Complex numbers practice · Mixed difficulty

Option B ($z=\frac{{1}}{{1+i x}}+i(\frac{{1}}{{1-i x}})$) is correct because it satisfies the condition that $f(x)$ and $g(x)$ are real-valued functions in the range (0,1) for $x \in (0,1)$.

Option C (-2) is correct because the expression simplifies to -2.

Option C (5) is correct because it represents the modulus of the complex number $z$ that satisfies the given equation, found by algebraic manipulation and simplification.

Option B (48) is correct because by substituting $z = x + iy$ into the given conditions, we find the region in the complex plane and maximize $|z+1|^2$.

Option C (1) is correct because the magnitude of the given expression simplifies to 1 when $\frac{{5{z^2}}}{{11{z_1}}}$ is purely imaginary.

Option D ($\frac{1}{4} + \frac{9}{4}i$) is correct because it is the result of simplifying the given complex number expression through algebraic manipulation.

Option A is correct because if $|z+1| < |z-2|$, then $w=3z+2+i$ satisfies $|w+1| < |w-8|$.

Option D is correct because all three statements are correct. They define the conjugate, modulus, and multiplicative inverse of complex numbers.

Option C (${\alpha _2},{\alpha _1},{\alpha _3}$) is correct because it represents the increasing order of the moduli of the given complex numbers.

Option C (Interior and boundary of an ellipse) is correct because the given inequality represents the locus of a point such that the sum of its distances from two fixed points is less than or equal to a constant, which defines the interior and boundary of an ellipse.

Option C ($2 - \sqrt 2$) is correct because it is the simplified form of the given expression after rationalizing the nested square roots.

Option C ($e^{i a / 2}$) is correct because the product of complex numbers in polar form simplifies to $e^{i \alpha / 2}$ after applying trigonometric identities and limit properties.

Option D (-1) is correct because the product of the complex numbers simplifies to cos(π) + i sin(π), which equals -1.

Option D (-n) is correct because the summation simplifies to -n using trigonometric identities and series summation formulas.

Option D (y+ix) is correct because by substituting z = y + ix and using the given condition x² + y² = 1, the expression simplifies to y + ix.

Option C (15) is correct because the centroid of an equilateral triangle coincides with the circumcenter, and thus the sum of the vertices is three times the circumcenter, leading to |z1 + z2 + z3| = 15.

Option B (0) is correct because the product of the roots of the given function, as n approaches infinity, is 0.

Option C ($a: 10$) is correct because it correctly relates the radius of the locus of $5z-1$ to the ratio of $|z_1|$ and $|z_2|$.

Option D (A Circle) is correct because the imaginary part condition leads to an equation of the form $16{x^2} + 16{y^2} + 16x + 3y + 4 = 0$, which represents a circle.

Option C (A circle) is correct because the given equation can be rearranged into the general form of a circle equation in the complex plane.

Option B (A circle) is correct because the given condition leads to the equation of a circle when $z$ is expressed as $x+iy$.

Option B (6,0) is correct because the greatest and least values of $|z+1|$ are found by interpreting the given inequality geometrically as a circle and finding the distances from the center to the point -1.

Option A ($\frac{{a^{2015}}}{{z-1}}$) is correct because the series is a geometric progression with common ratio z, and using the formula for the sum of an infinite geometric progression, along with the fact that $z^{2015} = -a^{2015}$, we arrive at the answer.

Option D (1) is correct because the magnitude of the given expression simplifies to 1 when $\frac{{2{z_1}}}{{3{z_2}}}$ is purely imaginary.

Option C ($\tan ^{-1}\left(\frac{{B}}{{A}}\right)$) is correct because the argument of a product of complex numbers is the sum of their arguments.

Option A (4) is correct because solving the system of equations Re(z²) = 0 and |z| = 2 yields four distinct solutions.

Option D ($i$) is correct because by converting the sum to a geometric series with complex exponentials and simplifying, the result is $i$.

Option A (0) is correct because simplifying the expression i^-39 results in 0 + i, so the value of 'a' is 0.

Option B (1) is correct because the continued product of the four values simplifies to cis(4$\pi$), which equals 1.

Option C is correct. By using the properties of complex numbers and limits, the expression simplifies to ${e^{i{\alpha }/2}}$.