Question

The imaginary part of $sinh(x+iy)$ is

• A. $cosxsinhy$
• B. $-coshxsiny$
• C. $coshxsiny$
• D. None of these

Answer 1

Answer: (C)

$\sinh (x+iy)=\frac{1}{i}\sin [i(x+iy)]$
$[\because \sinh (\theta )=i\sin i\theta ]$
$=-i\sin (ix-y)$
$=-i[\sin (ix)\cos y-\cos (ix)\sin y]$
$=-i[i\sin x.\cos y-\cosh x\sin y]$
$=-i^2\sin x.\cos y+i\cosh x\sin y$
$=\sinh x\cos y+i\cosh x\sin y$
$\therefore$ Imaginary part of $\sinh (x+iy)=\cosh x\sin y$