Complex numbers practice · Mixed difficulty

Option C ($\sin^{-1}(\frac{1}{\sqrt{3}})$) is correct because when the given expression is purely imaginary, its real part must be zero, leading to the equation $\sin^2 \theta = \frac{1}{3}$, which gives $\theta = \sin^{-1}(\frac{1}{\sqrt{3}})$.

Option A (a straight line) is correct because the given condition implies that the distance from z to the origin is equal to the distance from z to the point i/3, which defines a straight line.

Option A (0) is correct because the real part of $w$ is 0, which is derived using the property that $|z| = 1$ and $w = \frac{{z - 1}}{{z + 1}}$.

Option C ($+ 1\;\mathrm{or}\; - 1$) is correct because by using the given condition and trigonometric identities, we can show that $xyz$ can be either +1 or -1.

Option B (1) is correct because when the given ratios are equal to k, taking the logarithm of xyz results in 0, which means xyz equals 1.

Option C ($1+i \cot (\pi / 2 n)$) is correct because it is derived by simplifying the geometric series and using trigonometric identities.

Option B (3) is correct because simplifying the given equation using the properties of cube roots of unity leads to ω^n = 1, which is satisfied when n is a multiple of 3, and the least positive integral value is 3.

Option C (From an equilateral triangle) is correct because the distance between any two cube roots of unity is the same, forming an equilateral triangle.

Option A ($X$-axis) is correct because the condition |z-ai| = |z+ai| implies that z is equidistant from ai and -ai, which means z lies on the x-axis (y=0).

Option B (3) is correct because it is the radius of the circle represented by the given equation, calculated after converting the equation to the standard form of a circle.

Option D ($-1$) is correct because the expression simplifies to $-1$ using the properties of complex cube roots of unity.

Option D ($pqs = {s^2} + {q^2}r$) is correct because substituting a real root into the equation and separating real and imaginary parts leads to this relationship.

Option A ($\frac{{1 - x}}{2}$) is correct because performing partial fraction decomposition and solving for f(x) yields this result.

Option A ($2{\left( { - 1} \right)^n},$ when $n$ is a multiple of 3) is correct because solving the given equation yields $z = -\omega$ or $z = -\omega^2$, and substituting these values into $z^n + z^{-n}$ gives the stated result when n is a multiple of 3.

Option A (0) is correct because when ω is substituted into the function, the terms simplify to 1 + ω + ω^2, which equals 0.

The correct answer is D (${x^2} + x + 1 = 0$) because ${\alpha ^{19}}$ and ${\beta ^7}$ simplify to $\omega$ and ${\omega ^2}$ respectively, which are the roots of the given equation.

Option B ($\{0,3\}$) is correct because the function f(n) evaluates to either 0 or 3 depending on whether n is a multiple of 3 or not.

Option D (Less than one) is correct because the absolute value of the sum is less than the sum of the absolute values, and since each |a_k| < 1 and the sum of λ_k is 1, the result is less than 1.

None of these is the correct answer. The correct answer is $\sqrt{\frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}}$ because by taking the modulus squared of both sides of the equation, we can find ${x^2} + {y^2}$.

Option D ($2\,\cos \,n\alpha$) is correct because it is derived by using De Moivre's theorem and simplifying the expression.

Option B ($\theta$) is correct because using the property $|z|^2 = z \bar{z} = 1$, we can simplify the expression and find that the argument of the simplified expression is equal to $\theta$.

Option A (a straight line) is correct because the equation simplifies to x=0, which represents a straight line on the complex plane.

Option D (${z_1} + {z_2} + {z_3}$) is correct because in a triangle with the circumcenter at the origin, the orthocenter is the sum of the vertices' complex numbers.

Option D (3) is correct because by substituting y=0 into the given equation and solving for x, we find that the modulus of z is 3.

Option B ($\frac{4}{3}$) is correct because solving the equation ${z^2} + 3\overline z = 0$ gives 4 solutions, and the sum of the infinite geometric series with the common ratio $\frac{1}{4}$ is $\frac{4}{3}$.

Option B ($-\bar{\omega}$) is the correct answer because given the conditions $|z|=|w|$ and $\operatorname{Arg} z + \operatorname{Arg} w = \pi$, z can be expressed as $-\bar{\omega}$.

Option B (${z_2} = 2,{z_3} = 1 - i\sqrt 3$) is the correct answer because it accurately calculates the other two vertices of the equilateral triangle by rotating the given vertex by 2π/3 and 4π/3.

Option C (A pair of straight lines) is correct because the equation can be factored into two linear equations representing straight lines.

Option C ($-\frac{8}{25}-\frac{31}{25} i$) is correct because simplifying the complex fraction and then finding its multiplicative inverse yields this result.

Option D (None of the above) is correct because the given complex numbers are collinear, and therefore cannot form a triangle.