Question

Find the principal value of the following expressions. Write your answer in the form a + ib.

(a)
`(-1)^(√(2) )`

(b)
`sin^(2) i`

(c)
`cosh(√(2)+3i )`

Answer 1

(a) -0.266255 -0.963903i

(b) -1.381098

(c) -2.156385 +0.273077i

Explanation:

A suitable calculator is very handy for such questions. See attached.

__

Euler's relation and the definitions of sin and cosh in exponential terms are helpful.

(a)

`(-1)^(√(2))=e^(i\pi√(2))=\cos{(√(2)\cdot180^(\circ))}+i\sin{(√(2)\cdot180^(\circ))}\\\\\approx\cos{254.558^(\circ)}+i\sin{254.558^(\circ)}\\\\\approx\boxed{-0.26625534-0.96390253i}`

__

(b)

`\sin^2{i}=(-i\sinh{(-1)})^2=-\sinh^2{(-1)}\approx\boxed{-1.38109785}`

__

(c)

`\cosh{(√(2)+3i)}=(e^(√(2)+3i)+e^(-(√(2)+3i)))/(2)\\\\=(e^(√(2))(cos(3)+isin(3))+e^(-√(2))(cos((-3))+isin((-3))))/(2)\\\\=cos((3))\cosh{√(2)}+isin((3))\sinh{√(2)}\\\\=\boxed{-2.15638538+0.27307665i}`

_____

The useful relations are ...

`e^(i\theta)=cos(\theta)+isin(\theta)\\\\\sin(\theta)=(e^(i\theta)-e^(-i\theta))/(2i)=-i\sinh{(i\theta)}\\\\\cosh{(\theta)}=(e^(\theta)+e^(-\theta))/(2),\ \sinh{(\theta)}=(e^(\theta)-e^(-\theta))/(2)`