Complex numbers practice · Mixed difficulty

Option D ($\{z:|z|=1, z \neq 1\}$) is correct because the condition leads to $|z|=1$ and $z \neq 1$ due to the purely real constraint.

Option B (1 and 0) is correct because the given complex number simplifies to 1, which has a modulus of 1 and an argument of 0.

Option B ($3 \sqrt{2}$) is correct because the locus of z is a circle, and $|z-7-9i|$ represents the distance from the center of the circle to the point (7, 9).

Option B is correct because the equation represents a circle only if the square of the magnitude of 'a' is greater than 'b'.

Option A (6) is correct because solving the given equations for the complex number $\alpha = x + iy$ yields a real part of 6.

Option D (lie in the interval (1,2)) is correct because the minimum value of |z + 1/2| lies between | |z| - 1/2 | and |z| + 1/2, and since |z| >= 2, the minimum value is greater than 3/2 and less than 5/2, thus in the interval (1,2).

Option B ($1 / 2|z|^2$) is correct. By converting the complex vertices to Cartesian coordinates and calculating the triangle's area, we find it equals $1/2 |z|^2$.

Option B is correct. The product simplifies to a closed form expression involving z_0 and its powers, ultimately leading to the given result.

Option D (1) is correct because solving the equations $f(1)=2$ and $f(i^2)=0$ gives $f(x) = x^2 + x$, and solving $f(z+z^{-1})=0$ leads to $|z|=1$.

Option B is correct because the least value of $|z+1|$ is 0 and the greatest value is 6, given the condition $|z+4| \leq 3$.

Option A ($2i$) is correct because simplifying the summation using geometric series properties results in $2i$.

Option C ($a = b = c$) is correct because the quadratic equation must have roots that are also roots of the cubic equation, leading to the conclusion that the quadratic is a scaled version of a factor of the cubic.

Option A (1) is correct because by manipulating the given equation and using properties of complex numbers, we find that $\left| {{z_2}} \right|$ must be equal to 1.

Option A ($\frac{{\pi}}{{2}}$) is correct because the argument of the complex number is $\pm \frac{{\pi}}{{2}}$, and we take the positive value.

Option C ($\frac{{1}}{{2}}$) is correct because after simplifying the expression using properties of cube roots of unity, the sine of the resulting angle is 1/2.

Option D (None of these) is correct because after simplifying using the properties of cube roots of unity, the expression evaluates to 4, which is not among the given options.

Option D (2) is correct because when (z - alpha)/(z + alpha) is purely imaginary and |z| = 2, solving for alpha gives alpha = 2.

Option C (4 + 7i) is correct because it is the value obtained after expanding and simplifying the given determinant.

Option A (0) is correct because the given condition implies that $z_1$ and $z_2$ are orthogonal, leading to the expression simplifying to 0.

Option D ($\left| {{z_k}} \right| = \left| {{z_{k + 1}}} \right|$) is correct because all nth roots of unity lie on the unit circle, hence their magnitudes are equal to 1.

Option D (none of these) is correct because the given expression does not simplify to any of the provided options. The correct expansion involves terms with individual complex numbers and their conjugates.

Option D ($e^{-r \sin \theta}$) is correct because it is the result of calculating the magnitude of the complex exponential function.

Option B (1) is correct because the minimum value is the distance between the origin and the point $(\cos \alpha, \sin \alpha)$, which is 1.

Option C (The perpendicular bisector of $AB$) is correct because the equation $\left| {z - 3 + i} \right| = \left| {z - 2 - i} \right|$ represents the locus of a point $P$ such that its distance from $A$ and $B$ are equal, which is the perpendicular bisector of $AB$.

The correct answer is D ($\frac{{ - 1 - \sqrt 3 i}}{2}$) because it is one of the complex cube roots of unity.

Option B ($1,1$) is correct because by using the property $1 + \omega = -\omega^2$ and simplifying, we find that $A = 1$ and $B = 1$.

The correct answer is Origin because the equation of the line passing through the given points simplifies to an equation that always passes through the origin.

Option A (Circle of radius $\;\frac{{\sqrt 5 }}{2}$) is correct because when $\frac{{z + i}}{{z + 2}}$ is purely imaginary, the locus of z is a circle with the specified radius.

Option A ($x+y=1$) is correct because substituting $z=x+iy$ into the given equation and simplifying leads to the equation of the line $x+y=1$.

Option D (0) is correct because when expanding $(1+i)^6$ and $(1-i)^6$ and summing them, the terms cancel out.