Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 7

Complex differentiability requires the same limit for increments along the real and imaginary directions. Real increments give $f^{\prime }=u_x+i{v}_x$. Imaginary increments giveEquating real and imaginary parts yields the Cauchy-Riemann equationsThey hold throughout the open region where $f$ is holomorphic. Conversely, continuous first real partial derivatives satisfying these equations on an open set ensure holomorphy there.

For the real part specified in the PDF,Integrating $v_y=u_x$ givesSubstitution into $v_x=-u_y$ shows $h^{\prime }(x)=0$. ThusIndeed $\cos (x+iy)=\cos x\cosh y-i\sin x\sinh y$, so the real part of $z\cos z$ is exactly the prescribed expression. The arbitrary imaginary constant is the only freedom on a connected domain.

Write $g=u+iv$ and assume its domain is open and connected. If the constant modulus is zero then $g=0$. Otherwise let $u^2+v^2=c^2>0$. Differentiating this identity and using the Cauchy-Riemann equations givesThese are a linear system for $u_x,v_x$ with coefficient matrix $\left(\begin{array}{cc}u& v\\ v& -u\end{array}\right)$, whose determinant is $-c^2\ne 0$. Therefore $u_x=v_x=0$, and the Cauchy-Riemann equations also give $u_y=v_y=0$. It follows thatThis proves the constant-modulus holomorphic function result directly. On a disconnected domain the same conclusion holds separately on each connected component, but the constants can differ.

Partial fraction decomposition give $h(z)=1/(z-i)+1/(z+i)$. Let $w=z-1$. Each fraction has a geometric expansion, soEquivalently, its coefficient is $(-1{)}^j{2}^{(1-j)/2}\cos ((j+1)\pi /4)$. The first terms areThe linear and fifth-order coefficients vanish. Both geometric series converge for $|w|<\sqrt{2}$, and $h$ has genuine simple poles at $z=\pm i$, each distance $\sqrt{2}$ from the centre. Thus the Taylor series has