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.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: