement is correct, let's analyse each option given and also carefully observe that $\boldsymbol{p}(\boldsymbol{x})=(\mathbf{1}+\boldsymbol{x})^n$ where $n \geq 3$ and $a, b$ are nonzero real numbers.Option A: "I $a+i b$ is a root of $p(x)$, then $n$ must be even."The roots of $p(x)=(1+x)^n$ are determined by setting $(1+x)^n=0$. Since $n \geq 3$ and $(1+x)^n=0$ when $1+x=0$, the only root is $x=-1$. For a complex root like $a+i b$ to be a root, $(1+a+i b)^n$ must be 0 , which can only happen if $1+a+i b=0$, leading to $a=-1$ and $b=0$. However, since $b$ is nonzero, this condition cannot be satisfied. Thus, $a+i b$ cannot be a root.Therefore, Option A is incorrect.Option B: "If $\boldsymbol{a}$ is a root of $\boldsymbol{p}(\boldsymbol{x})$, then $\boldsymbol{n}$ must be odd."If $a$ is a real root, then $(1+a)^n=0$ which leads to $1+a=0$ or $a=-1$. This is true for any $n \geq 3$, not necessarily odd. Therefore, Option B is incorrect.Option C: "If $a+i b$ is a root of $p(x)$, then $(a+i b)^2$ can never be a root of $p(x)$."As established earlier, $(1+x)^n=0$ only gives real roots. Thus, if $a+i b$ were a root, $1+a+$ $i b$ would need to equal zero, and $a=-1, b=0$. This makes the question of whether $(a+i b)^2$ can be a root irrelevant because no complex roots exist for $p(x)$.Therefore, Option C is irrelevant.Option D: "If $a$ is a root of $p(x)$, then $a$ must be an integer."If $a$ is a root, $(1+a)^n=0$ leads to $a=-1$. Since -1 is an integer, this statement holds true.Option D is correct.

Let $\omega$ be a complex root of the quadratic polynomial $x^2+x+1$. The value of$$ \left(\omega+\f...

Revision Note on $\omega$ :$\omega$ is a primitive cube root of unity with the following key properties:$\omega^3=1$$\omega+\omega^2=-1$$$ \left(\omega+\frac{1}{\omega}\right) \cdot\left(\omega^2+\frac{1}{\omega^2}\right) \cdots\left(\omega^{100}+\frac{1}{\omega^{100}}\right) $$Let us simplify a few terms by replacing 1 with $\omega^3$ :$$ \begin{aligned} \left(\omega+\frac{1}{\omega}\right) & =\left(\omega+\frac{\omega^3}{\omega}\right)=\left(\omega+\omega^2\right)=-1 \\ \left(\omega^2+\frac{1}{\omega^2}\right) & =\left(\omega^2+\frac{\omega^3}{\omega^2}\right)=\left(\omega+\omega^2\right)=-1 \\ & \left(\omega^3+\frac{1}{\omega^3}\right)=1+1=2 \end{aligned} $$$$ \left(\omega^4+\frac{1}{\omega^4}\right)=\left(\omega^3(\omega)+\frac{1}{\omega^3(\omega)}\right)=\left(\omega+\frac{1}{\omega}\right)=-1 $$This will repeat in a cycle of 3 terms; therefore, we can rewrite the original equation as theand so on following:$$ \begin{gathered} \left(\omega+\frac{1}{\omega}\right) \cdot\left(\omega^2+\frac{1}{\omega^2}\right) \cdots\left(\omega^{100}+\frac{1}{\omega^{100}}\right) \\ =\left[\left(\omega+\frac{1}{\omega}\right) \cdot\left(\omega^2+\frac{1}{\omega^2}\right) \cdot\left(\omega^3+\frac{1}{\omega^3}\right)\right]^{33}\left(\omega^{99} \omega+\frac{1}{\omega^{99} \omega}\right) \\ =[-1 \times-1 \times 2]^{33} \times-1 \\ =-2^{33} \end{gathered} $$

Consider the following subset of the $X Y$-plane.$$ S=\left\{\left(|z-\mathrm{i} z|,|z|^2\right): \q...

Given the subset $ S $ in the $ XY $-plane:$ S = \left\{\left(|z-\mathrm{i} z|, |z|^2\right): z \text{ is a complex number}\right\} $we need to determine the nature of this subset.Step-by-step Breakdown:Identify Coordinates:$ X $-coordinate: $ |z - \mathrm{i} z| $$ Y $-coordinate: $ |z|^2 $Let $ z = x + \mathrm{i} y $:Compute $ X $-coordinate as:$ |(x + \mathrm{i} y) - \mathrm{i}(x + \mathrm{i} y)| = |x + \mathrm{i} y - \mathrm{i} x - \mathrm{i}^2 y| $$ = |x + \mathrm{i} y - \mathrm{i} x + y| = |x + y + \mathrm{i}(y - x)| $$ = \sqrt{(x + y)^2 + (y - x)^2} $$ = \sqrt{x^2 + y^2 + 2xy + x^2 + y^2 - 2xy} = \sqrt{2(x^2 + y^2)} $Compute $ Y $-coordinate as:$ |x + \mathrm{i} y|^2 = x^2 + y^2 $Establish the Relationship:We have:$ X = \sqrt{2(x^2 + y^2)} $$ Y = x^2 + y^2 $Thus:$ X^2 = 2(x^2 + y^2) $$ \frac{X^2}{2} = x^2 + y^2 = Y $The equation $\frac{X^2}{2} = Y$ describes a parabola in the $XY$-plane, which confirms that the correct characterization of the subset $ S $ is that it is a parabola.

Let $z_1, z_2$, and $z_3$ be complex numbers satisfying the following conditions$$ 2=\left|2 z_1\rig...

We are given complex numbers $ z_1, z_2, z_3 $ satisfying$ 2 = |2z_1| = |z_2 - 1| = |z_3 + 1| = \left|\frac{1}{z_1} + \frac{1}{z_2 - 1} + \frac{1}{z_3 + 1}\right|. $We want to find $\; |4z_1 + z_2 + z_3|.$Step 1. Simplify magnitudesFrom $ |2z_1| = 2 \Rightarrow |z_1| = 1.$So $ z_1 $ lies on the unit circle.Also:$ |z_2 - 1| = 2, \quad |z_3 + 1| = 2. $Hence,$z_2$ lies on a circle of radius 2 centered at $1$,$z_3$ lies on a circle of radius 2 centered at $-1$.Step 2. Introduce new variablesLet’s define$ a = 2z_1, \quad b = z_2 - 1, \quad c = z_3 + 1. $Then by the given magnitudes,$ |a| = |b| = |c| = 2. $We can express the condition on reciprocals:$ 2 = \left|\frac{1}{z_1} + \frac{1}{z_2 - 1} + \frac{1}{z_3 + 1}\right| = \left|\frac{1}{z_1} + \frac{1}{b} + \frac{1}{c}\right|. $But $\frac{1}{z_1} = \frac{2}{a}$. So the condition becomes:$ 2 = \left|\frac{2}{a} + \frac{1}{b} + \frac{1}{c}\right|. $Step 3. Note all have magnitude 2Let’s divide everything by 2 to normalize to unit circle magnitudes.Let’s write:$ \frac{a}{2} = e^{i\alpha}, \quad \frac{b}{2} = e^{i\beta}, \quad \frac{c}{2} = e^{i\gamma}. $Then $a = 2e^{i\alpha}$, etc.The reciprocal condition becomes:$ 2 = \left|\frac{2}{2 e^{i\alpha}} + \frac{1}{2 e^{i\beta}} + \frac{1}{2 e^{i\gamma}}\right| = \left| e^{-i\alpha} + \frac{1}{2}e^{-i\beta} + \frac{1}{2}e^{-i\gamma}\right|. $Multiply both sides by 2 to simplify the numeric coefficient? Wait, better to keep as is.We have:$ \left| e^{-i\alpha} + \frac{1}{2}e^{-i\beta} + \frac{1}{2}e^{-i\gamma}\right| = 2. $Step 4. Magnitude constraint boundingEach term has magnitude 1, ½, ½ respectively, so the maximum possible magnitude is $1 + ½ + ½ = 2.$To achieve equality $=2$,all three exponentials must have the same argument, i.e.$ e^{-i\alpha}, e^{-i\beta}, e^{-i\gamma} \text{ have same direction.} $Hence they must be all proportional with positive real coefficients.That means $ e^{-i\alpha} = e^{-i\beta} = e^{-i\gamma} $, up to integer multiple of $2\pi$.Step 5. So we can take them equal (since only magnitudes matter)Let $ e^{-i\alpha} = e^{-i\beta} = e^{-i\gamma} = e^{-i\theta} $.Then the condition is automatically satisfied:$ \left| e^{-i\theta} + \frac{1}{2}e^{-i\theta} + \frac{1}{2}e^{-i\theta}\right| = |2 e^{-i\theta}| = 2. $OK.Step 6. Return to $a,b,c$Since $ e^{i\alpha}, e^{i\beta}, e^{i\gamma}$ all are equal, say $ e^{i\theta}. $Then:$ a = 2 e^{i\theta}, \quad b = 2 e^{i\theta}, \quad c = 2 e^{i\theta}. $Step 7. Go back to $z_1, z_2, z_3$From definitions:$ a = 2z_1 \Rightarrow z_1 = e^{i\theta}, $$ b = z_2 - 1 \Rightarrow z_2 = 1 + 2e^{i\theta}, $$ c = z_3 + 1 \Rightarrow z_3 = -1 + 2e^{i\theta}. $Step 8. Compute the required expressionWe need $ |4z_1 + z_2 + z_3| $.Substitute:$ 4z_1 + z_2 + z_3 = 4e^{i\theta} + (1 + 2e^{i\theta}) + (-1 + 2e^{i\theta}) = 4e^{i\theta} + 1 + 2e^{i\theta} - 1 + 2e^{i\theta}. $Simplify:$ 4z_1 + z_2 + z_3 = (4 + 2 + 2)e^{i\theta} + (1 - 1) = 8 e^{i\theta}. $Hence$ |4z_1 + z_2 + z_3| = |8 e^{i\theta}| = 8. $✅Final Answer:$ \boxed{8}. $✅ OptionA. 8

Complex Numbers - IAT-IISER Maths PYQs

We are given complex numbers

z_1%2C%20z_2%2C%20z_3

satisfying

2%20%3D%20%7C2z_1%7C%20%3D%20%7Cz_2%20-%201%7C%20%3D%20%7Cz_3%20%2B%201%7C%20%3D%20%5Cleft%7C%5Cfrac%7B1%7D%7Bz_1%7D%20%2B%20%5Cfrac%7B1%7D%7Bz_2%20-%201%7D%20%2B%20%5Cfrac%7B1%7D%7Bz_3%20%2B%201%7D%5Cright%7C.

We want to find

%5C%3B%20%7C4z_1%20%2B%20z_2%20%2B%20z_3%7C.

Step 1. Simplify magnitudesFrom

%7C2z_1%7C%20%3D%202%20%5CRightarrow%20%7Cz_1%7C%20%3D%201.

z_1

lies on the unit circle.Also:

%7Cz_2%20-%201%7C%20%3D%202%2C%20%5Cquad%20%7Cz_3%20%2B%201%7C%20%3D%202.

Hence,

z_2

lies on a circle of radius 2 centered at

1

z_3

lies on a circle of radius 2 centered at

-1

.Step 2. Introduce new variablesLet’s define

a%20%3D%202z_1%2C%20%5Cquad%20b%20%3D%20z_2%20-%201%2C%20%5Cquad%20c%20%3D%20z_3%20%2B%201.

Then by the given magnitudes,

%7Ca%7C%20%3D%20%7Cb%7C%20%3D%20%7Cc%7C%20%3D%202.

We can express the condition on reciprocals:

2%20%3D%20%5Cleft%7C%5Cfrac%7B1%7D%7Bz_1%7D%20%2B%20%5Cfrac%7B1%7D%7Bz_2%20-%201%7D%20%2B%20%5Cfrac%7B1%7D%7Bz_3%20%2B%201%7D%5Cright%7C%20%3D%20%5Cleft%7C%5Cfrac%7B1%7D%7Bz_1%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%7D%5Cright%7C.

But

%5Cfrac%7B1%7D%7Bz_1%7D%20%3D%20%5Cfrac%7B2%7D%7Ba%7D

. So the condition becomes:

2%20%3D%20%5Cleft%7C%5Cfrac%7B2%7D%7Ba%7D%20%2B%20%5Cfrac%7B1%7D%7Bb%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%7D%5Cright%7C.

Step 3. Note all have magnitude 2Let’s divide everything by 2 to normalize to unit circle magnitudes.Let’s write:

%5Cfrac%7Ba%7D%7B2%7D%20%3D%20e%5E%7Bi%5Calpha%7D%2C%20%5Cquad%20%5Cfrac%7Bb%7D%7B2%7D%20%3D%20e%5E%7Bi%5Cbeta%7D%2C%20%5Cquad%20%5Cfrac%7Bc%7D%7B2%7D%20%3D%20e%5E%7Bi%5Cgamma%7D.

Then

a%20%3D%202e%5E%7Bi%5Calpha%7D

, etc.The reciprocal condition becomes:

2%20%3D%20%5Cleft%7C%5Cfrac%7B2%7D%7B2%20e%5E%7Bi%5Calpha%7D%7D%20%2B%20%5Cfrac%7B1%7D%7B2%20e%5E%7Bi%5Cbeta%7D%7D%20%2B%20%5Cfrac%7B1%7D%7B2%20e%5E%7Bi%5Cgamma%7D%7D%5Cright%7C%20%3D%20%5Cleft%7C%20e%5E%7B-i%5Calpha%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7De%5E%7B-i%5Cbeta%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7De%5E%7B-i%5Cgamma%7D%5Cright%7C.

Multiply both sides by 2 to simplify the numeric coefficient? Wait, better to keep as is.We have:

%5Cleft%7C%20e%5E%7B-i%5Calpha%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7De%5E%7B-i%5Cbeta%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7De%5E%7B-i%5Cgamma%7D%5Cright%7C%20%3D%202.

Step 4. Magnitude constraint boundingEach term has magnitude 1, ½, ½ respectively, so the maximum possible magnitude is

1%20%2B%20%C2%BD%20%2B%20%C2%BD%20%3D%202.

To achieve equality

%3D2

,all three exponentials must have the same argument, i.e.

e%5E%7B-i%5Calpha%7D%2C%20e%5E%7B-i%5Cbeta%7D%2C%20e%5E%7B-i%5Cgamma%7D%20%5Ctext%7B%20have%20same%20direction.%7D

Hence they must be all proportional with positive real coefficients.That means

e%5E%7B-i%5Calpha%7D%20%3D%20e%5E%7B-i%5Cbeta%7D%20%3D%20e%5E%7B-i%5Cgamma%7D

, up to integer multiple of

2%5Cpi

.Step 5. So we can take them equal (since only magnitudes matter)Let

e%5E%7B-i%5Calpha%7D%20%3D%20e%5E%7B-i%5Cbeta%7D%20%3D%20e%5E%7B-i%5Cgamma%7D%20%3D%20e%5E%7B-i%5Ctheta%7D

.Then the condition is automatically satisfied:

%5Cleft%7C%20e%5E%7B-i%5Ctheta%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7De%5E%7B-i%5Ctheta%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7De%5E%7B-i%5Ctheta%7D%5Cright%7C%20%3D%20%7C2%20e%5E%7B-i%5Ctheta%7D%7C%20%3D%202.

OK.Step 6. Return to

a%2Cb%2Cc

Since

e%5E%7Bi%5Calpha%7D%2C%20e%5E%7Bi%5Cbeta%7D%2C%20e%5E%7Bi%5Cgamma%7D

all are equal, say

e%5E%7Bi%5Ctheta%7D.

Then:

a%20%3D%202%20e%5E%7Bi%5Ctheta%7D%2C%20%5Cquad%20b%20%3D%202%20e%5E%7Bi%5Ctheta%7D%2C%20%5Cquad%20c%20%3D%202%20e%5E%7Bi%5Ctheta%7D.

Step 7. Go back to

z_1%2C%20z_2%2C%20z_3

From definitions:

a%20%3D%202z_1%20%5CRightarrow%20z_1%20%3D%20e%5E%7Bi%5Ctheta%7D%2C

b%20%3D%20z_2%20-%201%20%5CRightarrow%20z_2%20%3D%201%20%2B%202e%5E%7Bi%5Ctheta%7D%2C

c%20%3D%20z_3%20%2B%201%20%5CRightarrow%20z_3%20%3D%20-1%20%2B%202e%5E%7Bi%5Ctheta%7D.

Step 8. Compute the required expressionWe need

%7C4z_1%20%2B%20z_2%20%2B%20z_3%7C

.Substitute:

4z_1%20%2B%20z_2%20%2B%20z_3%20%3D%204e%5E%7Bi%5Ctheta%7D%20%2B%20(1%20%2B%202e%5E%7Bi%5Ctheta%7D)%20%2B%20(-1%20%2B%202e%5E%7Bi%5Ctheta%7D)%20%3D%204e%5E%7Bi%5Ctheta%7D%20%2B%201%20%2B%202e%5E%7Bi%5Ctheta%7D%20-%201%20%2B%202e%5E%7Bi%5Ctheta%7D.

Simplify:

4z_1%20%2B%20z_2%20%2B%20z_3%20%3D%20(4%20%2B%202%20%2B%202)e%5E%7Bi%5Ctheta%7D%20%2B%20(1%20-%201)%20%3D%208%20e%5E%7Bi%5Ctheta%7D.

Hence

%7C4z_1%20%2B%20z_2%20%2B%20z_3%7C%20%3D%20%7C8%20e%5E%7Bi%5Ctheta%7D%7C%20%3D%208.

✅Final Answer:

%5Cboxed%7B8%7D.

✅ OptionA. 8

Complex Numbers

High-Yield Trend

Questions
4 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Complex Numbers

High-Yield Trend

Questions 4 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Questions
4 MCQs