Question

Separate into real and imaginary parts of cos${\displaystyle {}^{-1}\left(\frac{3i}{4}\right)}$ 

 

Answer 1

Let a+ib= cos${\displaystyle {}^{-1}\left(\frac{3i}{4}\right)}$         .................(1) 

∴${\displaystyle \cos (a+ib)=\frac{3i}{4}}$

∴${\displaystyle \cos \left(a\right)\cos \left(ib\right)-\sin \left(a\right)\sin \left(ib\right)=\frac{3i}{4}}$ 

cos(a)cosh(b) – isin(a)sinh(b) =${\displaystyle 0+\frac{3i}{4}}$  ${\displaystyle \{\because \cos \left(ix\right)=\cosh \left(ix\right)=\sinh \left(x\right)\}}$

Comparing Real and Imaginary terms on both sides, 

${\displaystyle \cos \left(a\right)\cosh \left(b\right)=0}$           ...............(2) ${\displaystyle \& -\sin \left(a\right)\sin j\left(b\right)=\frac{3}{4}}$......(3) 

From (2), cos(a)=0 or cosh(b)=0, 

∴${\displaystyle a=\frac{\pi }{2}}$                  ....................(4)

${\displaystyle \text{From}\left(3\right)\&\left(4\right),-\sin \left(\frac{\pi }{2}\right)\sin \left(b\right)=\frac{3}{4}}$ 

∴ ${\displaystyle 1.\sinh \left(b\right)=\frac{-3}{4}}$ 

∴ b= sinh${\displaystyle {}^1\left(\frac{-3}{4}\right)}$  

= ${\displaystyle \log [\left(\frac{-3}{4}\right)+\sqrt{{\left(\frac{-3}{4}\right)}^2+1}]}$  ${\displaystyle \{\because {\sinh }^{-1}z=\log (z+\sqrt{z^2+1})\}}$

= ${\displaystyle \mathrm{Log}[\left(\frac{-3}{4}\right)+\sqrt{\frac{9}{16}+1}]}$ 

= ${\displaystyle \log [\left(\frac{-3}{4}\right)+\frac{5}{4}]}$

=${\displaystyle \log \frac{1}{2}}$ 

=${\displaystyle {\log 2}^{-1}}$ 

∴ ${\displaystyle b=-\log 2}$                            ........(5) 

Substituting (4) & (5) in (1), ${\displaystyle {\cos }^{-1}\left(\frac{3i}{4}\right)=\frac{\pi }{2}-i\log 2}$

Comparing Real and Imaginary terms on both sides, 

Real part${\displaystyle =a=\frac{\pi }{2}}$ 

Imaginary part${\displaystyle =b=-\log 2}$