💡Okay, so I have this problem where I'm given the inverse of matrix A, which is A⁻¹, and I need to find the inverse of A squared, which is (A²)⁻¹. Hmm, let me think about how to approach this.First, I remember that when dealing with inverses of matrices, there are some properties that might help. For example, I recall that the inverse of a product of matrices is the product of their inverses in reverse order. So, if I have two matrices B and C, then (BC)⁻¹ = C⁻¹B⁻¹. Maybe I can use something similar here.Since I need to find (A²)⁻¹, which is the inverse of A multiplied by itself, I wonder if I can express this in terms of A⁻¹. Let me write it out:(A²)⁻¹ = (A * A)⁻¹Using the property I just remembered, this should be equal to A⁻¹ * A⁻¹, right? Because the inverse of a product is the product of the inverses in reverse order. So, (A * A)⁻¹ = A⁻¹ * A⁻¹. That makes sense.Therefore, (A²)⁻¹ = (A⁻¹)². So, if I can compute the square of A⁻¹, that should give me the inverse of A squared. Cool, so I just need to square the given matrix A⁻¹.Given that A⁻¹ is:A⁻¹ = | 3 -1 | | 2 1 |I need to compute (A⁻¹)², which is A⁻¹ multiplied by itself. Let me set that up.So, let me denote A⁻¹ as matrix B for simplicity:B = | 3 -1 | | 2 1 |Then, (A⁻¹)² = B * B. Let's compute this matrix multiplication step by step.Matrix multiplication works by taking the dot product of the rows of the first matrix with the columns of the second matrix. Since both matrices are 2x2, the result will also be a 2x2 matrix.Let me compute each element of the resulting matrix:First element (top-left): (3)(3) + (-1)(2) = 9 - 2 = 7Second element (top-right): (3)(-1) + (-1)(1) = -3 - 1 = -4Third element (bottom-left): (2)(3) + (1)(2) = 6 + 2 = 8Fourth element (bottom-right): (2)(-1) + (1)(1) = -2 + 1 = -1Putting it all together, the resulting matrix is:| 7 -4 || 8 -1 |So, (A⁻¹)² = | 7 -4 | | 8 -1 |Therefore, since (A²)⁻¹ = (A⁻¹)², this matrix is the inverse of A squared.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First element: 3*3 + (-1)*2 = 9 - 2 = 7. Correct.Second element: 3*(-1) + (-1)*1 = -3 -1 = -4. Correct.Third element: 2*3 + 1*2 = 6 + 2 = 8. Correct.Fourth element: 2*(-1) + 1*1 = -2 + 1 = -1. Correct.Okay, all the elements check out. So, the inverse of A squared is indeed:| 7 -4 || 8 -1 |I think that's the final answer. Let me just recap the steps to ensure I followed the correct process.1. Recognize that (A²)⁻¹ is equal to (A⁻¹)².2. Square the given inverse matrix A⁻¹ by performing matrix multiplication.3. Compute each element of the resulting matrix step by step.4. Verify the calculations to ensure accuracy.Everything seems to add up, so I feel confident about this solution.**Final Answer**The inverse of (mathbf{A}^2) is (boxed{begin{pmatrix} 7 & -4 8 & -1 end{pmatrix}}).

💡Okay, so I have this trapezoid (ABCD) where (AD) is parallel to (BC). On the side (CD), there's a point (M). From vertex (A), they drop a perpendicular (AH) onto segment (BM). It's given that (AD = HD). I need to find the length of (AD), and I know that (BC = 16), (CM = 8), and (MD = 9).First, let me try to visualize the trapezoid. Since (AD) is parallel to (BC), (ABCD) is a trapezoid with bases (AD) and (BC). Point (M) is on the leg (CD), and it's divided into segments (CM = 8) and (MD = 9). So, the entire length of (CD) is (8 + 9 = 17).Now, from point (A), we drop a perpendicular to (BM), and the foot of this perpendicular is point (H). It's given that (AD = HD). That seems important. So, (H) is a point on (BM) such that the segment (HD) is equal in length to (AD).I think I need to use some properties of trapezoids and similar triangles here. Since (AD) is parallel to (BC), maybe there are some similar triangles formed by the intersection of (BM) and (AH).Let me consider triangles (BCM) and (KDM). Wait, where did (K) come from? Maybe I need to clarify that. Perhaps (K) is another point? Hmm, maybe I should think about the triangles formed by the perpendicular (AH).Since (AH) is perpendicular to (BM), triangle (AHM) is a right triangle. Also, since (AD = HD), triangle (ADH) is isosceles with (AD = HD). Maybe I can use some properties of isosceles triangles here.Let me try to assign coordinates to the trapezoid to make it easier. Let's place point (D) at the origin ((0, 0)). Since (AD) is parallel to (BC), and assuming the trapezoid is oriented with (AD) and (BC) horizontal, point (A) would be at ((x, 0)), point (D) at ((0, 0)), point (C) at ((c, h)), and point (B) at ((b, h)), where (h) is the height of the trapezoid.Given that (CM = 8) and (MD = 9), point (M) divides (CD) into segments of 8 and 9. So, if (C) is at ((c, h)) and (D) is at ((0, 0)), then point (M) can be found using the section formula. The coordinates of (M) would be:[M = left( frac{9 cdot c + 8 cdot 0}{8 + 9}, frac{9 cdot h + 8 cdot 0}{8 + 9} right) = left( frac{9c}{17}, frac{9h}{17} right)]Now, the equation of line (BM) can be found. Point (B) is at ((b, h)) and point (M) is at (left( frac{9c}{17}, frac{9h}{17} right)). The slope of (BM) is:[m_{BM} = frac{frac{9h}{17} - h}{frac{9c}{17} - b} = frac{-8h/17}{(9c - 17b)/17} = frac{-8h}{9c - 17b}]The equation of line (BM) is then:[y - h = frac{-8h}{9c - 17b}(x - b)]Now, the perpendicular from (A) to (BM) will have a slope that is the negative reciprocal of (m_{BM}). So, the slope of (AH) is:[m_{AH} = frac{9c - 17b}{8h}]Point (A) is at ((x, 0)), so the equation of (AH) is:[y - 0 = frac{9c - 17b}{8h}(x - x_A)]Wait, I'm getting confused with the coordinates. Maybe I should assign specific coordinates to simplify. Let me set (D) at ((0, 0)), (C) at ((17, h)) because (CD = 17), but actually, (CD) is a leg, not necessarily horizontal. Hmm, maybe assigning coordinates is complicating things.Let me try another approach. Since (AD = HD), triangle (ADH) is isosceles with (AD = HD). So, point (H) lies somewhere on (BM) such that (HD = AD). Maybe I can use the properties of similar triangles or coordinate geometry.Alternatively, maybe using vectors or coordinate geometry is the way to go. Let me try coordinate geometry again, but more carefully.Let me place point (D) at ((0, 0)). Since (AD) is parallel to (BC), and assuming (AD) is horizontal, point (A) will be at ((a, 0)). Point (C) is somewhere above (D), say at ((c, h)), and point (B) is above (A), at ((b, h)). So, (AD) is from ((0, 0)) to ((a, 0)), and (BC) is from ((b, h)) to ((c, h)).Given that (CM = 8) and (MD = 9), point (M) divides (CD) in the ratio (CM:MD = 8:9). So, coordinates of (M) are:[M = left( frac{9c + 8 cdot 0}{8 + 9}, frac{9h + 8 cdot 0}{8 + 9} right) = left( frac{9c}{17}, frac{9h}{17} right)]Now, equation of line (BM): points (B(b, h)) and (M(9c/17, 9h/17)).Slope of (BM):[m_{BM} = frac{9h/17 - h}{9c/17 - b} = frac{-8h/17}{(9c - 17b)/17} = frac{-8h}{9c - 17b}]Equation of (BM):[y - h = frac{-8h}{9c - 17b}(x - b)]Now, equation of (AH): perpendicular to (BM), so slope is reciprocal and opposite sign:[m_{AH} = frac{9c - 17b}{8h}]Point (A) is at ((a, 0)), so equation of (AH):[y = frac{9c - 17b}{8h}(x - a)]Intersection point (H) is where (AH) meets (BM). So, we can solve the two equations:1. (y - h = frac{-8h}{9c - 17b}(x - b))2. (y = frac{9c - 17b}{8h}(x - a))Substitute equation 2 into equation 1:[frac{9c - 17b}{8h}(x - a) - h = frac{-8h}{9c - 17b}(x - b)]This seems complicated, but maybe we can find relations between (a), (b), (c), and (h). Since (AD) is parallel to (BC), the sides (AB) and (CD) are the legs. The length of (AD) is (a), and (BC) is 16, so the length of (BC) is (c - b = 16).So, (c = b + 16).Also, since (AD = HD), and (H) is a point in the plane, we can express (HD) in terms of coordinates.Point (H) has coordinates ((x, y)), so:[HD = sqrt{(x - 0)^2 + (y - 0)^2} = sqrt{x^2 + y^2}][AD = a]Given (AD = HD), so:[a = sqrt{x^2 + y^2}][a^2 = x^2 + y^2]But (H) lies on (BM), so we can express (x) and (y) in terms of (b), (c), (h), and (a). This seems too involved. Maybe there's a better approach.Wait, since (AD) is parallel to (BC), and (M) divides (CD) in the ratio 8:9, maybe triangles (BCM) and (KDM) are similar? Wait, where is point (K)? Maybe I misread that.Alternatively, since (AD) is parallel to (BC), the triangles (ABM) and (HDM) might be similar? Not sure.Wait, another idea: Since (AH) is perpendicular to (BM), and (AD = HD), maybe triangle (ADH) is isosceles right triangle? Not necessarily, unless angles are 45 degrees, which we don't know.Alternatively, maybe using coordinate geometry is the way to go, but I need to set up the equations properly.Let me try again with coordinates:Let me place (D) at (0, 0), (C) at (17, 0) because (CD = 17), but wait, (CD) is a leg, not necessarily horizontal. Hmm, maybe I should place (AD) horizontally.Let me set (D) at (0, 0), (A) at (a, 0), (C) at (c, h), and (B) at (b, h). Then, (AD = a), (BC = 16), so (c - b = 16).Point (M) divides (CD) into (CM = 8) and (MD = 9), so coordinates of (M) are:[M = left( frac{9c + 8 cdot 0}{17}, frac{9h + 8 cdot 0}{17} right) = left( frac{9c}{17}, frac{9h}{17} right)]Equation of (BM): from (B(b, h)) to (M(9c/17, 9h/17)).Slope of (BM):[m_{BM} = frac{9h/17 - h}{9c/17 - b} = frac{-8h/17}{(9c - 17b)/17} = frac{-8h}{9c - 17b}]Equation of (BM):[y - h = frac{-8h}{9c - 17b}(x - b)]Equation of (AH): perpendicular to (BM), so slope is reciprocal and opposite sign:[m_{AH} = frac{9c - 17b}{8h}]Point (A) is at (a, 0), so equation of (AH):[y = frac{9c - 17b}{8h}(x - a)]Find intersection (H) by solving:1. (y - h = frac{-8h}{9c - 17b}(x - b))2. (y = frac{9c - 17b}{8h}(x - a))Substitute equation 2 into equation 1:[frac{9c - 17b}{8h}(x - a) - h = frac{-8h}{9c - 17b}(x - b)]Multiply both sides by (8h(9c - 17b)) to eliminate denominators:[(9c - 17b)^2 (x - a) - 8h^2 (9c - 17b) = -64h^2 (x - b)]This is getting too complicated. Maybe I need to find relations between (a), (b), (c), and (h).We know (c = b + 16) because (BC = 16).Also, from the trapezoid, the height (h) can be found using the area or other properties, but we don't have the area.Wait, maybe using vectors. Let me try vectors.Let me denote vectors with position vectors from (D) as the origin.So, (D = (0, 0)), (A = (a, 0)), (B = (b, h)), (C = (c, h)).Point (M) is on (CD), which is from (C(c, h)) to (D(0, 0)), divided in the ratio (CM:MD = 8:9). So, position vector of (M):[M = frac{9C + 8D}{17} = left( frac{9c}{17}, frac{9h}{17} right)]Vector (BM = M - B = left( frac{9c}{17} - b, frac{9h}{17} - h right) = left( frac{9c - 17b}{17}, frac{-8h}{17} right))Vector (AH) is perpendicular to (BM), so their dot product is zero.Let (H = (x, y)). Then vector (AH = H - A = (x - a, y - 0) = (x - a, y)).Dot product (AH cdot BM = 0):[(x - a)left( frac{9c - 17b}{17} right) + y left( frac{-8h}{17} right) = 0]Multiply both sides by 17:[(x - a)(9c - 17b) - 8h y = 0]Also, since (H) lies on (BM), we can express (H) as a point on line (BM). Parametrize (BM):Let (t) be a parameter such that when (t = 0), we are at (B), and (t = 1), we are at (M). So,[H = B + t(M - B) = (b, h) + t left( frac{9c - 17b}{17}, frac{-8h}{17} right)]So,[x = b + t cdot frac{9c - 17b}{17}][y = h + t cdot frac{-8h}{17}]Now, substitute (x) and (y) into the dot product equation:[(x - a)(9c - 17b) - 8h y = 0]Substitute (x) and (y):[left( b + t cdot frac{9c - 17b}{17} - a right)(9c - 17b) - 8h left( h + t cdot frac{-8h}{17} right) = 0]Simplify:First term:[left( (b - a) + t cdot frac{9c - 17b}{17} right)(9c - 17b)][= (b - a)(9c - 17b) + t cdot frac{(9c - 17b)^2}{17}]Second term:[-8h left( h - frac{8h t}{17} right)][= -8h^2 + frac{64h^2 t}{17}]Combine both terms:[(b - a)(9c - 17b) + t cdot frac{(9c - 17b)^2}{17} - 8h^2 + frac{64h^2 t}{17} = 0]Group terms with (t):[t left( frac{(9c - 17b)^2}{17} + frac{64h^2}{17} right) + (b - a)(9c - 17b) - 8h^2 = 0]Solve for (t):[t = frac{8h^2 - (b - a)(9c - 17b)}{frac{(9c - 17b)^2 + 64h^2}{17}}][t = frac{17[8h^2 - (b - a)(9c - 17b)]}{(9c - 17b)^2 + 64h^2}]Now, we also know that (AD = HD). Since (AD = a) and (HD = sqrt{x^2 + y^2}), we have:[a = sqrt{x^2 + y^2}][a^2 = x^2 + y^2]Substitute (x) and (y) from the parametrization:[a^2 = left( b + t cdot frac{9c - 17b}{17} right)^2 + left( h + t cdot frac{-8h}{17} right)^2]This is another equation involving (t), (a), (b), (c), and (h). It's getting very complicated. Maybe I need to find another relation.Wait, since (AD) is parallel to (BC), the sides (AB) and (CD) are the legs. The length of (AB) can be found, but we don't have it. Maybe using the fact that (AD = HD), which is a key condition.Alternatively, maybe using similar triangles. Since (AD) is parallel to (BC), triangles (ABM) and (HDM) might be similar. Let me check.If triangles (ABM) and (HDM) are similar, then:[frac{AB}{HD} = frac{BM}{HM} = frac{AM}{DM}]But I don't know (AB) or (AM). Hmm.Wait, another idea: Since (AH) is perpendicular to (BM), and (AD = HD), maybe reflecting point (A) over (BM) gives point (D). Because reflection over a line preserves distances, so if (AH) is the perpendicular, then (H) is the midpoint of (AD) if (AD = HD). Wait, no, because (AD = HD) doesn't necessarily mean (H) is the midpoint unless (AH) is also equal.Wait, no, reflection would require more conditions. Maybe not the right approach.Alternatively, since (AD = HD), triangle (ADH) is isosceles with (AD = HD). So, the median from (H) to (AD) is also the altitude. But I don't know if that helps.Wait, maybe using coordinate geometry again, but let's assume some values to simplify. Let me assume (h = 1) to make calculations easier, then scale back if needed.But without knowing (h), it's hard to proceed. Maybe I need to find (h) in terms of other variables.Alternatively, using the fact that (AD = HD), and (H) lies on (BM), maybe we can set up equations based on distances.Let me try to write down all knowns:1. (AD = a)2. (BC = 16), so (c - b = 16)3. (CM = 8), (MD = 9), so (M) divides (CD) in 8:94. (AH perp BM)5. (AD = HD)We need to find (a).Let me try to express everything in terms of (a), (b), (c), and (h), then find relations.From the earlier parametrization, we have expressions for (x) and (y) in terms of (t), and (t) in terms of other variables. Then, using (a^2 = x^2 + y^2), we can get an equation.But this seems too involved. Maybe I need to find another approach.Wait, another idea: Since (AD) is parallel to (BC), the triangles (ADM) and (BCM) might be similar. Let me check.Triangles (ADM) and (BCM):- (AD) is parallel to (BC), so corresponding angles are equal.- Therefore, triangles (ADM) and (BCM) are similar.Yes, that seems correct.So, similarity ratio:[frac{AD}{BC} = frac{AM}{BM} = frac{DM}{CM}]Given (DM = 9), (CM = 8), so:[frac{AD}{16} = frac{9}{8}][AD = 16 cdot frac{9}{8} = 18]Wait, that seems too straightforward. Is that correct?Let me verify. If triangles (ADM) and (BCM) are similar, then the ratio of corresponding sides is equal. Since (AD) corresponds to (BC), and (DM) corresponds to (CM), the ratio is (AD/BC = DM/CM).Given (DM = 9), (CM = 8), so (AD/16 = 9/8), hence (AD = (9/8)*16 = 18).Yes, that makes sense. So, the length of (AD) is 18.I think that's the answer. The key was recognizing the similar triangles (ADM) and (BCM) because of the parallel sides, leading to the ratio which gives (AD = 18).

💡Okay, so I need to prove Euler's theorem for different cases of ( k ). Euler's theorem states that if ( a ) and ( k ) are coprime, then ( a^{varphi(k)} equiv 1 mod k ), where ( varphi(k) ) is Euler's totient function. I remember that ( varphi(k) ) counts the number of integers less than ( k ) that are coprime to ( k ).Let me start with part (a), where ( k = p ) and ( p ) is a prime number. I know that for a prime ( p ), ( varphi(p) = p - 1 ) because all numbers less than ( p ) are coprime to ( p ). So, I need to show that ( a^{p-1} equiv 1 mod p ). Wait, isn't that exactly Fermat's Little Theorem? Yes, Fermat's Little Theorem states that if ( p ) is a prime and ( a ) is not divisible by ( p ), then ( a^{p-1} equiv 1 mod p ). Since ( a ) and ( p ) are coprime, ( a ) is not divisible by ( p ), so Fermat's Little Theorem applies directly here. Therefore, ( a^{varphi(p)} - 1 ) is divisible by ( p ). That seems straightforward.Moving on to part (b), where ( k = pq ) and ( p ) and ( q ) are prime numbers. I recall that ( varphi(pq) = varphi(p)varphi(q) = (p-1)(q-1) ) because ( p ) and ( q ) are distinct primes. So, I need to show that ( a^{(p-1)(q-1)} equiv 1 mod pq ). Since ( a ) is coprime to both ( p ) and ( q ), by Fermat's Little Theorem, ( a^{p-1} equiv 1 mod p ) and ( a^{q-1} equiv 1 mod q ). Now, I need to combine these two congruences. Maybe I can use the Chinese Remainder Theorem here. If ( a^{(p-1)(q-1)} equiv 1 mod p ) and ( a^{(p-1)(q-1)} equiv 1 mod q ), then by the Chinese Remainder Theorem, ( a^{(p-1)(q-1)} equiv 1 mod pq ). That makes sense. So, ( a^{varphi(pq)} - 1 ) is divisible by ( pq ).For part (c), ( k = p^2 ) where ( p ) is a prime number. I know that ( varphi(p^2) = p^2 - p = p(p - 1) ). So, I need to show that ( a^{p(p - 1)} equiv 1 mod p^2 ). Since ( a ) is coprime to ( p^2 ), ( a ) is not divisible by ( p ). From Fermat's Little Theorem, ( a^{p - 1} equiv 1 mod p ). But I need to lift this to modulo ( p^2 ). Maybe I can use the concept of lifting the exponent or consider expanding ( a^{p(p - 1)} ) using the binomial theorem. Let me think. If ( a^{p - 1} = 1 + kp ) for some integer ( k ), then ( a^{p(p - 1)} = (1 + kp)^p ). Expanding this using the binomial theorem:[(1 + kp)^p = 1 + binom{p}{1}kp + binom{p}{2}(kp)^2 + dots + (kp)^p]All terms beyond the second term have factors of ( p^2 ), so modulo ( p^2 ), this simplifies to:[1 + pkp equiv 1 + k p^2 equiv 1 mod p^2]Therefore, ( a^{p(p - 1)} equiv 1 mod p^2 ). Hence, ( a^{varphi(p^2)} - 1 ) is divisible by ( p^2 ).Finally, part (d), where ( k = p^lambda ) and ( p ) is a prime number, ( lambda geq 1 ). I know that ( varphi(p^lambda) = p^lambda - p^{lambda - 1} = p^{lambda - 1}(p - 1) ). So, I need to show that ( a^{p^{lambda - 1}(p - 1)} equiv 1 mod p^lambda ).This seems like a generalization of part (c). Perhaps I can use induction on ( lambda ). For ( lambda = 1 ), it's just Fermat's Little Theorem. For ( lambda = 2 ), it's part (c). Assume it's true for ( lambda = n ), that is, ( a^{p^{n - 1}(p - 1)} equiv 1 mod p^n ). Now, for ( lambda = n + 1 ), I need to show ( a^{p^{n}(p - 1)} equiv 1 mod p^{n + 1} ).Using the result from ( lambda = n ), ( a^{p^{n - 1}(p - 1)} = 1 + kp^n ) for some integer ( k ). Then, raising both sides to the power ( p ):[(a^{p^{n - 1}(p - 1)})^p = (1 + kp^n)^p]Expanding the right-hand side using the binomial theorem:[1 + binom{p}{1}kp^n + binom{p}{2}(kp^n)^2 + dots + (kp^n)^p]All terms beyond the second term have factors of ( p^{2n} ), which is at least ( p^{n + 1} ) since ( n geq 1 ). Therefore, modulo ( p^{n + 1} ), this simplifies to:[1 + pkp^n equiv 1 + k p^{n + 1} equiv 1 mod p^{n + 1}]Thus, ( a^{p^{n}(p - 1)} equiv 1 mod p^{n + 1} ), completing the induction step. Therefore, by induction, Euler's theorem holds for ( k = p^lambda ).I think I've covered all the cases by either directly applying known theorems or using induction and the binomial theorem to lift the exponent. Each step seems logical, and I don't see any gaps in the reasoning.

💡Okay, so I need to figure out how much water Liam has left after using some for his science experiment. He starts with 3 gallons and uses 5/4 gallons. Hmm, let me think about this step by step.First, I know that 3 gallons is a whole number, and 5/4 gallons is a fraction. To subtract these, it might be easier if both numbers are in the same form. Maybe I can convert the whole number 3 into a fraction with the same denominator as 5/4, which is 4.So, if I have 3 gallons, and I want to express that as a fraction with denominator 4, I can think of it as 3 times 4 over 4, right? That would be 12/4. Okay, so 3 gallons is the same as 12/4 gallons.Now, I need to subtract the amount he used, which is 5/4 gallons, from the total he had, which is now 12/4 gallons. So, subtracting these two fractions: 12/4 minus 5/4. Since the denominators are the same, I can just subtract the numerators.12 minus 5 is 7, so that gives me 7/4. Therefore, Liam has 7/4 gallons of water left. Wait, 7/4 is an improper fraction. If I wanted to express that as a mixed number, it would be 1 and 3/4 gallons. But the question asks for the fraction of a gallon, so 7/4 is the appropriate answer.Let me double-check my steps to make sure I didn't make a mistake. Starting with 3 gallons, converting to 12/4 gallons, subtracting 5/4 gallons, which gives 7/4 gallons. Yep, that seems correct. So, I'm confident that 7/4 is the right answer.

💡First, I need to determine the speed of the train. When the train crosses a signal pole, it covers its own length of 300 meters in 18 seconds. So, the speed of the train is:[text{Speed} = frac{text{Distance}}{text{Time}} = frac{300 text{ meters}}{18 text{ seconds}} = 16.67 text{ meters/second}]Next, when the train crosses the platform, it covers the combined length of the train and the platform in 45 seconds. Let ( L ) be the length of the platform. The total distance covered is:[text{Total Distance} = 300 text{ meters} + L]Using the speed calculated earlier:[300 + L = 16.67 times 45][300 + L = 750][L = 750 - 300][L = 450 text{ meters}]Therefore, the length of the platform is 450 meters.

💡Okay, so I have this problem here about sets and complex numbers. Let me try to figure it out step by step. The problem says:Given the sets ( M = {1, z(1+i)} ), where ( i ) is the imaginary unit, and ( N = {3,4} ). If ( M cup N = {1,2,3,4} ), then the point corresponding to the complex number ( z ) on the complex plane is located in which quadrant? The options are A: first quadrant, B: second quadrant, C: third quadrant, D: fourth quadrant.Alright, so let's break this down. First, ( M ) is a set containing two elements: 1 and ( z(1+i) ). ( N ) is a set containing 3 and 4. The union of ( M ) and ( N ) is given as ( {1,2,3,4} ). That means when we combine all the elements of ( M ) and ( N ), we get the numbers 1, 2, 3, and 4.So, ( M cup N = {1,2,3,4} ). Since ( M ) already has 1 and ( z(1+i) ), and ( N ) has 3 and 4, the union must include all these elements. Therefore, ( z(1+i) ) must be 2 because that's the only number missing from ( M ) to make the union complete as ( {1,2,3,4} ).So, ( z(1+i) = 2 ). Now, we need to solve for ( z ). To do that, I can divide both sides of the equation by ( (1+i) ). So,( z = frac{2}{1+i} ).But dividing by a complex number isn't straightforward, so I remember that to simplify expressions like this, I can multiply the numerator and the denominator by the complex conjugate of the denominator. The complex conjugate of ( 1+i ) is ( 1-i ). So,( z = frac{2}{1+i} times frac{1-i}{1-i} ).Multiplying the numerators: ( 2 times (1 - i) = 2 - 2i ).Multiplying the denominators: ( (1 + i)(1 - i) ). I remember that ( (a + b)(a - b) = a^2 - b^2 ), so here it would be ( 1^2 - i^2 ). Since ( i^2 = -1 ), this becomes ( 1 - (-1) = 1 + 1 = 2 ).So now, the expression simplifies to:( z = frac{2 - 2i}{2} ).Dividing both terms in the numerator by 2:( z = 1 - i ).Okay, so ( z = 1 - i ). Now, I need to figure out where this complex number is located on the complex plane. In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis. So, a complex number ( a + bi ) is plotted at the point ( (a, b) ).In this case, ( z = 1 - i ) can be written as ( 1 + (-1)i ), so the real part is 1 and the imaginary part is -1. Therefore, the point is at ( (1, -1) ).Now, let's recall the quadrants on the coordinate plane:- First quadrant: Both x and y are positive.- Second quadrant: x is negative, y is positive.- Third quadrant: Both x and y are negative.- Fourth quadrant: x is positive, y is negative.Looking at the point ( (1, -1) ), the x-coordinate is positive, and the y-coordinate is negative. That places it in the fourth quadrant.So, the correct answer should be D: the fourth quadrant.Wait, let me just double-check my steps to make sure I didn't make a mistake.1. I started with ( M cup N = {1,2,3,4} ).2. Since ( M = {1, z(1+i)} ) and ( N = {3,4} ), the union must include all elements from both sets.3. The only missing element from ( M ) is 2, so ( z(1+i) = 2 ).4. Solved for ( z ) by dividing both sides by ( 1+i ), which required multiplying numerator and denominator by the conjugate ( 1-i ).5. Simplified to get ( z = 1 - i ).6. Plotted ( z ) on the complex plane at ( (1, -1) ), which is in the fourth quadrant.Everything seems to check out. I don't think I made any mistakes here. So, yeah, the answer should be D.**Final Answer**boxed{D}