Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 9

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

Since the arguments of $-1$ are $(2k+1)\pi$, the complex logarithm has values $\boxed{\log (-1)=i(2k+1)\pi }$ for $k\in \mathbb{Z}$. The principal value is $\boxed{i\pi }$.

The modulus of $i$ is one and its arguments are $\pi /2+2k\pi$. Thus $\boxed{\log i=i(\pi /2+2k\pi )}$, with principal value $\boxed{i\pi /2}$.

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

Using a value of the complex logarithm, $i^i=\exp (i\log i)=\exp [-(\pi /2+2k\pi )]$. Thus every logarithmic-branch value is real and positive:The principal value is $\boxed{e^{-\pi /2}}$.

The hyperbolic cosine and hyperbolic sine are defined by $\boxed{\cosh x=(e^x+e^{-x})/2,\sinh x=(e^x-e^{-x})/2}$. Their quotient is $\tanh x=\sinh x/\cosh x$.

Expand the two squares using the exponential definitions:

The exponential definitions show that $\sinh$ is odd and $\cosh$ is even. Their quotient therefore satisfies $\boxed{\tanh (-x)=-\tanh x}$.

Differentiating the exponential definition gives $\boxed{(\cosh x{)}^{\prime }=(e^x-e^{-x})/2=\sinh x}$.

The fifth power of the odd function $\tanh x$ is odd and continuous. Its integral over the symmetric interval is therefore $\boxed{{\int }_{-1000}^{1000}{\tanh }^{5}xdx=0}$.

The derivative of $\ln \cosh x$ is $\tanh x$. Since $\cosh (\ln 2)=(2+1/2)/2=5/4$,