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Euler's formula and its version with $-\theta$ giveSquaring $e^{i\theta }=\cos \theta +i\sin \theta$ and comparing real and imaginary parts with $e^{2i\theta }$ yields the double-angle formulas

The exponential function definitions of the hyperbolic functions and Euler's formula giveDividing these identities wherever $\cos x\ne 0$ gives $\boxed{\tanh (ix)=i\tan x}$. The two undivided identities hold for every complex number $x$; the quotient identity excludes its poles.

By Euler's formula, $\boxed{e^{2i}=\cos 2+i\sin 2}$, with the angle measured in radians.

For the positive real base ten, use its real natural logarithm. Then Euler's formula gives

Euler's formula givesCubing the expression for the sine, and grouping the $e^{3ix}$ and $e^{-3ix}$ terms, givesThe required equation therefore becomes $\sin x(1-4{\sin }^{2}x)=0$. The complete real solution set is

Euler's formula gives $e^{iz}=e^{-y}(\cos x+i\sin x)$. Its real part and imaginary part are

Use Euler's formula to evaluate the two exponentials: $e^{i\pi /2}=i$ and $\sqrt{2}{e}^{i\pi /4}=1+i$. Thus $z=1+2i$ lies in the first quadrant, givingIts principal argument is approximately $1.10715$ radians.

Using the exponential definition of the hyperbolic cosine and Euler's formula,Therefore

The exponential definitions of hyperbolic sine and hyperbolic cosine, together with Euler's formula, give $\sinh (iu)=i\sin u$ and $\cosh (iu)=\cos u$. The argument of the complex logarithm therefore reduces to $i\sin (\pi /2)+\cos (9\pi /2)=i$.

Using the principal logarithm, $\log i=i\pi /2$, soIf no branch of the complex logarithm is selected, its complete set of values is $i(\pi /2+2\pi k)$, $k\in \mathbb{Z}$. The real part remains zero, while the imaginary part is $\pi /2+2\pi k$. The original paper does not explicitly state a branch convention.

Powers of the imaginary unit repeat in blocks $i,-1,-i,1$, each with sum zero. The first $44$ terms comprise eleven complete blocks; the last is $i^{45}=i$. Applying Euler's formula to the resulting complex exponential gives $e^i=\cos 1+i\sin 1$. HenceThe angles here are in radians.

Let $a=\pi /3$, $b=2\pi /3$ and take $n$ equal subintervals of width $h=(b-a)/n=\pi /(3n)$. Since sine is continuous, its integral is the limit of the left-endpoint Riemann sumsUsing Euler's formula, the corresponding complex sum is a finite geometric series:Since $(e^{ih}-1)/h\to i$,Taking imaginary parts givesOne can see the same limit entirely within the real sums: the finite trigonometric-sum identity obtained from this geometric sum givesThus the evaluation is explicitly by a Riemann sum, as required.

Since $x=\cos y$ with real $y$, necessarily $-1\le x\le 1$. Use Euler's formula and set $w=e^{iy}\ne 0$. Thenso the quadratic equation gives $w=x\pm i\sqrt{1-x^2}$. Let $q=x+i\sqrt{1-x^2}$. It lies on the upper half of the unit circle, so $q=e^{i\theta }$ with $0\le \theta \le \pi$. Choose the complex logarithm value $\ln q=i\theta$, taking $\ln (-1)=i\pi$ at $x=-1$. The two roots are $e^{\pm i\theta }$, so $e^{iy}=e^{\pm i\theta }$ gives $y=\pm \theta +2n\pi$.

Because $i\ln q=-\theta$, this set of real solutions is exactlyThe formula includes $x=\pm 1$: the two sign families then coincide after adjusting $n$. No real $y$ exists if $|x|>1$.

Start with the exponential function definition of the hyperbolic sine and use Euler's formula:Identifying the hyperbolic sine and hyperbolic cosine in the final line gives

The force law separates into two identical harmonic oscillator equations:For positive $m,k$, the general real solution isSince $\mathrm{Re}(e^{i\alpha })=\cos \alpha$ by Euler's formula, these are exactly the stated real parts of complex exponentials. The amplitudes $x_0,y_0$ and phases ${\varphi }_1,{\varphi }_2$ are fixed by the four initial position and velocity components. This is two-dimensional isotropic harmonic oscillator flow: the same angular frequency acts in every direction.

A nonzero complex number has polar formwhere $r$ is its modulus and $\theta$ is an argument. The equality uses Euler's formula; sine and cosine recover the imaginary part and real part from polar coordinates. For $z=3+4i$, take $r=5$ and $\theta =\arctan (4/3)\approx 0.927295$ radians, so $\cos \theta =3/5$ and $\sin \theta =4/5$.

If $w=s{e}^{i\varphi }\ne 0$, thenThus multiplication multiplies moduli and adds arguments modulo $2\pi$; division divides moduli and subtracts arguments modulo $2\pi$. Since $1+i=\sqrt{2}{e}^{i\pi /4}$, multiplying $3+4i$ by $1+i$ gives $-1+7i$, of modulus $5\sqrt{2}$ and argument $\theta +\pi /4\approx 1.712693$ radians. One formula simultaneously describes its new size and direction.

Repeated multiplication now becomes especially simple: $(r{e}^{i\theta }{)}^n=r^n{e}^{in\theta }$ for a positive integer $n$. This explains the evenly spaced arguments of the roots of a complex number. For example, $(3+4i{)}^2=-7+24i$ has modulus $25$ and principal argument $2\theta \approx 1.854590$ radians. The Cartesian calculation and the polar form provide independent checks on both its location and its size. Choose polar form when the task concerns multiplication, division, or repeated turns; use real parts and imaginary parts for addition.