Question bank — Exponential form z = re^(iθ)

False. For an imaginary exponent $|e^{i\theta }|=\sqrt{{\cos }^2\theta +{\sin }^2\theta }=1$ for every $\theta$; it only rotates on the unit circle, it never grows.

False. By convention $r=|z|\ge 0$; a negative sign is absorbed into the angle since $-r{e}^{i\theta }=r{e}^{i(\theta +\pi )}$.

False. For $z=0$ we have $r=0$ but $\theta$ is undefined — the origin has no direction — so $z=0$ has no unique exponential form.

False. Adding $2\pi$ is one full turn, landing on the same point: $\cos$ and $\sin$ have period $2\pi$, so $e^{i\theta }=e^{i(\theta +2\pi k)}$ for any integer $k$. The argument is only unique up to $2\pi$.

False. Moduli multiply, so if both moduli are less than $1$ (e.g. $r_1=r_2=0.5$) the product's modulus $r_1{r}_2=0.25$ is smaller. Multiplication scales; it does not always enlarge.

True. Conjugation flips the sign of the imaginary part, which reflects the arrow across the real axis — same length $r$, opposite angle, so the argument becomes $-\theta$.

False. Dividing $1=1\cdot e^{i0}$ by $z$ subtracts the angle: $\frac{1}{z}=\frac{1}{r}{e}^{-i\theta }$. The modulus reciprocates and the angle flips sign.

False. It follows directly from Euler: $e^{i\pi }=\cos \pi +i\sin \pi =-1+0i=-1$. Geometrically it says "spin the unit arrow half a turn ($\pi$) and you land at $-1$" — see Euler's identity e^(iπ)+1=0.

Error is the quadrant. $x<0,y<0$ puts $z$ in quadrant III, but $\arctan 1=\frac{\pi }{4}$ points into quadrant I. Correct angle: $\frac{\pi }{4}-\pi =-\frac{3\pi }{4}$ (or $\frac{5\pi }{4}$), giving $z=\sqrt{2}{e}^{-i3\pi /4}$.

Error: exponents of the same base add, they don't multiply. The correct rule is $e^{i{\theta }_1}{e}^{i{\theta }_2}=e^{i({\theta }_1+{\theta }_2)}$, which is exactly why arguments add on multiplication.

Error: the modulus is raised to the power, not multiplied. De Moivre gives $(r{e}^{i\theta }{)}^n=r^n{e}^{in\theta }$; only the angle is multiplied by $n$.

Error: confusing modulus with value. $|e^{i\theta }|=1$ means the length is 1, but the number itself moves around the unit circle; e.g. $e^{i\pi }=-1\ne 1$.

Error: it is derived. Feed $u=i\theta$ into the Taylor series of $e^u$; the real terms rebuild $\cos \theta$ and the imaginary terms rebuild $\sin \theta$ — see Taylor and Maclaurin series. It is a theorem, not a guess.

Error: $z=3$ sits on the positive real axis, so its argument is simply $\theta =0$. A zero imaginary part is a perfectly valid direction, not a missing one.

Because $|e^{i\theta }|=1$: multiplying by a unit-modulus number leaves lengths unchanged and only adds $\theta$ to the argument. Stretching would need a modulus $\ne 1$. See Complex multiplication as rotation and scaling.

Because $\tan$ repeats every $\pi$, so $\arctan$ can only return values in $(-\frac{\pi }{2},\frac{\pi }{2})$ and cannot distinguish quadrant II from IV or III from I. The signs of $x,y$ tell you which half-turn to add.

Because $z^n=r^n{e}^{in\theta }$ turns a power into "power the length, multiply the angle" — no messy expansion. $(1+i{)}^8=(\sqrt{2}{)}^8{e}^{i2\pi }=16$ in two moves via De Moivre's theorem.

Because $\frac{e^{i{\theta }_1}}{e^{i{\theta }_2}}=e^{i{\theta }_1}{e}^{-i{\theta }_2}=e^{i({\theta }_1-{\theta }_2)}$; dividing is multiplying by the reciprocal, which flips the second angle's sign, so angles subtract.

Taking the $n$-th root shares the angle among $n$ solutions differing by $\frac{2\pi }{n}$ (from the $2\pi$-periodicity of the argument), all with the same modulus $r^{1/n}$ — so they space evenly, as in Roots of unity.

Same content, but the exponential lets us use the exponent law $e^a{e}^b=e^{a+b}$ directly, so multiplication, division, and powers become effortless. See Polar form (trigonometric) r(cosθ + i sinθ).

It lies on the negative real axis, so $\theta =\pi$ (or equivalently $-\pi$); the modulus is $5$, giving $z=5e^{i\pi }$.

It points straight up, so $\theta =\frac{\pi }{2}$ and $r=2$: $z=2e^{i\pi /2}$. Here $\arctan (y/x)$ is useless because $x=0$ (division by zero) — you read the angle from the axis directly.

No. If $r=1$ the point stays exactly one unit from the origin for every $\theta$, tracing the whole unit circle as $\theta$ runs over an interval of length $2\pi$.

Yes, they are the same number — a full extra turn returns to the same direction. This is why the principal argument is restricted (usually to $(-\pi ,\pi ]$) to make it unique.

A negative angle just means a clockwise turn: $e^{-i\pi /2}=\cos (-\frac{\pi }{2})+i\sin (-\frac{\pi }{2})=-i$, pointing straight down. Negative arguments are fully legal.

The point slides toward the origin, so $z\to 0$. But the direction $\theta$ becomes meaningless at the limit, which is exactly why $\arg (0)$ is undefined.