Polar Coordinates and Complex Numbers — Free Trigonometry Review Games.

x = 5*cos(0) = 5, y = 5*sin(0) = 0. So (5, 0).

r is the radial distance from the origin (pole).

r = 3, theta = pi/2 (on the positive y-axis).

Modulus = \(\sqrt{9+16} = \sqrt{25} = 5\).

By definition, i^2 = -1.

\(x = 4\cos(\pi/3) = 4 \cdot (1/2) = 2\). \(y = 4\sin(\pi/3) = 4 \cdot (\sqrt{3}/2) = 2\sqrt{3}\).

\(r = \sqrt{1+1} = \sqrt{2}\). \(\theta = \pi - \pi/4 = 3\pi/4\) (Q2).

\(2\cos(60) = 2 \cdot (1/2) = 1\). \(2\sin(60) = 2 \cdot (\sqrt{3}/2) = \sqrt{3}\). So \(1 + i\sqrt{3}\).

1 - i^2 = 1 - (-1) = 2.

The point (-1, 0) is on the negative x-axis, so the angle is pi.

De Moivre's theorem raises modulus to n and multiplies argument by n.

\(1+i\) has \(r=\sqrt{2}\), \(\theta=\pi/4\). \((\sqrt{2})^4(\cos(\pi)+i\sin(\pi)) = 4(-1+0i) = -4\).

r = constant is a circle centered at the origin.

r = 2*cos(theta) is a circle with diameter 2, centered at (1,0).

The nth roots of a number always produce exactly n distinct complex roots. Cube roots of 8: three roots.

By definition, polar angles are measured counterclockwise from the polar axis, which coincides with the positive x-axis in the rectangular plane. "The positive y-axis" is wrong because that direction corresponds to \(\theta=\pi/2\), not the starting reference. Remembering that the polar axis matches the positive x-axis is essential for correctly converting between coordinate systems.

Using \(x=r\cos\theta\) and \(y=r\sin\theta\), \(x=6\cos\pi=-6\) and \(y=6\sin\pi=0\), giving \((-6,0)\). The choice "\((6,0)\)" is wrong because it ignores that \(\cos\pi=-1\), not \(1\). Always evaluate both trig functions at the given angle before multiplying by \(r\).

Since the point lies on the positive x-axis, \(r=\sqrt{4^2+0^2}=4\) and \(\theta=0\), matching \((4,0)\). "\((4,\pi)\)" is incorrect because that angle points to \((-4,0)\), not \((4,0)\). Points on the positive x-axis always have polar angle zero.

The modulus is \(\sqrt{a^2+b^2}=\sqrt{0^2+5^2}=5\), the distance of the point from the origin. "25" is wrong because that is \(5^2\), not the square root taken in the modulus formula. The modulus always measures straight-line distance from the origin in the complex plane.

Since \(i^2=-1\), \(i^3=i^2\cdot i=-1\cdot i=-i\). "\(-1\)" is wrong because that is \(i^2\), one power short of \(i^3\). Memorizing the four-cycle \(i, -1, -i, 1\) helps evaluate any power of \(i\) quickly.

By definition of polar form, \(r\) scales the unit vector at angle \(\theta\) and equals \(|z|\), the distance from the origin to the point. "The argument (angle)" is incorrect because that role belongs to \(\theta\), not \(r\). Keeping \(r\) (magnitude) and \(\theta\) (direction) conceptually separate prevents confusion when converting forms.

The polar axis is the ray from the pole along \(\theta=0\), which is identical to the positive x-axis in rectangular coordinates. "The positive y-axis" is wrong because that ray corresponds to \(\theta=\pi/2\), a quarter turn from the polar axis. This identification underlies every polar-to-rectangular conversion formula.

Here \(x=2\cos(\pi/2)=0\) and \(y=2\sin(\pi/2)=2\), giving the point \((0,2)\). "\((2,0)\)" is incorrect because it swaps the roles of sine and cosine at this angle. At \(\theta=\pi/2\) the point always lies straight up the y-axis a distance \(r\) from the origin.

Since \(i^4=(i^2)^2=(-1)^2=1\), the powers of \(i\) complete a full cycle every four exponents. "\(-1\)" is wrong because that equals \(i^2\), not \(i^4\). Recognizing that \(i^4=1\) lets you simplify any large power of \(i\) by reducing the exponent modulo 4.

By the distance formula, \(|a+bi|=\sqrt{a^2+b^2}\) is exactly the distance from the origin to the point in the complex plane. "The angle the point makes with the positive real axis" is wrong because that describes the argument, a different quantity. Distinguishing modulus (length) from argument (direction) is fundamental to polar form of complex numbers.

Since the point lies on the negative x-axis, \(r=3\) and \(\theta=\pi\) places the point correctly, giving \((3,\pi)\). "\((3,0)\)" is wrong because that angle points toward \((3,0)\) on the positive x-axis, the opposite direction. Points on the negative x-axis always have polar angle \(\pi\) when \(r\) is taken positive.

Replacing \(\theta\) with \(-\theta\) and getting the same equation means the graph mirrors across the polar axis, since reflecting a point over the x-axis negates its angle. "The line \(\theta=\pi/2\)" is wrong because that symmetry test instead replaces \(\theta\) with \(\pi-\theta\). Knowing which substitution tests which symmetry saves time when sketching polar graphs.

The point \((-4,0)\) lies on the negative real axis, which corresponds to an angle of \(\pi\) radians from the positive real axis. "0" is wrong because that angle describes points on the positive real axis, not the negative one. The argument always measures the counterclockwise angle from the positive real axis to the point.

Since \(\cos(\pi/4)=\sin(\pi/4)=\frac{\sqrt2}{2}\), both \(x\) and \(y\) equal \(3\cdot\frac{\sqrt2}{2}=\frac{3\sqrt2}{2}\). "\((3,3)\)" is wrong because it forgets to multiply by \(\cos(\pi/4)\) and \(\sin(\pi/4)\) instead of using \(r\) directly. At \(\theta=\pi/4\) the x and y coordinates are always equal since sine and cosine match there.

The modulus is \(r=\sqrt{2^2+(-2)^2}=2\sqrt2\), and since the point lies in the fourth quadrant, \(\theta=\arctan(-2/2)=-\pi/4\). "\((2\sqrt2,\pi/4)\)" is wrong because that angle places the point in the first quadrant instead of the fourth. Always check the quadrant of \((x,y)\) before finalizing the reference angle sign.

When multiplying complex numbers in polar form, moduli multiply (\(2\times3=6\)) and arguments add (\(30^\circ+60^\circ=90^\circ\)), giving \(6\,\text{cis}\,90^\circ\). "\(5\,\text{cis}\,90^\circ\)" is wrong because it adds the moduli instead of multiplying them. This multiply-magnitudes, add-angles rule is the core mechanism behind all polar multiplication problems.

Dividing complex numbers in polar form divides the moduli (\(6/2=3\)) and subtracts the arguments (\(120^\circ-40^\circ=80^\circ\)), giving \(3\,\text{cis}\,80^\circ\). "\(3\,\text{cis}\,160^\circ\)" is wrong because it adds the angles instead of subtracting them, which is the multiplication rule, not division. Subtracting angles for division is the mirror image of adding angles for multiplication.

The modulus is \(r=\sqrt{(-1)^2+(-\sqrt3)^2}=2\), and since the point lies in the third quadrant with reference angle \(\pi/3\), the argument is \(\pi+\pi/3=4\pi/3\). "\(r=2,\ \theta=\pi/3\)" is wrong because it ignores that both coordinates are negative, placing the point in the third quadrant rather than the first. Always locate the correct quadrant before adding or subtracting the reference angle from \(\pi\).

Since \(\cos150^\circ=-\frac{\sqrt3}{2}\) and \(\sin150^\circ=\frac12\), multiplying by \(r=4\) gives \(-2\sqrt3+2i\). "\(2\sqrt3+2i\)" is wrong because it uses the wrong sign for cosine at \(150^\circ\), which lies in the second quadrant where cosine is negative. Recalling reference angles and quadrant signs is essential when converting between polar and rectangular complex number forms.

For a rose curve \(r=\cos(n\theta)\) with \(n\) odd, the number of petals equals \(n\), so \(n=3\) gives 3 petals. "6" is wrong because doubling applies only when \(n\) is even, not odd. Always check whether \(n\) is odd or even before deciding whether petals equal \(n\) or \(2n\).

For \(r=\sin(n\theta)\) with \(n\) even, the curve traces \(2n\) petals, so \(n=4\) produces 8 petals. "4" is wrong because that would only be correct if \(n\) were odd, but here \(n=4\) is even. The even-versus-odd rule for rose curves is a key detail students must check every time.

A cardioid arises when \(r=a+b\cos\theta\) with \(|a|=|b|\), so \(r=1+\cos\theta\) satisfies this condition exactly. "\(r=1+2\cos\theta\)" is wrong because \(|a|<|b|\) there, which produces a limaçon with an inner loop instead. Comparing the relative sizes of \(a\) and \(b\) is the key mechanism for classifying limaçon-family graphs.

The substitution \(\theta\to\pi-\theta\) reflects points across the vertical line \(\theta=\pi/2\), so invariance under this swap indicates y-axis symmetry. "Polar axis" is wrong because that symmetry test instead uses \(\theta\to-\theta\). Knowing each substitution test lets students quickly identify a curve's symmetry without plotting many points.

Using a negative \(r\) requires adding \(\pi\) to the original angle to land on the same point, giving \((-5,\pi/3+\pi)=(-5,4\pi/3)\). "\((-5,\pi/3)\)" is wrong because keeping the same angle with a negative radius reflects the point through the pole to a different location entirely. To convert between positive and negative \(r\) representations, always add or subtract \(\pi\) from the angle.

An equation of the form \(r=2a\sin\theta\) produces a circle of radius \(a\) centered at \((0,a)\), so here \(2a=4\) means radius \(2\) and diameter \(4\). "8" is wrong because it treats the coefficient itself as the diameter without recognizing the coefficient equals \(2a\), twice the radius. Recognizing \(r=2a\sin\theta\) or \(r=2a\cos\theta\) as circle equations is a common polar-graph shortcut.

The modulus is \(\sqrt{3^2+(-3)^2}=3\sqrt2\), and since the point lies in the fourth quadrant, the argument is \(-\pi/4\), giving \(3\sqrt2\,\text{cis}(-\pi/4)\). "\(3\sqrt2\,\text{cis}(\pi/4)\)" is wrong because that places the point in the first quadrant, but the negative imaginary part puts it in the fourth quadrant instead. Always verify both the modulus calculation and the correct quadrant sign for the argument.

Squaring in polar form squares the modulus (\((\sqrt2)^2=2\)) and doubles the angle (\(45^\circ\times2=90^\circ\)), giving \(2\,\text{cis}\,90^\circ=2(\cos90^\circ+i\sin90^\circ)=2i\). "\(2+2i\)" is wrong because it incorrectly assumes the angle doubling changes both real and imaginary parts equally rather than landing exactly on \(90^\circ\). Doubling the angle and squaring the modulus is a direct application of De Moivre's theorem for \(n=2\).

A lemniscate has the characteristic form \(r^2=a^2\cos(2\theta)\) or \(r^2=a^2\sin(2\theta)\), producing a figure-eight shape, so \(r^2=4\cos(2\theta)\) fits this pattern. "\(r=4\cos(2\theta)\)" is wrong because without squaring \(r\), this equation instead produces a four-petal rose. The presence of \(r^2\) rather than \(r\) is the defining signature of a lemniscate equation.

Since \(\sec\theta=1/\cos\theta\), the equation becomes \(r\cos\theta=4\), and because \(x=r\cos\theta\), this simplifies directly to \(x=4\). "\(y=4\)" is wrong because it mistakenly applies the sine relationship instead of the cosine one implied by secant. Recognizing \(r\cos\theta=x\) lets students quickly convert secant- or cosecant-based polar equations into simple vertical or horizontal lines.

Multiplying gives modulus \(2\times5=10\) and angle \(\pi/6+\pi/3=\pi/2\), so the product is \(10\,\text{cis}(\pi/2)=10(\cos(\pi/2)+i\sin(\pi/2))=10i\). "10" is wrong because it ignores the imaginary component entirely, treating the modulus as the final rectangular answer. After multiplying magnitudes and adding angles, always convert the resulting polar form back to rectangular by evaluating cosine and sine at the final angle.

Because the equation involves \(\sin\theta\), replacing \(\theta\) with \(\pi-\theta\) leaves \(\sin\theta\) unchanged, confirming symmetry about the line \(\theta=\pi/2\). "The polar axis (x-axis)" is wrong because that symmetry would require the equation to remain unchanged under \(\theta\to-\theta\), which fails here since \(\sin(-\theta)=-\sin\theta\). Equations built from sine typically show y-axis symmetry, while cosine-based equations typically show x-axis symmetry.

Since \(x^2+y^2=r^2\), the equation becomes \(r^2=16\), and taking the positive square root gives \(r=4\). "\(r=16\)" is wrong because it fails to take the square root after substituting \(r^2\) for \(x^2+y^2\). Recognizing \(x^2+y^2=r^2\) is the standard shortcut for converting circle equations centered at the origin into polar form.

The modulus of \(\sqrt3+i\) is \(2\) and its argument is \(\pi/6\), so raising to the 6th power gives modulus \(2^6=64\) and angle \(6\times\pi/6=\pi\), yielding \(64(\cos\pi+i\sin\pi)=-64\). "\(64\)" is wrong because it omits that the angle lands exactly on \(\pi\), where cosine equals \(-1\), not \(1\). De Moivre's theorem always requires tracking both the modulus raised to the power and the angle multiplied by the power before evaluating the trig functions.

Writing \(16=16\,\text{cis}(0)\), each fourth root has modulus \(16^{1/4}=2\) and angles \(\frac{0+2k\pi}{4}\) for \(k=0,1,2,3\), producing \(2,\ 2i,\ -2,\ -2i\). "\(4,\ 4i,\ -4,\ -4i\)" is wrong because it mistakenly uses the original modulus instead of taking its fourth root. Taking the \(n\)th root of the modulus and spacing angles evenly by \(2\pi/n\) is the general method for finding all \(n\)th roots of a complex number.

Rewriting in the form \(r=\frac{ed}{1-e\cos\theta}\) shows the eccentricity \(e=1\), and an eccentricity of exactly 1 always defines a parabola. "Ellipse" is wrong because ellipses require \(0<e<1\), not \(e=1\) as found here. Comparing the coefficient of \(\cos\theta\) to 1 in the standard polar conic form is the key step for identifying conic type.

Dividing numerator and denominator by 2 gives \(r=\frac{3}{1+1.5\cos\theta}\), revealing eccentricity \(e=1.5\), and since \(e>1\) the conic is a hyperbola. "Ellipse" is wrong because ellipses require \(e<1\), but here the coefficient of \(\cos\theta\) exceeds 1 after normalization. Always normalize the polar conic equation so the constant term in the denominator equals 1 before reading off the eccentricity.

Writing \(1=\text{cis}(0)\), the fourth roots have modulus 1 and angles \(\frac{2k\pi}{4}\) for \(k=0,1,2,3\), giving \(1, i, -1, -i\), the fourth roots of unity. "\(1,\ -1\)" is wrong because it only lists the two real roots and misses the two complex roots required to satisfy a degree-4 polynomial. A degree-\(n\) equation like \(z^n=1\) always has exactly \(n\) distinct complex roots evenly spaced around the unit circle.

The reciprocal of a complex number in polar form takes the reciprocal of the modulus and negates the argument, giving \(\frac{1}{2}\,\text{cis}(-40^\circ)\). "\(2\,\text{cis}(-40^\circ)\)" is wrong because it keeps the original modulus instead of inverting it. Reciprocals in polar form always invert the magnitude while flipping the sign of the angle.

A negative radius reflects the point through the pole, which is equivalent to adding \(\pi\) to the angle while making \(r\) positive, so \((-3,\pi/6)=(3,\pi/6+\pi)=(3,7\pi/6)\). "\((3,\pi/6)\)" is wrong because simply dropping the negative sign without adjusting the angle moves the point to the opposite side of the plane. Converting negative-\(r\) coordinates to positive-\(r\) form always requires adding \(\pi\) to the angle, not just changing the sign of \(r\).

The curve passes through the pole whenever \(\sin(2\theta)=0\), which occurs at \(\theta=0,\pi/2,\pi,3\pi/2\) within one full rotation, giving 4 pole crossings. "2" is wrong because it only counts half the solutions of \(\sin(2\theta)=0\) in the given interval. Setting \(r=0\) and solving for \(\theta\) is the standard method for finding where any polar curve passes through the origin.

The modulus of \(1-i\) is \(\sqrt2\) and its argument is \(-\pi/4\), so raising to the 8th power gives modulus \((\sqrt2)^8=16\) and angle \(8\times(-\pi/4)=-2\pi\), which is coterminal with \(0\), yielding \(16(\cos0+i\sin0)=16\). "\(-16\)" is wrong because it assumes the final angle lands on \(\pi\) rather than correctly reducing \(-2\pi\) to an angle coterminal with \(0\). Always reduce the resulting angle modulo \(2\pi\) before evaluating cosine and sine in De Moivre's theorem.

An inner loop forms when the constant term is smaller in magnitude than the coefficient of cosine, since \(r\) becomes negative for part of the domain, causing the curve to loop back through the pole; this occurs when \(|a|<|b|\). "\(|a|=|b|\)" is wrong because that condition instead produces a cardioid with a cusp at the pole rather than a full inner loop. Comparing \(|a|\) and \(|b|\) is the key test for classifying limaçons as having a dimple, a cusp, or an inner loop.

Since \(z_2^{-1}=2\,\text{cis}(-20^\circ)\), it follows that \(z_2=\frac{1}{2}\,\text{cis}(20^\circ)\), so \(z_1z_2=4\times\frac12\,\text{cis}(70^\circ+20^\circ)=2\,\text{cis}(90^\circ)=2i\). "2" is wrong because it ignores the imaginary component that results from the angle landing exactly at \(90^\circ\). This problem combines reciprocal and multiplication rules, both of which rely on manipulating moduli and arguments separately.

Dividing numerator and denominator by 3 gives \(r=\frac{2}{1+(2/3)\cos\theta}\), showing eccentricity \(e=2/3\), and since \(0<e<1\) the conic is an ellipse. "Hyperbola" is wrong because hyperbolas require \(e>1\), but the normalized coefficient here is less than 1. Normalizing the denominator's constant term to 1 is essential before comparing the eccentricity to the thresholds for ellipse, parabola, and hyperbola.

The number \(\frac{1+i}{\sqrt2}\) has modulus 1 and argument \(\pi/4\), so raising it to the 20th power gives angle \(20\times\pi/4=5\pi\), which reduces to \(\pi\) after subtracting \(4\pi\), yielding \(\cos\pi+i\sin\pi=-1\). "\(1\)" is wrong because it assumes the angle reduces to \(0\) rather than correctly reducing \(5\pi\) to the coterminal angle \(\pi\). When a number's modulus is 1, its powers always lie on the unit circle, so only the angle reduction matters for the final answer.