Question

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

(a)$(-1)^{\sqrt{2} }$

(b) $sin^{2} i$

(c)$cosh(\sqrt{2}+3i )$

Answer 1

Answer:

  (a) -0.266255 -0.963903i

  (b) -1.381098

  (c) -2.156385 +0.273077i

Step-by-step explanation:

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

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

(a)

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

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(b)

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

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(c)

  $\cosh{(\sqrt{2}+3i)}=\dfrac{e^{\sqrt{2}+3i}+e^{-(\sqrt{2}+3i)}}{2}\\\\=\dfrac{e^{\sqrt{2}}(\cos{3}+i\sin{3})+e^{-\sqrt{2}}(\cos{(-3)}+i\sin{(-3)})}{2}\\\\=\cos{(3)}\cosh{\sqrt{2}}+i\sin{(3)}\sinh{\sqrt{2}}\\\\=\boxed{-2.15638538+0.27307665i}$

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The useful relations are ...

  $e^{i\theta}=\cos{\theta}+i\sin{\theta}\\\\\sin(\theta)=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}=-i\sinh{(i\theta)}\\\\\cosh{(\theta)}=\dfrac{e^{\theta}+e^{-\theta}}{2},\ \sinh{(\theta)}=\dfrac{e^{\theta}-e^{-\theta}}{2}$