💡Okay, so I have this problem where I need to find the value of tan(α - π/6). The angle α has its vertex at the origin, its initial side is along the positive x-axis, and its terminal side passes through the point (-√3, 2). Hmm, let me try to visualize this.First, I remember that when an angle is in standard position (vertex at the origin, initial side on the positive x-axis), the terminal side can be anywhere on the coordinate plane. In this case, it's passing through (-√3, 2), which is in the second quadrant because the x-coordinate is negative and the y-coordinate is positive. So, angle α is in the second quadrant.I need to find tan(α - π/6). To do this, I think I should first find tan α and then use the tangent subtraction formula. The formula for tan(A - B) is (tan A - tan B)/(1 + tan A tan B). So, I need tan α and tan(π/6).Let me recall that tan(π/6) is 1/√3. So, I have that part. Now, I need to find tan α. Since the terminal side passes through (-√3, 2), I can use the definition of tangent, which is opposite over adjacent in the right triangle formed by the terminal side and the axes.Wait, but in this case, the point is (-√3, 2), so the x-coordinate is -√3 and the y-coordinate is 2. Since tangent is y/x, tan α should be 2/(-√3). That simplifies to -2/√3. But usually, we rationalize the denominator, so multiplying numerator and denominator by √3, we get -2√3/3.So, tan α is -2√3/3. Now, I can plug this into the tangent subtraction formula. Let me write that down:tan(α - π/6) = (tan α - tan(π/6)) / (1 + tan α tan(π/6))Substituting the known values:= (-2√3/3 - 1/√3) / (1 + (-2√3/3)(1/√3))Hmm, let me simplify numerator and denominator separately.First, the numerator:-2√3/3 - 1/√3I need to combine these terms. To do that, I should have a common denominator. The denominators are 3 and √3. The common denominator would be 3√3.So, let me rewrite each term:-2√3/3 = (-2√3 * √3)/(3 * √3) = (-2*3)/(3√3) = -6/(3√3) = -2/√3Wait, that seems like going in circles. Maybe another approach. Alternatively, I can express both terms with denominator 3√3.So:-2√3/3 = (-2√3 * √3)/(3 * √3) = (-2*3)/(3√3) = -6/(3√3) = -2/√3Similarly, 1/√3 = (1 * 3)/(√3 * 3) = 3/(3√3) = 1/√3Wait, that didn't help much. Maybe I should just compute the numerator as is.Numerator: (-2√3/3) - (1/√3)Let me compute this numerically to see what it is approximately. But maybe it's better to find a common denominator.Let me write 1/√3 as √3/3. Because 1/√3 = √3/3.So, numerator becomes:-2√3/3 - √3/3 = (-2√3 - √3)/3 = (-3√3)/3 = -√3Oh, that's simpler! So, the numerator simplifies to -√3.Now, the denominator:1 + (-2√3/3)(1/√3)First, compute the product:(-2√3/3)(1/√3) = (-2√3)/(3√3) = (-2)/(3) because √3 cancels out.So, denominator becomes:1 + (-2/3) = (3/3 - 2/3) = 1/3So, putting it all together:tan(α - π/6) = numerator / denominator = (-√3) / (1/3) = -√3 * 3 = -3√3Wait, that seems straightforward. So, the value is -3√3. Looking back at the options, option A is -3√3, so that must be the answer.But let me double-check my steps to make sure I didn't make a mistake.1. Found tan α by using the coordinates (-√3, 2): tan α = y/x = 2/(-√3) = -2/√3 = -2√3/3. That seems correct.2. Applied the tangent subtraction formula: (tan α - tan π/6)/(1 + tan α tan π/6). That's the correct formula.3. Substituted tan α = -2√3/3 and tan π/6 = 1/√3.4. Simplified the numerator:-2√3/3 - 1/√3. I converted 1/√3 to √3/3, so numerator became (-2√3 - √3)/3 = -3√3/3 = -√3. That seems right.5. Simplified the denominator:1 + (-2√3/3)(1/√3) = 1 - 2/3 = 1/3. Correct.6. Divided numerator by denominator: (-√3)/(1/3) = -3√3. Correct.So, all steps check out. Therefore, the answer is A: -3√3.**Final Answer**boxed{A}

💡Okay, so I need to find the domain of the function ( y = sqrt{16 - x^2} + frac{1}{sqrt{sin x}} ). Hmm, the domain is all the real numbers x for which the function is defined. Since this function has two parts, I should consider the domains of each part separately and then find their intersection.First, let's look at the square root function ( sqrt{16 - x^2} ). For a square root to be defined, the expression inside must be non-negative. So, I need:[ 16 - x^2 geq 0 ]Solving this inequality:[ 16 geq x^2 ][ x^2 leq 16 ][ -4 leq x leq 4 ]So, the domain for the first part is all real numbers from -4 to 4, inclusive.Now, moving on to the second part of the function: ( frac{1}{sqrt{sin x}} ). This is a bit more complicated because it involves both a square root and a reciprocal. Let's break it down.First, the expression inside the square root, ( sin x ), must be positive because we can't take the square root of a negative number, and also, since it's in the denominator, it can't be zero because division by zero is undefined. So, we need:[ sin x > 0 ]I remember that the sine function is positive in the first and second quadrants, which corresponds to angles between ( 0 ) and ( pi ) radians (or 0 to 180 degrees). But sine is periodic, so this repeats every ( 2pi ) radians. So, the general solution for ( sin x > 0 ) is:[ 2kpi < x < (2k + 1)pi ]where ( k ) is any integer.But since we're dealing with the domain of the entire function, which is limited by the first part to ( x ) between -4 and 4, we need to find the overlap between ( -4 leq x leq 4 ) and the intervals where ( sin x > 0 ).Let me visualize the sine function between -4 and 4. The sine function has a period of ( 2pi ), which is approximately 6.28. So, between -4 and 4, we're covering a bit less than one full period.Let's find the points where ( sin x = 0 ) within this interval because that's where the function changes sign. The zeros of sine occur at multiples of ( pi ), so approximately 3.14, 6.28, etc., and their negatives.Within -4 to 4, the zeros are at approximately -3.14 (which is -π), 0, and 3.14 (π). So, the intervals where ( sin x > 0 ) within -4 to 4 would be:1. Between -π and 0: Wait, no. Let me think. Sine is positive in the first and second quadrants, which is between 0 and π. But for negative angles, sine is negative in the third and fourth quadrants. So, between -π and 0, sine is negative.So, actually, within -4 to 4, the intervals where ( sin x > 0 ) are:- From 0 to π (approximately 0 to 3.14)- From -2π to -π (approximately -6.28 to -3.14), but since our domain is limited to -4, which is greater than -6.28, the overlapping interval would be from -4 to -π.Wait, is that correct? Let me double-check.Sine is positive in the first and second quadrants, which correspond to angles between 0 and π. For negative angles, sine is positive in the third and fourth quadrants? Wait, no. Actually, sine is positive in the first and second quadrants (0 to π) and negative in the third and fourth quadrants (π to 2π). But when dealing with negative angles, it's a bit different.Let me recall that ( sin(-x) = -sin x ). So, if x is negative, say between -π and 0, then ( sin x = -sin(-x) ). Since ( -x ) would be between 0 and π, ( sin(-x) ) is positive, so ( sin x ) is negative in that interval.Therefore, in the interval from -π to 0, ( sin x ) is negative, and from 0 to π, it's positive. Beyond π, from π to 2π, it's negative again.But our domain is limited to x between -4 and 4. So, let's see:- From -4 to -π: Let's calculate -4 in terms of π. Since π is approximately 3.14, -π is approximately -3.14. So, -4 is less than -π. Therefore, the interval from -4 to -π is from approximately -4 to -3.14.In this interval, what is the sign of ( sin x )? Let's pick a test point, say x = -3.5. ( sin(-3.5) = -sin(3.5) ). 3.5 radians is approximately 200 degrees, which is in the third quadrant where sine is negative. Therefore, ( sin(-3.5) = -sin(3.5) ) would be positive because ( sin(3.5) ) is negative, and the negative of a negative is positive. Wait, that's confusing.Wait, no. Let me think again. If x is between -π and 0, which is approximately -3.14 to 0, ( sin x ) is negative. But if x is less than -π, say between -2π and -π, which is approximately -6.28 to -3.14, then ( sin x ) is positive because it's equivalent to the sine of a positive angle in the first and second quadrants due to periodicity.But in our case, the domain is limited to x between -4 and 4. So, -4 is approximately -4, which is greater than -π (≈-3.14). Wait, no, -4 is less than -π. So, -4 is to the left of -π on the number line.Therefore, the interval from -4 to -π is from approximately -4 to -3.14. In this interval, x is between -4 and -π, which is equivalent to angles between -4 and -π. Since sine has a period of 2π, we can add 2π to x to find the equivalent positive angle.So, for x = -4, adding 2π gives approximately -4 + 6.28 = 2.28 radians, which is in the second quadrant where sine is positive. Similarly, for x = -π, adding 2π gives π, which is where sine is zero.Therefore, in the interval from -4 to -π, ( sin x ) is positive because it's equivalent to sine of angles between 2.28 and π, which are in the second quadrant where sine is positive.So, putting it all together:- The first part, ( sqrt{16 - x^2} ), requires ( -4 leq x leq 4 ).- The second part, ( frac{1}{sqrt{sin x}} ), requires ( sin x > 0 ), which in the interval ( -4 leq x leq 4 ) happens in two places: - From -4 to -π (approximately -4 to -3.14), because as we saw, ( sin x ) is positive here. - From 0 to π (approximately 0 to 3.14), where sine is positive.But wait, at x = 0, ( sin x = 0 ), so ( frac{1}{sqrt{sin x}} ) is undefined. Similarly, at x = π, ( sin x = 0 ), so it's also undefined there. Therefore, the intervals where ( sin x > 0 ) are open intervals: (-4, -π) and (0, π).However, we also need to consider the endpoints of the first part, which is closed at -4 and 4. But since at x = -4, ( sin(-4) ) is positive (as we saw earlier), but we need to check if x = -4 is included. Wait, at x = -4, ( sin(-4) ) is positive, so ( frac{1}{sqrt{sin(-4)}} ) is defined. Therefore, x = -4 is included.Similarly, at x = 4, ( sin(4) ) is approximately sin(4 radians) which is about -0.7568, which is negative. Therefore, ( sin(4) < 0 ), so ( frac{1}{sqrt{sin(4)}} ) is undefined. Therefore, x = 4 is not included.Wait, but earlier I thought that in the interval from -4 to -π, ( sin x ) is positive, so x = -4 is included because ( sin(-4) ) is positive. Let me confirm:( sin(-4) = -sin(4) ). Since ( sin(4) ) is negative (as 4 radians is in the third quadrant where sine is negative), ( sin(-4) = -sin(4) ) would be positive. Therefore, ( sin(-4) > 0 ), so x = -4 is included.Similarly, at x = -π, ( sin(-π) = 0 ), so ( frac{1}{sqrt{sin(-π)}} ) is undefined. Therefore, x = -π is not included.So, combining all this:- The domain for the first part is ( [-4, 4] ).- The domain for the second part is ( (-4, -π) cup (0, π) ).Therefore, the intersection of these two domains is ( [-4, -π) cup (0, π) ).Wait, but let me double-check. At x = -4, ( sin(-4) ) is positive, so it's included. At x = -π, ( sin(-π) = 0 ), so it's excluded. At x = 0, ( sin(0) = 0 ), so it's excluded. At x = π, ( sin(π) = 0 ), so it's excluded. At x = 4, ( sin(4) ) is negative, so it's excluded.Therefore, the domain is ( [-4, -π) cup (0, π) ).Looking at the options:A: ( (0,4] ) – This only includes the positive side up to 4, but we have to exclude 4 and include up to π, which is less than 4. Also, it doesn't include the negative side.B: ( [-4,-π] cup (0,4] ) – This includes x = -π, which is not allowed because ( sin(-π) = 0 ). Also, it includes x = 4, which is not allowed.C: ( [-π,0] ) – This is only the interval from -π to 0, but in this interval, ( sin x ) is negative, so it's not allowed.D: ( [-4,-π) cup (0,π) ) – This matches our conclusion. It includes x = -4, excludes x = -π, excludes x = 0, and excludes x = π. It also correctly limits the positive side up to π instead of 4.Therefore, the correct answer is D.

💡Alright, so I have this problem where I'm given two points, M(-1, 0) and N(1, 0), and a point P somewhere in the plane. The condition is that the dot products of certain vectors form an arithmetic sequence with a common difference less than zero. I need to find the locus of point P.First, I need to understand what the vectors are. The vectors mentioned are MP, MN, PM, PN, NM, and NP. Let me write down what each of these vectors is in terms of coordinates.Point M is (-1, 0), and point N is (1, 0). Let’s denote point P as (x, y). So, vector MP would be from M to P, which is (x - (-1), y - 0) = (x + 1, y). Similarly, vector MN is from M to N, which is (1 - (-1), 0 - 0) = (2, 0). Vector PM is from P to M, which is (-1 - x, 0 - y) = (-x -1, -y). Vector PN is from P to N, which is (1 - x, 0 - y) = (1 - x, -y). Vector NM is from N to M, which is (-2, 0), and vector NP is from N to P, which is (x - 1, y - 0) = (x - 1, y).Now, I need to compute the dot products of these vectors as specified. The problem states that the dot products MP · MN, PM · PN, and NM · NP form an arithmetic sequence with a common difference less than zero.Let me compute each of these dot products one by one.First, MP · MN. Vector MP is (x + 1, y) and vector MN is (2, 0). The dot product is (x + 1)*2 + y*0 = 2(x + 1) = 2x + 2.Second, PM · PN. Vector PM is (-x -1, -y) and vector PN is (1 - x, -y). The dot product is (-x -1)(1 - x) + (-y)(-y). Let me compute that:(-x -1)(1 - x) = (-x)(1 - x) -1*(1 - x) = -x + x^2 -1 + x = x^2 -1.Then, (-y)(-y) = y^2. So, PM · PN = x^2 -1 + y^2.Third, NM · NP. Vector NM is (-2, 0) and vector NP is (x - 1, y). The dot product is (-2)(x - 1) + 0*y = -2(x - 1) = -2x + 2.So, now I have the three dot products:1. MP · MN = 2x + 22. PM · PN = x^2 + y^2 - 13. NM · NP = -2x + 2These three form an arithmetic sequence with a common difference less than zero. So, the middle term should be the average of the first and the third terms, and the common difference should be negative.Let me denote the three terms as a, b, c. So, a = 2x + 2, b = x^2 + y^2 - 1, c = -2x + 2.Since it's an arithmetic sequence, b - a = c - b, and the common difference d = b - a = c - b < 0.So, first, let me set up the equation for the arithmetic sequence:b - a = c - bSubstituting the expressions:(x^2 + y^2 - 1) - (2x + 2) = (-2x + 2) - (x^2 + y^2 - 1)Simplify both sides:Left side: x^2 + y^2 -1 -2x -2 = x^2 + y^2 -2x -3Right side: -2x + 2 -x^2 - y^2 +1 = -x^2 - y^2 -2x +3So, setting left side equal to right side:x^2 + y^2 -2x -3 = -x^2 - y^2 -2x +3Let me bring all terms to one side:x^2 + y^2 -2x -3 + x^2 + y^2 +2x -3 = 0Combine like terms:2x^2 + 2y^2 -6 = 0Divide both sides by 2:x^2 + y^2 -3 = 0So, x^2 + y^2 = 3Okay, so that's one equation. Now, the other condition is that the common difference is less than zero. So, d = b - a < 0.Compute d = b - a = (x^2 + y^2 -1) - (2x + 2) = x^2 + y^2 -1 -2x -2 = x^2 + y^2 -2x -3We have from earlier that x^2 + y^2 = 3, so substitute that into d:d = 3 -2x -3 = -2xSo, d = -2x < 0Which implies that -2x < 0 => x > 0Therefore, the locus of point P is the set of points (x, y) such that x^2 + y^2 = 3 and x > 0.So, it's the right half of the circle centered at the origin with radius sqrt(3).Wait, let me double-check the calculations to make sure I didn't make any mistakes.First, the dot products:MP · MN: (x+1, y) · (2, 0) = 2(x+1) = 2x + 2. That seems correct.PM · PN: (-x -1, -y) · (1 - x, -y) = (-x -1)(1 - x) + (-y)(-y). Let's compute (-x -1)(1 - x):Multiply term by term: (-x)(1) + (-x)(-x) + (-1)(1) + (-1)(-x) = -x + x^2 -1 + x = x^2 -1. Then, (-y)(-y) = y^2. So, total is x^2 + y^2 -1. Correct.NM · NP: (-2, 0) · (x -1, y) = (-2)(x -1) + 0*y = -2x + 2. Correct.Then, setting up the arithmetic sequence condition:b - a = c - b(x^2 + y^2 -1) - (2x + 2) = (-2x + 2) - (x^2 + y^2 -1)Simplify left side: x^2 + y^2 -1 -2x -2 = x^2 + y^2 -2x -3Right side: -2x + 2 -x^2 - y^2 +1 = -x^2 - y^2 -2x +3Set equal:x^2 + y^2 -2x -3 = -x^2 - y^2 -2x +3Bring all terms to left:x^2 + y^2 -2x -3 +x^2 + y^2 +2x -3 = 0Simplify: 2x^2 + 2y^2 -6 = 0 => x^2 + y^2 = 3. Correct.Then, common difference d = b - a = x^2 + y^2 -2x -3. Since x^2 + y^2 = 3, substitute: 3 -2x -3 = -2x. So, d = -2x < 0 => x > 0. Correct.Therefore, the locus is the right half of the circle x^2 + y^2 = 3.I think that's solid. So, the final answer is the right half of the circle centered at the origin with radius sqrt(3).**Final Answer**The locus of point ( P ) is the right half of the circle centered at the origin with radius ( sqrt{3} ). So, the final answer is boxed{x^2 + y^2 = 3} with ( x > 0 ).However, since the problem asks for the locus, and not just the equation, it's better to describe it as the right half of the circle. But in the required format, I think the equation suffices with the condition.But to match the previous answer, I think the box should contain the equation and the condition. But in the initial problem, the user might expect just the equation. Wait, in the previous response, the assistant concluded with the equation inside the box. Maybe I should do the same.But actually, the locus is the set of points satisfying x² + y² = 3 with x > 0. So, to represent this, I can write it as x² + y² = 3, x > 0. But in the box, maybe just the equation, since the condition is part of the locus.Alternatively, since the problem is about the locus, which is a geometric figure, the right half of the circle. So, perhaps the answer is the right half of the circle x² + y² = 3.But in the initial problem, the assistant wrote the conclusion as the right half of the circle. So, maybe I should follow that.But the user instruction says to put the final answer within boxed{}.So, perhaps I should write the equation with the condition inside the box.But LaTeX-wise, it's a bit tricky. Maybe just the equation, and mention the condition separately.But in the initial problem, the assistant wrote:boxed{text{The locus of point } P text{ is the right half of the circle centered at the origin with radius } sqrt{3}}.But the user might expect a mathematical equation in the box.Alternatively, since the equation is x² + y² = 3, and the condition is x > 0, perhaps write both.But I think the standard way is to describe the locus, so maybe the sentence inside the box is acceptable.But in the initial problem, the assistant wrote the sentence inside the box. So, to be consistent, I think I'll do the same.**Final Answer**boxed{text{The locus of point } P text{ is the right half of the circle centered at the origin with radius } sqrt{3}}.

💡Okay, so I have this equation here: (frac{z^4 - 1}{z^3 + z^2 - 2z} = 0). I need to find the number of complex solutions to this equation. Hmm, let me think about how to approach this.First, I remember that for a fraction to be zero, the numerator has to be zero while the denominator is not zero. So, the equation (frac{z^4 - 1}{z^3 + z^2 - 2z} = 0) implies that (z^4 - 1 = 0) and (z^3 + z^2 - 2z neq 0). That makes sense because if the denominator were zero, the expression would be undefined, right?So, let me start by solving the numerator equation: (z^4 - 1 = 0). This is a quartic equation, and I know it can be factored. Let me try factoring it.I remember that (z^4 - 1) is a difference of squares, so it can be written as ((z^2)^2 - 1^2), which factors into ((z^2 - 1)(z^2 + 1)). Now, each of these can be factored further. The first term, (z^2 - 1), is another difference of squares, so it factors into ((z - 1)(z + 1)). The second term, (z^2 + 1), is a sum of squares, which factors into ((z - i)(z + i)) because (i) is the imaginary unit where (i^2 = -1).So, putting it all together, the numerator factors into:[(z - 1)(z + 1)(z - i)(z + i)]This means the solutions to (z^4 - 1 = 0) are (z = 1), (z = -1), (z = i), and (z = -i). So, four solutions in total.But wait, I can't just stop here because the denominator also plays a role. The denominator is (z^3 + z^2 - 2z), and I need to make sure that none of the solutions I found make the denominator zero. If any of them do, those solutions would be excluded because division by zero is undefined.So, let me factor the denominator as well. The denominator is (z^3 + z^2 - 2z). I can factor out a (z) from each term:[z(z^2 + z - 2)]Now, let's factor the quadratic (z^2 + z - 2). I need two numbers that multiply to (-2) and add up to (1). Hmm, 2 and (-1) work because (2 times (-1) = -2) and (2 + (-1) = 1). So, the quadratic factors into:[(z + 2)(z - 1)]Therefore, the denominator factors into:[z(z + 2)(z - 1)]So, the denominator is zero when (z = 0), (z = -2), or (z = 1).Now, going back to the solutions of the numerator: (z = 1), (z = -1), (z = i), and (z = -i). I need to check if any of these make the denominator zero.Looking at (z = 1): Plugging into the denominator, we get (1(1 + 2)(1 - 1) = 1 times 3 times 0 = 0). So, (z = 1) makes the denominator zero, which means (z = 1) is not a valid solution for the original equation.Checking the other solutions:- (z = -1): Plugging into the denominator, (-1(-1 + 2)(-1 - 1) = -1 times 1 times (-2) = 2), which is not zero. So, (z = -1) is a valid solution.- (z = i): Plugging into the denominator, (i(i + 2)(i - 1)). Let's compute this step by step. First, (i + 2 = 2 + i), and (i - 1 = -1 + i). Multiplying these together: ((2 + i)(-1 + i)). Let's expand this: [ 2(-1) + 2(i) + i(-1) + i(i) = -2 + 2i - i + i^2 = -2 + i -1 = -3 + i ] Then multiply by (i): [ i(-3 + i) = -3i + i^2 = -3i -1 ] Which is (-1 - 3i), definitely not zero. So, (z = i) is a valid solution.- (z = -i): Similarly, plugging into the denominator, (-i(-i + 2)(-i - 1)). Let's compute this: First, (-i + 2 = 2 - i), and (-i - 1 = -1 - i). Multiply these together: [ (2 - i)(-1 - i) = 2(-1) + 2(-i) - i(-1) - i(-i) = -2 - 2i + i + i^2 = -2 - i -1 = -3 - i ] Then multiply by (-i): [ -i(-3 - i) = 3i + i^2 = 3i -1 ] Which is (-1 + 3i), not zero. So, (z = -i) is also a valid solution.So, out of the four solutions from the numerator, (z = 1) is excluded because it makes the denominator zero. The remaining three solutions, (z = -1), (z = i), and (z = -i), are all valid.Wait, but just to make sure I didn't make a mistake in my calculations, especially with the complex numbers. Let me double-check the denominator evaluations for (z = i) and (z = -i).For (z = i):Denominator: (i^3 + i^2 - 2i). Let's compute each term:- (i^3 = (i^2)(i) = (-1)(i) = -i)- (i^2 = -1)- So, denominator becomes (-i + (-1) - 2i = -1 - 3i), which is not zero.For (z = -i):Denominator: ((-i)^3 + (-i)^2 - 2(-i)). Compute each term:- ((-i)^3 = (-i)(-i)(-i) = (-i)^2(-i) = (-1)(i^2)(-i) = (-1)(-1)(-i) = -i)Wait, hold on, that seems a bit convoluted. Alternatively, ((-i)^3 = - (i)^3 = -(-i) = i). Hmm, maybe I should compute it step by step:- ((-i)^3 = (-i) times (-i) times (-i))First, ((-i) times (-i) = i^2 = -1)Then, (-1 times (-i) = i)So, ((-i)^3 = i)- ((-i)^2 = (-i) times (-i) = i^2 = -1)- So, denominator becomes (i + (-1) - 2(-i) = i -1 + 2i = -1 + 3i), which is not zero.Okay, so both (z = i) and (z = -i) result in non-zero denominators. Therefore, they are valid solutions.So, summarizing:- Numerator solutions: (z = 1), (z = -1), (z = i), (z = -i)- Denominator zero at: (z = 0), (z = -2), (z = 1)- Excluding (z = 1), the valid solutions are (z = -1), (z = i), (z = -i)Therefore, there are three complex solutions to the equation.Wait, but hold on a second. The problem is asking for the number of complex solutions. I know that in complex analysis, every non-constant polynomial equation of degree (n) has exactly (n) roots in the complex plane, counting multiplicities. But in this case, we're dealing with a rational function, not a polynomial. So, does that affect the number of solutions?Hmm, the equation is (frac{z^4 - 1}{z^3 + z^2 - 2z} = 0). As I thought earlier, this reduces to solving (z^4 - 1 = 0) with (z^3 + z^2 - 2z neq 0). So, it's not a polynomial equation, but the solutions are still the roots of the numerator that don't make the denominator zero. So, in this case, the number of solutions is equal to the number of roots of the numerator that are not roots of the denominator.Since the numerator is a quartic (degree 4) and the denominator is a cubic (degree 3), and they share a common root at (z = 1), which is excluded, the number of solutions is 4 - 1 = 3.Therefore, the number of complex solutions is 3.I think that's solid. I don't see any mistakes in my reasoning. I factored both numerator and denominator correctly, checked each root, and confirmed that only (z = 1) is excluded. So, yeah, three solutions.**Final Answer**The number of complex solutions is boxed{3}.

💡First, I need to find the midpoint ( N ) of segment ( overline{AB} ) with points ( A(4, 3) ) and ( B(16, 21) ). To do this, I'll use the midpoint formula, which averages the x-coordinates and the y-coordinates of points ( A ) and ( B ).Next, I'll reflect segment ( overline{AB} ) over the y-axis. Reflecting a point over the y-axis changes the sign of its x-coordinate while keeping the y-coordinate the same. So, point ( A(4, 3) ) will become ( A'(-4, 3) ) and point ( B(16, 21) ) will become ( B'(-16, 21) ).After reflecting the points, I'll find the midpoint ( N' ) of the reflected segment ( overline{A'B'} ) using the same midpoint formula. Finally, I'll calculate the sum of the coordinates of point ( N' ) to get the final answer.

💡To determine when all four runners will meet again, we need to find the least common multiple (LCM) of the times it takes for each pair of runners to meet. 1. **Calculate the relative speeds between each pair of runners:** - Runner 1 and Runner 2: (6.5 - 5 = 1.5) m/s - Runner 1 and Runner 3: (7.5 - 5 = 2.5) m/s - Runner 1 and Runner 4: (8 - 5 = 3) m/s2. **Determine the time it takes for each pair to meet:** - For Runner 1 and Runner 2: ( frac{600}{1.5} = 400 ) seconds - For Runner 1 and Runner 3: ( frac{600}{2.5} = 240 ) seconds - For Runner 1 and Runner 4: ( frac{600}{3} = 200 ) seconds3. **Find the LCM of these times to determine when all runners will meet again:** - LCM of 400 and 240 is 1200 - LCM of 1200 and 200 is 1200Therefore, all runners will meet again after 1200 seconds.