Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 1 / 1A (Vectors and Matrices)

The quadratic formula givesso both roots of a polynomial have real part $-b/2$. Writing $z=x+iy$, the complex exponential function satisfies $e^z=e^x(\cos y+i\sin y)$, and hence its modulus is $e^x$. ThereforeIntersecting this condition with the allowed parameter interval gives the answer for either root:The endpoint $b=2$ is included: the repeated root of a polynomial is $z=-1$, whose exponential has modulus $e^{-1}<1$.

Write the complex number as $z=x+iy$. Since $iz=-y+ix$, the modulus of the complex exponential givesThis equals one precisely when $y=0$. The quadratic formula shows that the imaginary part of either root is $\pm \sqrt{4-b^2}/2$. Consequently the roots are real exactly at the two endpoints:Both endpoints satisfy the condition. The repeated roots are respectively $1$ and $-1$; their multiples by $i$ have exponentials on the unit circle.

Using the exponential definition of the hyperbolic cosine and the complex exponential function,ThusFor real $x$, the hyperbolic sine vanishes only at $x=0$. Also $|y|\le 1<\pi$, so the only zero of sine available to $y$ is $y=0$. The product therefore vanishes exactly when $b=0$ or $4-b^2=0$:There are three possible parameter values. At $b=0$ the roots are $\pm i$ and $\cosh (\pm i)=\cos 1$ is real; at $b=\pm 2$ the roots themselves are real.