Exercises — Euler's formula — e^(iθ) = cos θ + i sin θ (proof via Taylor series)

Substitute $\theta =0$: $e^{i0}=\cos 0+i\sin 0=1+i(0)=1.$ Angle $0$ means the clock hand points along the positive real axis, so the tip is at the point $1$. Answer: $e^{i0}=1$.

$e^{i\pi }=\cos \pi +i\sin \pi =(-1)+i(0)=-1.$ A half-turn ($\pi$ rad $=180^{\circ }$) lands the hand on the negative real axis, at $-1$. Rearranging: $e^{i\pi }+1=0$ — this is Euler's identity. Answer: $e^{i\pi }=-1$.

$|e^{i\theta }|=\sqrt{{\cos }^2\theta +{\sin }^2\theta }=\sqrt{1}=1.$ The modulus is the distance from $0$; ${\cos }^2+{\sin }^2=1$ is the Pythagorean identity, so the tip is always distance $1$ from the centre — it never leaves the unit circle. Answer: $1$.

$e^{i3\pi /2}=\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}=0+i(-1)=-i.$ $\frac{3\pi }{2}$ rad is three quarter-turns; from $1$ that lands at the bottom of the circle, the point $-i$. Answer: $-i$.

Modulus: $r=\sqrt{(-1{)}^2+1^2}=\sqrt{2}$. Now the argument. The real part is negative and the imaginary part is positive, so $z$ sits in the second quadrant (upper-left) — look at the reference figure. The reference angle whose tangent is $\left|\frac{1}{-1}\right|=1$ is $\frac{\pi }{4}$. In quadrant II the true angle is $\pi -\frac{\pi }{4}=\frac{3\pi }{4}$. $z=\sqrt{2}{e}^{i3\pi /4}.$ Check: $\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)=\sqrt{2}\left(-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)=-1+i$. ✔

By De Moivre, the power multiplies the angle: ${\left(e^{i\pi /6}\right)}^3=e^{i3\pi /6}=e^{i\pi /2}=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=i.$ Answer: $i$ (i.e. $a=0,b=1$).

Square the exponential form: ${\cos }^2\theta ={\left(\frac{e^{i\theta }+e^{-i\theta }}{2}\right)}^2=\frac{e^{i2\theta }+2e^{i\theta }{e}^{-i\theta }+e^{-i2\theta }}{4}.$ The middle term: $e^{i\theta }{e}^{-i\theta }=e^{i(\theta -\theta )}=e^0=1$, so $2\cdot 1=2$. The outer pair recombines into a cosine: $e^{i2\theta }+e^{-i2\theta }=2\cos 2\theta$. Hence ${\cos }^2\theta =\frac{2\cos 2\theta +2}{4}=\frac{1+\cos 2\theta }{2}.■$

Recall $\mathrm{Re}$ picks off the part with no $i$, and $\mathrm{Re}\left(e^{ik\theta }\right)=\cos k\theta$. So read off each term: $\mathrm{Re}(e^{i\theta }+e^{i2\theta }+e^{i3\theta })=\cos \theta +\cos 2\theta +\cos 3\theta .$ Answer: $\cos \theta +\cos 2\theta +\cos 3\theta$.

Using the exponent rule, $\frac{1}{e^{i\theta }}=e^{-i\theta }=\cos (-\theta )+i\sin (-\theta )$. Since cosine is even ($\cos (-\theta )=\cos \theta$) and sine is odd ($\sin (-\theta )=-\sin \theta$), $e^{-i\theta }=\cos \theta -i\sin \theta .$ The conjugate of $e^{i\theta }=\cos \theta +i\sin \theta$ flips the sign of the imaginary part, giving exactly $\cos \theta -i\sin \theta$. All three expressions match. $■$ Geometrically, reciprocal and conjugate both reflect the point across the real axis here, because the modulus is $1$.

Polar form of $1-i$: modulus $r=\sqrt{1^2+(-1{)}^2}=\sqrt{2}$. Real part positive, imaginary part negative → quadrant IV (lower-right). Reference angle $\arctan \frac{1}{1}=\frac{\pi }{4}$, so $\theta =-\frac{\pi }{4}$. Thus $1-i=\sqrt{2}{e}^{-i\pi /4}$. $(1-i{)}^6=(\sqrt{2}{)}^6{e}^{-i6\pi /4}=8e^{-i3\pi /2}.$ Now $e^{-i3\pi /2}=\cos (-\frac{3\pi }{2})+i\sin (-\frac{3\pi }{2})=0+i(1)=i.$ $(1-i{)}^6=8i.$ Answer: $8i$.

Write $8=8e^{i0}$, but remember the angle is only fixed up to full turns: $8=8e^{i2\pi k}$ for any integer $k$. Seek $z=r{e}^{i\varphi }$ with $z^3=r^3{e}^{i3\varphi }=8e^{i2\pi k}$.

• Modulus: $r^3=8\Rightarrow r=2$.
• Angle: $3\varphi =2\pi k\Rightarrow \varphi =\frac{2\pi k}{3}$, taking $k=0,1,2$ for three distinct roots.

$k=0:z=2e^{i0}=2.$ $k=1:z=2e^{i2\pi /3}=2(-\frac{1}{2}+i\frac{\sqrt{3}}{2})=-1+i\sqrt{3}.$ $k=2:z=2e^{i4\pi /3}=2(-\frac{1}{2}-i\frac{\sqrt{3}}{2})=-1-i\sqrt{3}.$ Answer: $2,-1+i\sqrt{3},-1-i\sqrt{3}$. These sit at the corners of an equilateral triangle on the circle of radius $2$ — see Roots of Unity.

De Moivre: $(\cos \theta +i\sin \theta {)}^3=\cos 3\theta +i\sin 3\theta$. Expand the left with the binomial: ${\cos }^3\theta +3{\cos }^2\theta (i\sin \theta )+3\cos \theta (i\sin \theta {)}^2+(i\sin \theta {)}^3.$ Using $i^2=-1,i^3=-i$: $=({\cos }^3\theta -3\cos \theta {\sin }^2\theta )+i(3{\cos }^2\theta \sin \theta -{\sin }^3\theta ).$ Match real parts: $\cos 3\theta ={\cos }^3\theta -3\cos \theta {\sin }^2\theta$. Now replace ${\sin }^2\theta =1-{\cos }^2\theta$: $\cos 3\theta ={\cos }^3\theta -3\cos \theta (1-{\cos }^2\theta )=4{\cos }^3\theta -3\cos \theta .$ Answer: $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$.

This is a geometric series with first term $1$ and common ratio $q=e^{i\theta }$. Since $q\ne 1$, $S=\frac{1-q^n}{1-q}=\frac{1-e^{in\theta }}{1-e^{i\theta }}.$ To simplify, factor a half-angle out of top and bottom (the standard trick): $1-e^{i\alpha }=e^{i\alpha /2}(e^{-i\alpha /2}-e^{i\alpha /2})=e^{i\alpha /2}(-2i\sin \frac{\alpha }{2})$. Applying with $\alpha =n\theta$ on top and $\alpha =\theta$ on the bottom: $S=\frac{e^{in\theta /2}(-2i\sin \frac{n\theta }{2})}{e^{i\theta /2}(-2i\sin \frac{\theta }{2})}=e^{i(n-1)\theta /2}\frac{\sin \frac{n\theta }{2}}{\sin \frac{\theta }{2}}.$ Taking the real part (and $\mathrm{Re}{e}^{i(n-1)\theta /2}=\cos \frac{(n-1)\theta }{2}$): ${\sum }_{k=0}^{n-1}\cos k\theta =\frac{\sin \frac{n\theta }{2}}{\sin \frac{\theta }{2}}\cos \frac{(n-1)\theta }{2}.$ Verification ($n=3,\theta =\frac{\pi }{2}$): direct sum $\cos 0+\cos \frac{\pi }{2}+\cos \pi =1+0-1=0$; formula gives $\frac{\sin (3\pi /4)}{\sin (\pi /4)}\cos (\pi /2)=\frac{\sqrt{2}/2}{\sqrt{2}/2}\cdot 0=0$. ✔

Differentiate the right-hand side of Euler's formula term by term (this is legal — the series converges everywhere, from Taylor & Maclaurin Series): $\frac{d}{d\theta }(\cos \theta +i\sin \theta )=-\sin \theta +i\cos \theta .$ Factor out $i$: $-\sin \theta +i\cos \theta =i(\cos \theta +i\sin \theta )$ because $i\cos \theta +i^2\sin \theta =i\cos \theta -\sin \theta$. ✔ Hence $\frac{d}{d\theta }{e}^{i\theta }=i{e}^{i\theta }.$ Interpretation: the velocity of the moving point is its position multiplied by $i$, i.e. rotated $90^{\circ }$. A velocity always perpendicular to the radius, of the same length, is exactly uniform motion around a circle — this is Multiplication as Rotation & Scaling made kinetic. $■$

Put $\theta =ix$ into the exponential form of cosine: $\cos (ix)=\frac{e^{i(ix)}+e^{-i(ix)}}{2}=\frac{e^{-x}+e^x}{2}=\cosh x.■$ (We used $i\cdot i=-1$, so $e^{i(ix)}=e^{-x}$.) This is the bridge in Hyperbolic Functions — $\cosh,\sinh$ vs $\cos,\sin$: rotating the angle into the imaginary direction turns circular functions into hyperbolic ones.

Which single fact makes all of L4 and L5 work? Answer: An angle is only defined up to $2\pi$, and multiplying complex numbers adds their angles ($e^{iA}{e}^{iB}=e^{i(A+B)}$) — powers, roots, and sums all follow from this.