Question : Find sinh(x+iy) formula
Solution :
Let the complex number as \[z=sinh\left(x+iy\right)\] We know that \[\sin{\left(iz\right)}=isinh\left(z\right)\] So we shall multiply and divide by i we get \[z=\frac{isinh\left(x+iy\right)}{i}=\frac{\sin{\left(i\left(x+iy\right)\right)}}{i}=\frac{\sin{\left(ix-y\right)}}{i}\] Now we shall use formula of sin(A-B) =sin(A)cos(b) -cos(A) sin(B) \[z=\frac{\sin{\left(ix-y\right)}}{i}=\frac{\sin{\left(ix\right)}\cos{\left(y\right)}-\cos{\left(ix\right)}\sin{\left(y\right)}}{i}\] Now we shall use \[Cos\left(ix\right)=\cosh{\left(x\right)}\ ,\ \ \sin{\left(ix\right)}=isinh\left(x\right)\] \[z=\frac{isinh{\left(x\right)}\cos{\left(y\right)}-\cosh{\left(x\right)}\sin{\left(y\right)}}{i}\ =\sinh{\left(x\right)}\cos{\left(y\right)}-\frac{1}{i}\cosh{\left(x\right)}\sin{\left(y\right)}\] \[z=\sinh{\left(x\right)}\cos{\left(y\right)}+i\cosh{\left(x\right)}\sin{\left(y\right)}\]