Complex numbers practice · Mixed difficulty

Option C is correct because A Δ B represents the symmetric difference, which includes elements in B but not in A.

Option D (12) is correct because the sum can be simplified using the properties of complex roots of unity.

Option D (1) is correct because by using the change of base formula for logarithms and simplifying the sum, the result is 1.

Option D ($\frac{3}{4}$) is correct because if $a = \frac{3}{4}$, then z would be real, contradicting the condition that the imaginary part of z is non-zero.

Option B (9) is correct because, after simplifying using the properties of cube roots of unity, the expression evaluates to 9.

Option D ($i$) is correct because the sum is simplified using De Moivre's theorem and the properties of geometric progressions.

Option C (1) is correct because the expression simplifies to 1 using the property that if $|z|=1$, then $\bar{z} = \frac{1}{z}$.

Option A (0) is correct because the given equations lead to an impossible condition, indicating no solution for z.

The correct answer is D (${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)\,$) because the real part of the complex number must be zero for it to be purely imaginary.

Option A ($x = 4n$, where $n$ is any positive integer) is correct because it represents the general solution for x that satisfies the equation, derived by simplifying the complex fraction and using the properties of imaginary units.

Option C ($-\frac{1}{2}(\sqrt{3}-i)$) is correct because after simplifying the given expression using trigonometric identities and De Moivre's theorem, the result matches this option.

Option B ($\cos \frac{\pi }{8} + i\sin \frac{\pi }{8}$) is correct because it is derived from solving the equation $i{z^4} + 1 = 0$ using De-Moivre’s theorem.

Option A ($\frac{1}{2}$) is correct because when the degree of the numerator and denominator are the same, we perform polynomial long division to find the constant term $A$.

Option C (25) is correct because it uses the logarithmic property to simplify the expression to 5 squared, which equals 25.

Option D ($1 / 2-\sqrt{3}$) is the correct answer. By equating the given complex numbers and solving for 'a', we arrive at the solution after rationalizing the denominator.

Option A (0) is correct because the given condition implies that z_1/z_2 is a complex cube root of unity, which leads to the equation z_1² + z_2² - z_1z_2 = 0.

Option D ($4z_1+z_2+z_3=0$) is correct because the midpoint of AG being the origin leads to the given relationship between the complex numbers.

Option B ($z$ is purely real) is correct because the conjugate of the determinant is equal to the determinant itself, implying that $z$ is a real number.

Option B is correct because by cubing the given expression and equating real and imaginary parts, we can find the value of k.

Option B (0) is correct because substituting $\alpha = \omega$ and using the properties of cube roots of unity ($\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$) simplifies the expression to 0.

Option B (-z) is correct because simplifying the determinant using properties of $\omega$ leads to $k = -z$.

Option C (-1) is correct. The infinite product of complex numbers in polar form simplifies to cos(π) + isin(π), which equals -1.

Option A (1) is correct because the product simplifies to 1 using the properties of complex cube roots of unity.

The correct answer is C (${m_3} < {m_2} < {m_4} < {m_1}$) because calculating the moduli and comparing them yields this order.

Option D ($- 2$) is correct because after substituting $z = x - iy$ and ${z^{1/3}} = p + iq$ and simplifying, the expression $\left( {\frac{x}{p} + \frac{y}{q}} \right)/\left( {{p^2} + {q^2}} \right)$ evaluates to -2.

Option B ($\frac{2 \pi}{3}$ or $2 \pi$ or $\frac{10 \pi}{3}$) is correct because it represents the possible values of the argument of $z^{2/3}$ when $z=-1$.

Option A ($1 - i, - 1 + i$) is correct because these are the vertices obtained by rotating the given vertex by 90 and 270 degrees around the origin.

Option A (Circle) is correct because if $|z|=1$, then $2/z$ also lies on a circle, specifically a circle with radius 2.

Option B (Less than 1) is correct because the magnitude of the weighted sum is less than 1, given the constraints on the magnitudes of a_i and the non-negative weights lambda_i.

Option C (a hyperbola) is correct because the equation $z^2 + \bar{z}^2 = 2$ simplifies to $x^2 - y^2 = 1$, which represents a hyperbola.