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Chapter 7: Problem 5

Show that \(\sin (x+i y)=\sin x \cosh y+i \cos x \sinh y\) using trigonometric identities and the exponential forms of these functions.

Short Answer

Expert verified
Through converting the trigonometric function to exponential forms and applying hyperbolic definitions, it has been shown that \(\sin (x+i y)=\sin x \cosh y+i \cos x \sinh y\).

Step by step solution

01

Recall the Exponential Form of Sine Function

To begin, we must remember how to write the sine function in exponential form using Euler's formula. Applying Euler's formula gives the exponential form of sine as: \(\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}\).
02

Substitution of \(x + iy\) into the Exponential Form of sine

Let's substitute \(x + iy\) into the exponential form of sine. Hence, \(\sin(x + iy) = \frac{e^{i(x +iy)} - e^{-i(x +iy)}}{2i}\).
03

Simplify the Exponent

We simplify the exponent by multiplying \(i\) through the brackets, which results in \(e^{ix-y} - e^{-ix+y}\). Therefore \(\sin(x + iy) = \frac{e^{ix-y} - e^{-ix+y}}{2i}\).
04

Use the Properties of Hyperbolic Functions and Simplify Further

Recall that the hyperbolic cosine, \(\cosh y\), and hyperbolic sine, \(\sinh y\), are given by \(\cosh y = \frac{e^y + e^{-y}}{2}\) and \(\sinh y = \frac{e^y - e^{-y}}{2}\) respectively. Hence, \(e^y\) and \(e^{-y}\) can be expressed in terms of \(\cosh y\) and \(\sinh y\). The expression \(e^{ix-y} - e^{-ix+y}\) can therefore be written as \(2i\sin x \cosh y - 2 \cos x \sinh y\) given that \(\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}\) and \(\cos(x) = \frac{e^{ix} + e^{-ix}}{2}\). Hence, \(\sin(x + iy) = i\sin x \cosh y - \cos x \sinh y\).
05

Simplify the Expression

Rearrange the simplified expression to obtain \(\sin(x + iy) = \sin x \cosh y + i \cos x \sinh y\).

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Most popular questions from this chapter

. For the following, determine if the given point is a removable singularity, an essential singularity, or a pole (indicate its order). a. \(\frac{1-\cos z}{z^{2}}, \quad z=0\) b. \(\frac{\sin 2}{z^{2}}, \quad z=0\) c. \(\frac{z^{2}-1}{(z-1)^{2}}, \quad z=1\). d. \(z e^{1 / z}, \quad z=0\). e. \(\cos \frac{\pi}{\pi-\pi}, \quad z=\pi\)

Write the following in standard form. a. \((4+5 i)(2-3 i)\) b. \((1+i)^{3}\) c. \(\frac{5+3 i}{1-i}\).

Write the following in rectangular form, \(z=a+i b\). a. \(4 e^{i \pi / 6}\) b. \(\sqrt{2} e^{5 i \pi / 4}\) c. \((1-i)^{100} .\)

Find the principal value of \(i^{i}\). Rewrite the base, \(i\), as an exponential first.

Show that for \(g\) and \(h\) analytic functions at \(z_{0}\), with \(g\left(z_{0}\right) \neq 0, h\left(z_{0}\right)=0\), and \(h^{\prime}\left(z_{0}\right) \neq 0\) $$ \operatorname{Res}\left[\frac{g(z)}{h(z)} ; z_{0}\right]=\frac{g\left(z_{0}\right)}{h^{\prime}\left(z_{0}\right)} $$
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