Q.If $4x + i(3x - y) = 3 + i(-6)$, where $x$ and $y$ are real numbers, then find the values of $x$ and $y$.

We are asked to count the number of points in $A \cap B$.

Set $A$ is defined by two inequalities involving distances in the complex plane.

Set $B$ is a finite grid of integer-coordinate points with $y$ between $-3$ and $3$ inclusive, and both $x$ and $y$ natural numbers (here $\mathbb{N}$ includes 0).

The intersection is therefore a small set of lattice points we can list once we know which $x$-values are allowed.

1. Interpret the conditions for $A$

The first condition: $|z-4| < |z-2|$.

This says the distance from $z$ to 4 is less than the distance to 2.

Geometrically, the set of points closer to 4 than to 2 is the half‑plane to the right of the perpendicular bisector of the segment joining 2 and 4.

The midpoint is 3, so the bisector is the vertical line $x = 3$.

Since we want points closer to 4 (the larger number), we take the side where $x > 3$.

The second condition: $|z-7| > |z-3|$.

This says the distance to 7 is greater than the distance to 3.

The midpoint of 3 and 7 is 5, so the perpendicular bisector is $x = 5$.

We want points farther from 7 than from 3, which is the side where $x < 5$.

Thus $A$ is the vertical strip:

2. Intersect with $B$

Set $B$ is:

The key here is to see what the given information really means geometrically. When you have two complex numbers $w_1 = z_1 - \alpha$ and $w_2 = z_2 - \alpha$, the condition $|w_1| = |w_2|$ says they lie on a circle centered at the origin — they have the same magnitude. The argument condition says the angle from the positive real axis to $w_1$ plus the angle to $w_2$ equals $90^\circ$. That sum being $\pi/2$ is the signature of two vectors that are perpendicular, but with a specific orientation.

Let’s work it out step by step.

1. Let $w_1 = z_1 - \alpha$ and $w_2 = z_2 - \alpha$. Then $|w_1| = |w_2|$ and $\operatorname{Arg}(w_1) + \operatorname{Arg}(w_2) = \frac{\pi}{2}$.

2. Write $w_1$ and $w_2$ in polar form. Since they have equal magnitude, let $|w_1| = |w_2| = r > 0$. Then

where $\theta_1 = \operatorname{Arg}(w_1)$ and $\theta_2 = \operatorname{Arg}(w_2)$.

3. The argument condition gives $\theta_1 + \theta_2 = \frac{\pi}{2}$.

4. Now consider the ratio we need:

5. We know $\theta_1 + \theta_2 = \frac{\pi}{2}$, so $\theta_1 - \theta_2 = \theta_1 - \left(\frac{\pi}{2} - \theta_1\right) = 2\theta_1 - \frac{\pi}{2}$. That doesn’t look constant — but wait, we haven’t used the fact that the ratio itself must be a fixed number independent of the specific $\theta_1$. The trick is to notice that the ratio’s argument is $\theta_1 - \theta_2$, and from $\theta_1 + \theta_2 = \frac{\pi}{2}$ we can solve for $\theta_2 = \frac{\pi}{2} - \theta_1$, so

That still depends on $\theta_1$ — unless there’s an additional constraint we missed.

Actually, the ratio $\frac{w_1}{w_2}$ has argument $\theta_1 - \theta_2$. But we also know that $\frac{w_2}{w_1}$ has argument $\theta_2 - \theta_1 = -(\theta_1 - \theta_2)$. The sum condition alone doesn’t fix the difference. However, there is a classic result: if $|w_1| = |w_2|$ and $\operatorname{Arg}(w_1) + \operatorname{Arg}(w_2) = \frac{\pi}{2}$, then $w_1$ and $w_2$ are such that $w_1 = i w_2$ or $w_1 = -i w_2$? Let’s test.

Suppose $w_1 = i w_2$. Then $|w_1| = |i||w_2| = |w_2|$, good. And $\operatorname{Arg}(w_1) = \operatorname{Arg}(i) + \operatorname{Arg}(w_2) = \frac{\pi}{2} + \theta_2$, so $\theta_1 + \theta_2 = \frac{\pi}{2} + 2\theta_2$, which is not constant — so that’s not forced.

But the problem asks for $\frac{z_1 - \alpha}{z_2 - \alpha}$ as a specific value, meaning it must be independent of the particular $z_1, z_2$ satisfying the conditions. The only way that happens is if the ratio is constant for all such pairs. Let’s solve directly.

Let $\frac{w_1}{w_2} = k$. Then $w_1 = k w_2$. Since $|w_1| = |w_2|$, we have $|k| = 1$. Also, $\operatorname{Arg}(w_1) = \operatorname{Arg}(k) + \operatorname{Arg}(w_2)$. So $\theta_1 = \phi + \theta_2$, where $\phi = \operatorname{Arg}(k)$. Then $\theta_1 + \theta_2 = \phi + 2\theta_2 = \frac{\pi}{2}$. For this to hold for all possible $\theta_2$ (since $z_1, z_2$ can vary), $\phi$ must be $\frac{\pi}{2}$ and $2\theta_2 = 0$? That can’t be — so the ratio is not constant for all pairs? But the problem expects a single answer.

So pick a convenient pair. Let $\theta_1 = 0$. Then $\theta_2 = \frac{\pi}{2}$. Then $w_1 = r$, $w_2 = r i$. So $\frac{w_1}{w_2} = \frac{1}{i} = -i$. Alternatively, pick $\theta_1 = \frac{\pi}{4}$, then $\theta_2 = \frac{\pi}{4}$ as well? No, $\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$, so then $w_1 = r e^{i\pi/4}$, $w_2 = r e^{i\pi/4}$, ratio = 1, which doesn’t satisfy $|w_1| = |w_2|$? It does, but then the ratio is 1, not constant. So different choices give different ratios — meaning the problem must intend that the same pair satisfies both conditions, and the ratio is computed from that pair. But then the answer isn’t unique unless we use the fact that the argument sum is $\pi/2$ and magnitudes equal, which forces the two vectors to be symmetric about the line at $45^\circ$? Let’s derive properly.

Let $\theta_1 = \frac{\pi}{4} + \delta$, $\theta_2 = \frac{\pi}{4} - \delta$. Then $\theta_1 + \theta_2 = \frac{\pi}{2}$. Then $\frac{w_1}{w_2} = e^{i(2\delta)}$, which can be any point on the unit circle. So the ratio is not fixed — unless there’s an implicit assumption that $z_1$ and $z_2$ are distinct? Even then, $\delta$ can vary.

The concept first

Over the complex numbers, "$\sqrt{z}$" is not a single number: every non-zero $z$ has two square roots, differing by a sign. So an expression like $\sqrt{i}+\sqrt{-i}$ genuinely has several possible values, one for each choice of sign — and the question asks which of the printed numbers can never be produced by any such choice.

The cleanest route is polar form. Writing $z=r\,e^{i\theta}$,

Step-by-step

Step 1 — find both square roots of $i$. Since $i=e^{i\pi/2}$ (modulus $1$, argument $\dfrac{\pi}{2}$),

Verify: $\left(\dfrac{1+i}{\sqrt2}\right)^{2}=\dfrac{1+2i+i^{2}}{2}=\dfrac{2i}{2}=i\ \checkmark$

Step 2 — find both square roots of $-i$. Since $-i=e^{-i\pi/2}$,

Verify: $\left(\dfrac{1-i}{\sqrt2}\right)^{2}=\dfrac{1-2i+i^{2}}{2}=\dfrac{-2i}{2}=-i\ \checkmark$

Step 3 — combine, taking every sign choice. There are $2\times 2=4$ combinations, which pair up:

Same signs:

Opposite signs: …

Concept & Intuition

When raising a complex number to a fractional power, we get multiple values because the argument (angle) is only defined up to multiples of $2\pi$. The key is to first express the base in polar form, then apply the exponent, and finally list the distinct roots. Here, the exponent $5/6$ means we take the 6th root first (giving 6 candidates) and then raise to the 5th power. We only need to identify which of the given choices appears among those values.

Step-by-step solution

1. Write $-64i$ in polar form. The modulus is $| -64i | = 64$. The argument: $-64i$ lies on the negative imaginary axis, so its principal argument is $-\frac{\pi}{2}$ (or $\frac{3\pi}{2}$). We'll use $-\frac{\pi}{2}$ for convenience. Thus

2. Apply the exponent $5/6$. By De Moivre’s theorem for fractional powers,

where $k = 0, 1, 2, 3, 4, 5$ gives the six distinct values.

3. Simplify the modulus.

   $64^{5/6} = (2^6)^{5/6} = 2^{6 \cdot 5/6} = 2^5 = 32$.

   So every value has modulus 32.

4. Simplify the angle for a general $k$.

5. Find a value that matches one of the options. Compute for $k=0$: $\theta_0 = -\frac{5\pi}{12}$. That gives $32\left(\cos(-\frac{5\pi}{12}) + i\sin(-\frac{5\pi}{12})\right)$, which is not a nice standard angle. For $k=1$:

So one value is

That’s not among the options either (though it’s a negative version of (B) without the factor).

For $k=2$:

Reduce modulo $2\pi$: $\frac{35\pi}{12} - 2\pi = \frac{35\pi}{12} - \frac{24\pi}{12} = \frac{11\pi}{12}$.

So the value is $32\left(\cos\frac{11\pi}{12} + i\sin\frac{11\pi}{12}\right)$. That’s not a simple multiple-choice match.

For $k=3$:

$\theta_3 = -\frac{5\pi}{12} + 5\pi = -\frac{5\pi}{12} + \frac{60\pi}{12} = \frac{55\pi}{12}.$ …

The key insight is that powers of $i$ follow a simple repeating cycle:

So every fourth power brings us back to $1$. This means the sum of any consecutive block of four terms like $i^{4k} + i^{4k+1} + i^{4k+2} + i^{4k+3}$ is $1 + i - 1 - i = 0$. Therefore, the whole sum reduces to whatever remains after grouping complete cycles.

1. Identify the pattern of exponents

   The sum is $1 + i^2 + i^4 + i^6 + \dots + i^{2024}$.

   Notice the exponents are all even: $0, 2, 4, 6, \dots, 2024$.

   So we are summing $i^{2k}$ for $k = 0, 1, 2, \dots, 1012$.

   That’s $1013$ terms in total.

2. Simplify each term using $i^2 = -1$

   Since $i^{2k} = (i^2)^k = (-1)^k$, the sum becomes:

alternating between $+1$ and $-1$.

3. Count the number of terms

   From $k=0$ to $1012$ inclusive, there are $1013$ terms.

   Since $1013$ is odd, the alternating sum of an odd number of terms starting with $+1$ is $+1$.

   (For an even number of terms, the sum would be $0$.)

4. Check with a smaller example

   For $1 + i^2 = 1 + (-1) = 0$ (2 terms, even → 0). …

We are told that

is purely real. That means its imaginary part is zero. The key idea: for a complex number to be real, the numerator and denominator must be "in phase" — equivalently, the complex number equals its own conjugate. This gives a direct equation in $\cos\theta$ without having to rationalise first.

1. Set the fraction equal to its conjugate If $z$ is real, then $z = \overline{z}$. So

2. Cross-multiply

3. Expand both sides Left:

Right:

4. Equate and simplify

Cancel $1$ and $-2\cos^2\theta$ from both sides, leaving

Hence $\cos\theta = 0$.

5. Check the denominator-zero case

   Denominator $1-2i\cos\theta = 0$ would require $\cos\theta = -\frac{i}{2}$, impossible for real $\theta$. So no issue there.

6. But is $\cos\theta = 0$ the only solution?

   Let’s test $\cos\theta = -1$:

   $\frac{1 + i(-1)}{1 - 2i(-1)} = \frac{1 - i}{1 + 2i}.$ …

Let $\theta=\dfrac{2\pi}{9}$ and write $z=(1+\sin\theta)+i\cos\theta$. The denominator is $\bar z$, so the expression is $\left(\dfrac{z}{\bar z}\right)^n$.

Using $1+\sin\theta=2\cos^2\!\left(\tfrac\pi4-\tfrac\theta2\right)$ and $\cos\theta=2\sin\!\left(\tfrac\pi4-\tfrac\theta2\right)\cos\!\left(\tfrac\pi4-\tfrac\theta2\right)$,

Hence $\dfrac{z}{\bar z}=e^{2i\arg z}=e^{i(\pi/2-\theta)}$. With $\theta=\dfrac{2\pi}{9}$, …

Concept & Intuition

The polynomial $x^3+x^2+x+1=0$ is a geometric series: $1+x+x^2+x^3=0$. Multiplying by $(x-1)$ gives $x^4-1=0$, so the roots are the primitive 4th roots of unity (excluding $x=1$). That is, the roots are $i,\,-i,\,-1$. This observation makes every computation straightforward: we know the roots explicitly, so we can evaluate each sum directly rather than using general symmetric sums. The key is that each root satisfies $x^4=1$ and $x\neq1$, so $x^3=-x^2-x-1$, etc.

Step-by-step solution

1. Identify the roots The equation $x^3+x^2+x+1=0$ can be factored as $(x+1)(x^2+1)=0$, so the roots are

(Any permutation is fine; the symmetric sums are unchanged.)

2. Compute (i): $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$

Sum: $(-i) + i + (-1) = -1$.

So (i) = $-1$, which matches a in List II.

3. Compute (ii): $\alpha^3+\beta^3+\gamma^3$

Sum: $(-i)+i+(-1) = -1$.

So (ii) = $-1$, also matching a.

4. Compute (iii): $\alpha^4+\beta^4+\gamma^4$ Since each root satisfies $x^4=1$ (check: $i^4=1$, $(-i)^4=1$, $(-1)^4=1$),

So (iii) = $3$, which matches d.

5. Compute (iv): $(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2$ Use the identity:

First find $\alpha^2+\beta^2+\gamma^2$:

$i^2 = -1,\; (-i)^2 = -1,\; (-1)^2 = 1 \quad\Rightarrow\quad \text{sum} = -1-1+1 = -1.$ …

The key insight is that the cubic $3x^3 + x^2 - 7x + 3 = 0$ is a reciprocal equation of the type where coefficients read the same forward and backward (with a sign pattern). For $p=1$, $q=-7$, the equation becomes:

Notice the coefficients: $3, 1, -7, 3$. The first and last are equal, and the middle pair sum to something interesting. This suggests that if $r$ is a root, then $1/r$ is also a root. Let’s verify: divide the whole equation by $x^3$ (since $x=0$ is not a root — check: $3 \neq 0$), we get:

Multiply through by $x^3$ again, and you recover the original. So indeed, the set of roots is closed under taking reciprocals. One root is rational (by the Rational Root Theorem, possible candidates are $\pm 1, \pm 3, \pm 1/3$), and the other two are irrational and reciprocals of each other.

Let’s work through it step by step.

1. Find the rational root. Test $x = 1$: $3 + 1 - 7 + 3 = 0$. So $x = 1$ is a root. That means $(x-1)$ is a factor. Divide the cubic by $(x-1)$:

   Using synthetic division with coefficients $3, 1, -7, 3$:

   • Bring down $3$.
   • $1 \times 3 = 3$, add to $1$ gives $4$.
   • $1 \times 4 = 4$, add to $-7$ gives $-3$.
   • $1 \times (-3) = -3$, add to $3$ gives $0$.

   So the quotient is $3x^2 + 4x - 3$.

2. The quadratic factor $3x^2 + 4x - 3 = 0$ gives the other two roots. Its roots are:

So $\alpha = \frac{-2 + \sqrt{13}}{3}$ and $\beta = \frac{-2 - \sqrt{13}}{3}$ (or vice versa). Notice their product:

$\alpha \beta = \frac{(-2)^2 - (\sqrt{13})^2}{9} = \frac{4 - 13}{9} = \frac{-9}{9} = -1$ …