Complex numbers practice · Mixed difficulty

Option B ($2^{2n}$) is correct because the expression simplifies to $2^{2n}$ using the properties of cube roots of unity.

Option A (0) is correct because if the magnitudes are equal and the argument of their ratio is π, the complex numbers are additive inverses.

Option D (None of these) is correct. The square root of 6 + 4√3 cannot be expressed in the form √x + √y where x and y are rational numbers, as the discriminant becomes negative.

Option B (${\left| \alpha \right|^2} - ad > 0\\,\,\\mathrm{and}\\,\,a \in R - \\{ 0\\}$) is correct because it represents the condition for the given equation to represent a circle.

Option B (1) is correct because after manipulating the given equation and substituting into the expression, the result simplifies to 1.

Option C ($\left| {z + \bar z} \right| = 2$) is correct because the given condition implies that the real part of z is either 1 or -1, hence the magnitude of $z + \bar z$ is 2.

Option B ($- {110^0}$) is correct because it is the principal amplitude of the given complex number after simplification using trigonometric identities and properties of complex numbers.

Option A (1) is correct because the expression is simplified using the properties of cube roots of unity.

Option C ($\frac{{1}}{{3}}$) is correct because all roots of the equation lie outside the circle with radius $\frac{{2}}{{3}}$, which means they lie outside the circle with radius $\frac{{1}}{{3}}$ as well.

Option C (3) is correct because by manipulating the given equation and using the properties of complex conjugates, we can isolate and solve for the desired expression.

Option C ($| |{{z_1}} | - |{z_2}|| |$) is correct because when the arguments of two complex numbers are equal, the magnitude of their difference is the absolute difference of their magnitudes.

Option D ($\left(2 \sec \frac{3 \pi}{8}, \frac{3 \pi}{8}\right)$) is correct because it correctly identifies the modulus and argument of the complex number after simplification.

Option A ($x$-axis) is correct because the given condition implies that the distance from z to 2i is equal to the distance from z to -2i, which means z lies on the perpendicular bisector of the line segment joining 2i and -2i, which is the x-axis.

Options B and C are correct because they represent the possible scenarios where the product of the imaginary parts of the two complex numbers is negative, satisfying the given conditions.

Option C (4) is correct because by cubing $Z^{\frac{{1}}{{3}}}=a-ib$ and equating real and imaginary parts, we find that $\frac{{x}}{{a}}-\frac{{y}}{{b}} = 4(a^2 - b^2)$.

Option C (-256) is correct because it is the value of the expression ${\alpha ^{15}} + {\beta ^{15}}$ where $\alpha$ and $\beta$ are the roots of the given quadratic equation.

The polar form of the complex number $1+\sqrt{3} i$ is $2\left(\cos \frac{{\pi}}{{3}}+i \sin \frac{{\pi}}{{3}}\right)$ because the modulus is 2 and the argument is $\frac{{\pi}}{{3}}$.

Option A (5) is correct because simplifying the fraction z1/z2 results in (1/5)(9 + 12i), so the value of 'a' is 5.

Option A ($\pi$) is correct because when $\arg(z) < 0$, adding $\pi$ to the argument of $z$ gives the argument of $-z$, and the difference is $\pi$.

Option C (a straight line) is correct because the given condition translates to the perpendicular bisector of the line segment joining 0 and 5 in the complex plane.

Option B (2) is correct because the minimum value of |z1-z2| is found using the triangle inequality and given conditions.

Option C (1) is correct because using the given conditions |a|=|b|=|c|=1 and abc=a+b+c, we can manipulate the equations to show that bc+ca+ab = 1.

Option B (-1) is correct because the value of the given expression simplifies to $\omega + \omega^2$, which equals -1.

Option D (None of these) is correct because the correct formula leads to $\frac{{n(2^{n-1})}}{{2^n - 1}}$, which is not among the options.

Option C ($n$) is correct because the given expression is equal to the derivative of $z^n - 1 = 0$ evaluated at z=1, which results in n.

Option C (2$\cos(m\theta + n\phi)$) is correct because by using Euler's formula and simplifying the expression, we arrive at 2$\cos(m\theta + n\phi)$.

Option A (-1) is correct because the expression simplifies to $\omega + \omega^2$, which equals -1 due to the properties of complex roots of unity.

Option C (form an equilateral triangle) is correct because the given condition leads to $|z_1| = |z_2|$ and the angle between $z_1$ and $z_2$ being 60 degrees, thus forming an equilateral triangle with the origin.

Option A (0) is correct because simplifying the expression i⁹ + i¹⁹ results in 0 + 0i, so the value of 'a' is 0.

Option A ($|z|=4$) is correct because the equation represents a circle in the complex plane, and solving for $|z|$ gives the radius.