# Ngô Quốc Anh

## January 11, 2009

### A complex trigonometric equation

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247) — Ngô Quốc Anh @ 20:29

In NUS, department of mathematics, QE in semester 1 AY 08-09 has the following complex trigonometric equation

To solve the above equation, firstly we can see that $2 \sinh (iz) = 2(-i \sin (i^2z) = 2i \sin z$. Then the given equation can be transformed to $4 + \cos z = 2i \sin z$. Using the following formulas

$\displaystyle\cos z = \frac{{{e^{iz}} + {e^{ - iz}}}} {2}, \quad \sin z = \frac{{{e^{iz}} - {e^{ - iz}}}} {2i}$

we get the following equivalent equation

$\displaystyle 4 + \frac{{{e^{iz}} + {e^{ - iz}}}} {2} = 2i\frac{{{e^{iz}} - {e^{ - iz}}}} {{2i}}$,

i.e., $8 + 3{e^{ - iz}} = {e^{iz}}$. Denoting $t=e^{iz}$ one can easily compute that $t=4 \pm \sqrt {19}$. Thus, the solutions are

$\displaystyle z = \frac{{\ln \left| {4 + \sqrt {19} } \right| + i2n\pi }}{i}$

and

$\displaystyle z = \frac{{\ln \left| {4 - \sqrt {19} } \right| + i\left( {\pi + 2n\pi } \right)}} {i}$

where $n \in \mathbb{Z}$. If you wish to express the solutions in Cartesian form, then

$\displaystyle z = 2n\pi - i\ln \left| {4 + \sqrt {19} } \right|$

and

$\displaystyle z = (2n+1)\pi - i\ln \left| {4 - \sqrt {19} } \right|$

where $n \in \mathbb{Z}$ is the answer.