Question

Find each of the following in the $x+iy$ form and check your answers by computer. $$\sinh (\ln 2+\frac{i\pi }{3})$$

Short answer

$\sinh (\ln 2+\frac{i\pi }{3})=\frac{1}{2}+\frac{i\sqrt{3}}{2}$

Step-by-step solution

- Review the Hyperbolic Sine Function

The hyperbolic sine function is defined as $$sinh(z)=\frac{e^z-e^{-z}}{2}$$

- Substitute the Given Value

Substitute the value $z=\ln 2+\frac{i\pi }{3}$ into the hyperbolic sine function formula: $$sinh(\ln 2+\frac{i\pi }{3})=\frac{e^{\ln 2+\frac{i\pi }{3}}-e^{-(\ln 2+\frac{i\pi }{3})}}{2}$$

- Utilize Euler's Formula

Using Euler's formula: $e^{a+ib}=e^a(\cos b+i\sin b)$, express the exponentials: $$e^{\ln 2+\frac{i\pi }{3}}=e^{\ln 2}(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})\text{leads}to2(\frac{1}{2}+i\frac{\sqrt{3}}{2})=1+i\sqrt{3}$$

- Simplify the Inverse Exponential

Now simplify the other term: $$e^{-(\ln 2+\frac{i\pi }{3})}=e^{-\ln 2}(\cos (-\frac{\pi }{3})+i\sin (-\frac{\pi }{3}))\text{leads}to\frac{1}{2}(\frac{1}{2}-i\frac{\sqrt{3}}{2})=\frac{1}{2}-\frac{i\sqrt{3}}{2}$$

- Put it All Together

Combine the results and simplify: $$sinh(\ln 2+\frac{i\pi }{3})=\frac{(1+i\sqrt{3})-(\frac{1}{2}-\frac{i\sqrt{3}}{2})}{2}=\frac{(\frac{1}{2}+\frac{3i\sqrt{3}}{2})}{2}=\frac{1}{2}+\frac{i\sqrt{3}}{2}$$

- Verify with a Calculator or Software

Check the result using a computer or math software to ensure correctness.

Key concepts

hyperbolic sine function

The hyperbolic sine function, often denoted as $\sinh (z)$, is similar in some ways to the regular sine function from trigonometry. However, instead of being defined with respect to a circle, hyperbolic functions are related to hyperbolas. The formula for hyperbolic sine is given by:
$$\sinh (z)=\frac{e^z-e^{-z}}{2}$$ This means we take the exponential of $z$, subtract the exponential of $-z$, and then divide by 2.
Hyperbolic functions often appear in contexts where we deal with growth and decay, or in the solutions to certain differential equations. Their properties are very useful in both pure and applied mathematics.
When solving for $\sinh (z)$ in this example:
Given $z=\ln 2+\frac{i\pi }{3}$, we start by substituting $z$ into the formula.

Euler's formula

Euler's formula connects complex exponentiation with trigonometric functions. It states:
$$e^{a+ib}=e^a(\cos b+i\sin b)$$ This means that when dealing with a complex number in the exponent, we can break it down into its real and imaginary parts. The real part, $e^a$, acts as a magnitude, while the imaginary part, $b$, determines a rotation in the complex plane through $\cos b$ and $\sin b$.
In our problem, we use Euler's formula to simplify $e^{\ln 2+\frac{i\pi }{3}}$:
$$e^{\ln 2+\frac{i\pi }{3}}=e^{\ln 2}(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})$$ Since $e^{\ln 2}=2$, it follows that:
$$2(\frac{1}{2}+i\frac{\sqrt{3}}{2})=1+i\sqrt{3}$$

complex exponentiation

Complex exponentiation takes a complex number to a complex power. In general, exponentiation involves either a real base and an imaginary exponent, or vice versa.
In this exercise, the complex number $z=\ln 2+\frac{i\pi }{3}$ was raised to the power of the exponential base $e$.
This required us to use Euler's formula to simplify the expression and then take the hyperbolic sine of the resultant complex number:
We also calculated the inverse of the exponent by taking the negative of $z$:
$$e^{-(\ln 2+\frac{i\pi }{3})}=e^{-\ln 2}(\cos (-\frac{\pi }{3})+i\sin (-\frac{\pi }{3}))$$ Simplifying, we have:
$$\frac{1}{2}(\frac{1}{2}-i\frac{\sqrt{3}}{2})=\frac{1}{2}-\frac{i\sqrt{3}}{2}$$
Finally, combining all these results helped us simplify the given expression in the problem.