If z = (1 + i)/(1 - i), then z^4 equals
- 1 Correct answer.
- -1
- -i
- i
Explanation
Multiply numerator and denominator by the conjugate 1 + i: z = (1 + i)^2/((1 - i)(1 + i)) = 2i/2 = i. Hence z^4 = i^4 = (i^2)^2 = (-1)^2 = 1.
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Multiply numerator and denominator by the conjugate 1 + i: z = (1 + i)^2/((1 - i)(1 + i)) = 2i/2 = i. Hence z^4 = i^4 = (i^2)^2 = (-1)^2 = 1.
A decagon has 10 vertices, so the number of line segments joining pairs of vertices is C(10,2) = 45. Of these, 10 are sides of the polygon, leaving 45 - 10 = 35 diagonals, which agrees with n(n - 3)/2 for n = 10.
Write sin(3x)/(5x) as (3/5) · sin(3x)/(3x). Since sin(t)/t → 1 as t → 0, the limit is (3/5) · 1 = 3/5.
For any n × n matrix A, the identity det(adj A) = (det A)^(n-1) holds. Here n = 3 and det A = 5, so det(adj A) = 5^2 = 25.
The first card is a king with probability 4/52 = 1/13, and then the second is a king with probability 3/51 = 1/17. Since draws are without replacement, multiply: (1/13)(1/17) = 1/221.
Let I be the integral and replace x by π/2 - x. This gives I = ∫ cos x/(sin x + cos x) dx over the same limits, so 2I = ∫ 1 dx = π/2 and therefore I = π/4.
Expand the squared magnitude: |a + b|^2 = |a|^2 + 2a·b + |b|^2 = 9 + 0 + 16 = 25. Taking the positive square root gives |a + b| = 5.
By Vieta's formulas, α + β = 6 and αβ = 4. Then α^2 + β^2 = (α + β)^2 - 2αβ = 36 - 8 = 28.
The k-th term is 1/((2k - 1)(2k + 1)) = (1/2)[1/(2k - 1) - 1/(2k + 1)]. Summing telescopes to (1/2)[1 - 1/(2n + 1)] = (1/2)(2n/(2n + 1)) = n/(2n + 1).
For the line ax + by + c = 0, the reflection of (x1, y1) is (x1 - 2aD/(a^2 + b^2), y1 - 2bD/(a^2 + b^2)) where D = ax1 + by1 + c. Here D = 1 - 6 + 4 = -1 and a^2 + b^2 = 10, giving x = 1 + 2/10 = 6/5 and y = 2 - 6/10 = 7/5.
For z = x + iy with x > 0, the principal argument is arctan(y/x). Here x = 1 and y = √3, so arctan(√3) = π/3. The point lies in the first quadrant, so no quadrant adjustment is needed.
Multiplying the matrix by itself gives A^2 = [[0·0 + 1·(-1), 0·1 + 1·0], [-1·0 + 0·(-1), -1·1 + 0·0]] = [[-1, 0], [0, -1]] = -I. This matrix represents a 90° rotation, whose square is a 180° rotation.
On a fair die the possible prime outcomes are 2, 3 and 5, giving three favourable outcomes among six equally likely faces. Therefore the probability is 3/6 = 1/2. The number 1 is not prime.
Using the standard expansion cos x = 1 - x^2/2 + O(x^4), the numerator becomes x^2/2 plus higher-order terms. Dividing by x^2 and taking the limit gives 1/2. Equivalently, write 1 - cos x = 2 sin^2(x/2) and use lim(t→0) sin t/t = 1.
For parallel lines ax + by = c1 and ax + by = c2, the distance is |c1 - c2|/√(a² + b²). Here |15 - 5|/√(3² + 4²) = 10/5 = 2. The lines are distinct because their constant terms differ.
The scalar triple product is the determinant of the component rows: |1 1 0; 0 1 1; 1 0 1|. Expanding gives 1(1·1 - 1·0) - 1(0·1 - 1·1) + 0 = 1 + 1 = 2. It also equals the volume of the parallelepiped spanned by the three vectors.
The general term is C(10, r) x^(10-r) (1/x)^r = C(10, r) x^(10-2r). For the constant term the exponent must be zero, so 10 - 2r = 0 and r = 5. The coefficient is C(10, 5) = 252.
The word CUSAT has five distinct letters, with vowels U and A. Treat UA as one block, so the objects are the block, C, S and T: 4! arrangements. The vowels inside the block can be ordered in 2! ways, giving 4! × 2! = 48.
Use 1 + ω + ω² = 0, so 1 + ω = -ω² and 1 + ω² = -ω. Then 1 + ω - ω² = -2ω², whose cube is -8ω^6 = -8, and 1 - ω + ω² = -2ω, whose cube is -8ω³ = -8. The sum is -16.
Let I be the integral and substitute x → π - x. Adding the two expressions gives 2I = π ∫ from 0 to π of sin x/(1 + cos²x) dx. With u = cos x, this becomes π ∫ from -1 to 1 du/(1 + u²) = π(π/4 + π/4) = π²/2. Hence I = π²/4.
In the binomial expansion of (1 + x)^n, the coefficient of x^r is C(n, r). Here n = 7 and r = 3, so the coefficient is C(7, 3) = 35. The other listed values are binomial coefficients for different pairs of n and r.
The sum of the first n terms of an arithmetic progression is S_n = n/2[2a + (n - 1)d]. Substituting a = 1, n = 20 and S_20 = 400 gives 10(2 + 19d) = 400, so 2 + 19d = 40 and d = 2.
Using the double-angle identity sin 2θ = 2 sin θ cos θ, we get sin 15° cos 15° = (1/2) sin 30°. Since sin 30° = 1/2, the product equals 1/4.
Choosing and arranging 3 books from 5 means taking ordered selections, so the count is P(5, 3) = 5 × 4 × 3 = 60. The value C(5, 3) = 10 would count only the choices, without arranging the selected books.
Completing the square gives (x - 3)² + (y + 4)² = 9 + 16 + 11 = 36. This is a circle with centre (3, -4) and radius √36 = 6.
Factor x² - 16 = (x - 4)(x + 4) and rationalise √x - 2 = (x - 4)/(√x + 2). The quotient becomes (x + 4)(√x + 2), which tends to 8 × 4 = 32 as x tends to 4.
The total number of ways to draw 2 balls from 7 is C(7, 2) = 21. The number of ways to draw 2 red balls from 3 is C(3, 2) = 3, so the probability is 3/21 = 1/7.
Write x^x = e^(x ln x). Differentiating gives f'(x) = x^x(ln x + 1). At x = 1 this equals 1 × (0 + 1) = 1.
By the Cayley-Hamilton theorem, A satisfies A² - (trace A)A + (det A)I = 0. Hence A² - 5A + 3I is the zero matrix, whose determinant is 0.
Let w = (z - 1)/(z + 1). Since |z| = 1, we have conjugate(z) = 1/z. Taking conjugates gives conjugate(w) = (1/z - 1)/(1/z + 1) = (1 - z)/(1 + z) = -w, so the real part of w is zero. Therefore w lies on the imaginary axis.
A set with n elements has 2^n subsets in total. Proper subsets exclude the set itself, so the count is 2^6 − 1 = 63.
Parallel lines have proportional x- and y-coefficients, so the required line has the form 4x − 3y + k = 0. Substituting (2, 3) gives 8 − 9 + k = 0, hence k = 1.
Using sin A − sin B = 2 cos((A+B)/2) sin((A−B)/2), we get 2 cos 45° sin 30° = 2 × (√2/2) × (1/2) = √2/2.
Substitute u = x², so du = 2x dx. The integral becomes (1/2)∫ from 0 to 1 of e^u du = (1/2)(e − 1).
There are 4 kings and 13 hearts, but the king of hearts is counted twice. Thus the favourable count is 4 + 13 − 1 = 16, and the probability is 16/52 = 4/13.
The distance from (x₁, y₁, z₁) to ax + by + cz + d = 0 is |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²). Here the numerator is |2 − 2 + 6 − 5| = 1 and the denominator is √(4 + 1 + 4) = 3.
The modulus of a quotient is the quotient of the moduli, so |z| = |3 + 4i| / |1 + i|. This gives √(9 + 16) / √(1 + 1) = 5/√2.
The order is the highest derivative present, which is d²y/dx², so the order is 2. The degree is the exponent of that highest derivative when the equation is polynomial in derivatives, giving degree 3.
Using inclusion-exclusion, the number of onto functions from 5 balls to 3 boxes is 3^5 − C(3,1)2^5 + C(3,2)1^5 = 243 − 96 + 3 = 150.
Taking the squared magnitude of a + b + c = 0 gives 0 = |a|² + |b|² + |c|² + 2(a·b + b·c + c·a). Since each vector is a unit vector, 0 = 3 + 2S, so S = −3/2.
Substituting x = 2 into the equation gives 4 − 14 + k = 0, so k = 10. Equivalently, the sum of the roots is 7, so the other root is 5 and the product of the roots is 2 × 5 = 10.
The term containing x² comes from choosing r = 2 in the general binomial term C(4,r)(2x)^r(−3)^(4−r). This gives C(4,2)(2x)²(−3)² = 6 × 4x² × 9 = 216x². Hence the required coefficient is 216.
The expression has the form sin A cos B + cos A sin B, which equals sin(A + B). Here A = 30° and B = 60°, so the value is sin 90° = 1.
The first term is a = 4 and the common ratio is r = 1/2, with |r| < 1, so the series converges. Its sum is a/(1 − r) = 4/(1 − 1/2) = 8.
For independent events, P(A ∩ B) = P(A)P(B) = 0.4 × 0.5 = 0.2. Using the addition rule, P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7.
Dividing by 144 gives x²/16 + y²/9 = 1, so a² = 16 and b² = 9. Thus c² = a² − b² = 7, and the eccentricity is e = c/a = √7/4.
The scalar projection is (a · b)/|b|. Here a · b = 2 + 3 + 1 = 6 and |b| = √3, so the projection is 6/√3 = 2√3.
The word CUSAT has five distinct letters, with vowels U and A. Treat U and A as one block, giving four units: the block, C, S, and T. These can be arranged in 4! ways, and the two vowels can be arranged inside the block in 2! ways, giving 24 × 2 = 48.
Integrating by parts twice gives an antiderivative −e^(−x)(x² + 2x + 2). Evaluating from 0 to 1 gives −5/e − (−2) = 2 − 5/e.
Since 1 + ω + ω² = 0, we have 1 + ω = −ω². Therefore 1 + ω − ω² = −2ω², whose cube is (−2)³(ω²)³ = −8ω⁶ = −8, because ω³ = 1.