A) \[cos\text{ }x\text{ }sinh\text{ }y\]
B) \[-cosh\text{ }x\text{ }sin\text{ }y\]
C) \[cosh\text{ }x\text{ }sin\text{ }y\]
D) None of these
Correct Answer: C
Solution :
\[\sinh (x+iy)=\frac{1}{i}\sin [i(x+iy)]\] \[[\because \sinh (\theta )=i\sin i\theta ]\] \[=-i\sin (ix-y)\] \[=-i[\sin (ix)\cos y-\cos (ix)\sin y]\] \[=-i[i\sin x.\cos y-\cosh x\sin y]\] \[=-{{i}^{2}}\sin x.\cos y+i\cosh x\sin y\] \[=\sinh x\cos y+i\cosh x\sin y\] \[\therefore \]Imaginary part of\[\sinh (x+iy)=\cosh x\sin y\]
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