Question

Separate into real and imaginary parts of cos`"^-1((3i)/4)` 

 

Answer 1

Let a+ib= cos`"^-1((3i)/4)`         .................(1) 

∴` cos(a+ib)=(3i)/4`

∴` cos (a) cos(ib)-sin (a) sin(ib)=(3i)/4` 

cos(a)cosh(b) – isin(a)sinh(b) =`0+(3i)/4`  `{∵ cos (ix)=cosh(ix)=sinh(x)}`

Comparing Real and Imaginary terms on both sides, 

`cos(a)cosh(b)=0`           ...............(2) `   &   -sin(a)sinj(b)=3/4`......(3) 

From (2), cos(a)=0 or cosh(b)=0, 

∴` a=pi/2`                  ....................(4)

`"From" (3) & (4), -sin(pi/2)sin(b)=3/4` 

∴ `1.sinh(b)=(-3)/4` 

∴ b= sinh`"^1((-3)/4)`  

= `log[((-3)/4)+sqrt(((-3)/4)^2+1)]`  `{∵ sinh"^-1z=log (z+sqrt(z^2+1))}`

= `Log [((-3)/4)+sqrt(9/16+1)]` 

= `log[((-3)/4)+5/4]`

=`log  1/2` 

=` log2^-1` 

∴ `b=-log2`                            ........(5) 

Substituting (4) & (5) in (1), `cos^-1((3i)/4)=pi/2-i log2`

Comparing Real and Imaginary terms on both sides, 

Real part`=a= pi/2` 

Imaginary part`=b=-log2`