VCAA Specialist Mathematics Algebra, number and structure

This corresponds to the polynomial $(z-2)(z^2+9)$, which has a root of $z=2$ instead of $z=-2$.

This corresponds to the polynomial $(z+3)(z^2+4)$, which has roots of $-3$ and $\pm 2i$ rather than $-2$ and $\pm 3i$.

This corresponds to the polynomial $(z-3)(z^2-4)$, which has real roots $3$, $2$, and $-2$.

This corresponds to the polynomial $(z+2)(z^2-9)$, which has real roots $-2$, $3$, and $-3$ instead of the required complex roots.

The coefficient $q=i$ is non-real. If the polynomial had complex conjugate roots, the third root would have to be real to make the constant term $1$, which would force all coefficients to be real.

The coefficient $p=i$ is non-real. The complex conjugate root theorem requires all coefficients to be real for non-real roots to occur in conjugate pairs.

Both $p=i$ and $q=i$ are non-real coefficients. A polynomial with non-real coefficients and a real constant term cannot have complex conjugate roots.

Incorrect. This negative value might result from a sign error when multiplying by the complex conjugate or evaluating the denominator.

Incorrect. While the expression $\frac{1+z}{1-z}$ has a real part of $0$ for $|z|=1$, the expression $\frac{1}{1-z}$ does not.

Incorrect. This value might be confused with the imaginary part for specific values of $z$ (like $z = e^{i\pi/3}$), but the real part is constantly $\frac{1}{2}$ for all valid $z$.

This option incorrectly multiplies the modulus by the exponent ($2 \times 3 = 6$) instead of raising the modulus to the power ($2^3 = 8$).

This answer results from two errors: multiplying the modulus by the exponent ($2 \times 3$) and incorrectly evaluating $\text{cis}(\pi)$ as $1$.

This option correctly calculates the new modulus ($2^3 = 8$) but incorrectly evaluates $\text{cis}(\pi)$ as $1$ instead of $-1$.

The roots are separated by an angle of $120^\circ$ from the origin, meaning the triangle's internal angles are all $60^\circ$, not $90^\circ$.

While an equilateral triangle is a special type of isosceles triangle, "equilateral" is the most precise and accurate description of the shape formed since all three sides are equal.

A scalene triangle has sides of all different lengths, but the distances between any two cube roots of unity are identical.

This only accounts for real values of $z$ (where $z^2 > 0$), but $z$ can also be purely imaginary (where $z^2 < 0$) and still have a real square.

This only accounts for purely imaginary values of $z$ (where $z^2 < 0$), missing the real values of $z$ (where $z^2 > 0$) which also have real squares.

This only represents positive purely imaginary numbers, which is an incomplete set of solutions since $z$ can be any real or purely imaginary number.

This only represents negative purely imaginary numbers, which is an incomplete set of solutions since $z$ can be any real or purely imaginary number.

This is the root already provided in the question. The question asks for "another" root, specifically the one implied by the properties of polynomials with real coefficients.

This is incorrect because the real part of the conjugate should remain $-2$. The complex conjugate of $x + yi$ is $x - yi$, so only the sign of the imaginary part changes.

This is incorrect because it changes the sign of the real part rather than the imaginary part. The correct conjugate of $-2 + 3i$ is $-2 - 3i$.

If $x > 0$ and $y = 1 - x$, the real part is positive and the imaginary part $y - 1 = -x$ is negative. This places the point in the fourth quadrant, giving an argument of $-\frac{\pi}{4}$ (or $\frac{7\pi}{4}$), not $\frac{3\pi}{4}$.

The equation $y = 1 + x$ implies $\frac{y - 1}{x} = 1$, which corresponds to an angle whose tangent is $1$. This would mean the argument is either $\frac{\pi}{4}$ or $-\frac{3\pi}{4}$, not $\frac{3\pi}{4}$.

If $y = 1 + x$ and $x > 0$, both the real part $x$ and imaginary part $y - 1$ are positive. This places the point in the first quadrant, giving an argument of $\frac{\pi}{4}$.

If $y = 1 + x$ and $x < 0$, both the real part $x$ and imaginary part $y - 1$ are negative. This places the point in the third quadrant, giving an argument of $-\frac{3\pi}{4}$ (or $\frac{5\pi}{4}$).

This would be the result if $(-i)^{14}$ evaluated to $-i$, which only happens for powers like $n=15$, not $n=14$.

This would be the result if $(-i)^{14}$ evaluated to $i$, which occurs for powers like $n=13$, but $(-i)^{14} = -1$.

This incorrectly applies the $-1$ multiplier only to the real part $a$, rather than distributing it to both terms to get $-a-ib$.

This assumes the transformation results in multiplying by $i$, but $(-i)^{14} = (i^2)^7 = (-1)^7 = -1$, not $i$.