Exercises — Exponential form z = re^(iθ)

WHAT: We use Euler's formula $e^{i\theta }=\cos \theta +i\sin \theta$ — the dictionary that turns an "angle instruction" into coordinates. WHY this tool: $e^{i\theta }$ is defined to sit on the unit circle at angle $\theta$; to get its $x+iy$ we just read off $\cos$ and $\sin$. Here $\theta =\pi /2$ (a quarter turn, 90°). Look at figure s01: a quarter turn lands the arrow straight up. $e^{i\pi /2}=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=0+i\cdot 1=i.$ Answer: $\boxed{i}$.

The form $r{e}^{i\theta }$ literally displays its two pieces:

• the number multiplying the exponential is $r=5$ (the length of the arrow),
• the number in the exponent (next to $i$) is $\theta =\pi /6$ (the turn). Answer: $r=5$, $\theta =\frac{\pi }{6}$.

$\left|e^{i\theta }\right|=\sqrt{{\cos }^2\theta +{\sin }^2\theta }=\sqrt{1}=1$. The Pythagorean identity guarantees the arrow always has length $1$ — it only turns, never grows. Answer: $1$.

Step 1 — modulus. $r=\sqrt{(\sqrt{3}{)}^2+1^2}=\sqrt{3+1}=2$. Why: $r$ is the arrow length $\sqrt{x^2+y^2}$. Step 2 — argument. $x>0,y>0$ ⇒ first quadrant, no adjustment needed. The reference angle is $\arctan \left(\frac{y}{x}\right)=\arctan \left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$. Why $\arctan (y/x)$: on the right triangle formed by the arrow, $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{y}{x}$, so $\theta$ is "the angle whose tan is $y/x$." Answer: $z=2e^{i\pi /6}$.

Step 1: $r=\sqrt{(-\sqrt{3}{)}^2+1^2}=\sqrt{4}=2$. Step 2: reference angle $=\arctan \left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$. Now $x<0,y>0$ ⇒ quadrant II. In quadrant II the true angle is $\pi$ minus the reference angle: $\theta =\pi -\frac{\pi }{6}=\frac{5\pi }{6}.$ Check figure s02: the quadrant-II arrow points up-and-left, so $\theta$ must be between $\pi /2$ and $\pi$ — and $5\pi /6$ is. Answer: $z=2e^{i5\pi /6}$.

Step 1: $r=2$ (same components' sizes). Step 2: reference $=\pi /6$; $x<0,y<0$ ⇒ quadrant III. In quadrant III we go past $\pi$ by the reference angle. To keep the principal argument in $(-\pi ,\pi ]$, we subtract: $\theta =-\pi +\frac{\pi }{6}=-\frac{5\pi }{6}.$ (Equivalently $+\frac{7\pi }{6}$, the same arrow.) Figure s02: the quadrant-III arrow points down-and-left. Answer: $z=2e^{-i5\pi /6}$.

Step 1: $r=2$. Step 2: reference $=\pi /6$; $x>0,y<0$ ⇒ quadrant IV. The arrow dips just below the positive real axis, so the argument is simply negative: $\theta =-\frac{\pi }{6}.$ Answer: $z=2e^{-i\pi /6}$.

These are the axis cases where $\arctan (y/x)$ either gives $0$ or divides by zero — you must reason from the picture.

• (a) $z=4$: on the positive real axis, $\theta =0$. $\Rightarrow 4e^{i0}=4$.
• (b) $z=-4$: on the negative real axis, half a turn, $\theta =\pi$. $\Rightarrow 4e^{i\pi }$.
• (c) $z=3i$: straight up, $\theta =\pi /2$. $\Rightarrow 3e^{i\pi /2}$.
• (d) $z=-3i$: straight down, $\theta =-\pi /2$. $\Rightarrow 3e^{-i\pi /2}$.

WHY exponential form: because exponents of the same base add, $e^a{e}^b=e^{a+b}$, so multiplying is just "multiply lengths, add turns." Moduli multiply: $2\times 2=4$. Arguments add: $\frac{5\pi }{6}+\left(-\frac{\pi }{6}\right)=\frac{4\pi }{6}=\frac{2\pi }{3}$. $z_1{z}_2=4e^{i2\pi /3}=4\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)=4\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)=-2+2\sqrt{3}i.$ Answer: $z_1{z}_2=4e^{i2\pi /3}=-2+2\sqrt{3}i$.

Division flips the rule: moduli divide, arguments subtract. $\frac{z_1}{z_2}=\frac{2}{2}{e}^{i\left(\frac{5\pi }{6}-\left(-\frac{\pi }{6}\right)\right)}=1\cdot e^{i\pi }=-1.$ Answer: $\frac{z_1}{z_2}=e^{i\pi }=-1$.

Step 1 — polar form: $1+i=\sqrt{2}{e}^{i\pi /4}$ (modulus $\sqrt{2}$, first-quadrant angle $\pi /4$). Step 2 — De Moivre: raising to a power powers the modulus and multiplies the angle: $z^n=r^n{e}^{in\theta }$. $(1+i{)}^6=(\sqrt{2}{)}^6{e}^{i\cdot 6\pi /4}=2^3{e}^{i3\pi /2}=8e^{i3\pi /2}.$ Step 3 — back to rectangular: $\cos \frac{3\pi }{2}=0,\sin \frac{3\pi }{2}=-1$, so $8(0-i)=-8i$. Answer: $(1+i{)}^6=8e^{i3\pi /2}=-8i$.

By Euler's formula $e^{i\pi }=\cos \pi +i\sin \pi =-1+i\cdot 0=-1$, so $e^{i\pi }+1=0$ — the famous identity. Geometrically: $\theta =\pi$ is exactly half a turn; the unit arrow starting at $1$ swings all the way to $-1$. Answer: $e^{i\pi }=-1$.

Step 1 — write the target in exponential form, keeping all its angles. The key insight: $8$ is the same arrow after $0$, one, two, … full turns. So $8=8e^{i(0+2\pi k)}$ for every integer $k$. Step 2 — take cube roots: modulus $\to 8^{1/3}=2$; argument $\to \frac{0+2\pi k}{3}$. $z_k=2e^{i2\pi k/3},k=0,1,2.$

• $k=0:2e^{i0}=2$.
• $k=1:2e^{i2\pi /3}=2\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)=-1+\sqrt{3}i$.
• $k=2:2e^{i4\pi /3}=2\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)=-1-\sqrt{3}i$. Why only $k=0,1,2$: $k=3$ gives angle $2\pi$, the same as $k=0$ — the roots repeat after three. These are $2\times$ the Roots of unity. Answer: $2,-1+\sqrt{3}i,-1-\sqrt{3}i$.

Step 1 — convert each base. $1+i=\sqrt{2}{e}^{i\pi /4}$; and $\sqrt{3}-i=2e^{-i\pi /6}$ (quadrant IV, from L2.4). Step 2 — power each (De Moivre).

• Numerator: $(1+i{)}^{10}=(\sqrt{2}{)}^{10}{e}^{i10\pi /4}=2^5{e}^{i5\pi /2}=32e^{i5\pi /2}$. Reduce the angle: $5\pi /2-2\pi =\pi /2$, so $=32e^{i\pi /2}$.
• Denominator: $(\sqrt{3}-i{)}^4=2^4{e}^{-i4\pi /6}=16e^{-i2\pi /3}$. Step 3 — divide: moduli divide, arguments subtract. $\frac{32}{16}{e}^{i\left(\frac{\pi }{2}-\left(-\frac{2\pi }{3}\right)\right)}=2e^{i\left(\frac{\pi }{2}+\frac{2\pi }{3}\right)}=2e^{i7\pi /6}.$ Answer: $2e^{i7\pi /6}$ (equivalently $2e^{-i5\pi /6}$).

Idea: $(\cos \theta +i\sin \theta {)}^3=e^{i3\theta }=\cos 3\theta +i\sin 3\theta$. Expand the left side and match real parts. Expand with the binomial theorem (write $c=\cos \theta ,s=\sin \theta$): $(c+is{)}^3=c^3+3c^2(is)+3c(is{)}^2+(is{)}^3=c^3+3i{c}^{2}s-3c{s}^2-i{s}^3.$ Real part: $c^3-3c{s}^2$. Set equal to $\cos 3\theta$: $\cos 3\theta =c^3-3c{s}^2=c^3-3c(1-c^2)=c^3-3c+3c^3=4c^3-3c.$ (Used $s^2=1-c^2$.) $■$ Answer: identity confirmed: $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$.

Step 1 — it's a geometric series with ratio $q=e^{i\theta }$: $S=\frac{1-q^n}{1-q}=\frac{1-e^{in\theta }}{1-e^{i\theta }}.$ Step 2 — the "half-angle factoring" trick. Pull out half the total angle from each factor so we can use $e^{i\alpha }-e^{-i\alpha }=2i\sin \alpha$: $1-e^{in\theta }=e^{in\theta /2}\left(e^{-in\theta /2}-e^{in\theta /2}\right)=e^{in\theta /2}\cdot (-2i)\sin \frac{n\theta }{2}.$ Similarly $1-e^{i\theta }=e^{i\theta /2}(-2i)\sin \frac{\theta }{2}$. Step 3 — divide: $S=\frac{e^{in\theta /2}\sin \frac{n\theta }{2}}{e^{i\theta /2}\sin \frac{\theta }{2}}=e^{i(n-1)\theta /2}\frac{\sin \frac{n\theta }{2}}{\sin \frac{\theta }{2}}.$ Step 4 — take real parts. The left side's real part is $\sum \cos (k\theta )$; the right side's real part uses $\mathrm{Re}{e}^{i(n-1)\theta /2}=\cos \frac{(n-1)\theta }{2}$: ${\sum }_{k=0}^{n-1}\cos k\theta =\frac{\sin \frac{n\theta }{2}}{\sin \frac{\theta }{2}}\cos \frac{(n-1)\theta }{2}.■$ Sanity check ($n=2,\theta =\pi /2$): LHS $=1+\cos \frac{\pi }{2}=1$. RHS $=\frac{\sin (\pi /2)}{\sin (\pi /4)}\cos (\pi /4)=\frac{1}{1/\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=\sqrt{2}\cdot \frac{1}{\sqrt{2}}=1$. ✓

Step 1 — convert numerator and denominator. $1+i\sqrt{3}=2e^{i\pi /3}$ (QI); $1-i=\sqrt{2}{e}^{-i\pi /4}$ (QIV). Step 2 — divide inside the brackets: moduli divide, arguments subtract. $\frac{2}{\sqrt{2}}{e}^{i(\pi /3-(-\pi /4))}=\sqrt{2}{e}^{i(\pi /3+\pi /4)}=\sqrt{2}{e}^{i7\pi /12}.$ Step 3 — raise to the 12th power (De Moivre). $z=(\sqrt{2}{)}^{12}{e}^{i\cdot 12\cdot 7\pi /12}=2^6{e}^{i7\pi }=64e^{i7\pi }.$ Step 4 — reduce the angle to the principal range $(-\pi ,\pi ]$: $7\pi -6\pi =\pi$. So $z=64e^{i\pi }=-64$. Answer: modulus $64$, principal argument $\pi$ (so $z=-64$).

Since $|z|=1$, write $z=e^{i\theta }$ (arrow of length $1$). Then $\frac{1}{z}=e^{-i\theta }$ (reciprocal ⇒ modulus $1/1=1$, argument negated). $z+\frac{1}{z}=e^{i\theta }+e^{-i\theta }=(\cos \theta +i\sin \theta )+(\cos \theta -i\sin \theta )=2\cos \theta .■$ The imaginary parts cancel exactly; this is why $e^{i\theta }+e^{-i\theta }=2\cos \theta$ is the standard cosine bridge.