Write sin(-1+2i) in the form u+iv
Write $\sin(-1+2i)$ in the form $u+iv$.
Explanation of Method
Use the standard identity $\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y$ (derived by substituting Euler's formula into the definition $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$), then evaluate numerically.
Step 1 · The identity for complex sine
$$ \sin(x+iy)=\sin x\cosh y+i\cos x\sinh y $$
(derived from $\sin z=\sin x\cos(iy)+\cos x\sin(iy)$, then $\cos(iy)=\cosh y,\ \sin(iy)=i\sinh y$.)
Step 2 · Apply with $x=-1,\ y=2$
$$ \sin(-1+2i)=\sin(-1)\cosh(2)+i\cos(-1)\sinh(2)=-\sin(1)\cosh(2)+i\cos(1)\sinh(2) $$
(using $\sin(-1)=-\sin(1)$, $\cos(-1)=\cos(1)$.)
Step 3 · Numerical evaluation
$$ \sin(1)\approx0.841471,\quad \cos(1)\approx0.540302,\quad \cosh(2)=\frac{e^2+e^{-2}}2\approx3.762196,\quad \sinh(2)=\frac{e^2-e^{-2}}2\approx3.626860 $$
$$ u=-\sin(1)\cosh(2)\approx-3.16578, \qquad v=\cos(1)\sinh(2)\approx1.95960 $$
Step 4 · Result
$$ \boxed{\sin(-1+2i)=-\sin(1)\cosh(2)+i\cos(1)\sinh(2)\approx-3.1658+1.9596i} $$
Verification
Cross-checked via the exponential definition $\sin z=\dfrac{e^{iz}-e^{-iz}}{2i}$ with $z=-1+2i$: $\ iz=-2-i$, so $e^{iz}=e^{-2}(\cos(-1)+i\sin(-1))$ and $e^{-iz}=e^2(\cos1+i\sin1)$; combining and dividing by $2i$ reproduces the same numerical value to high precision.
Result
Unlike the real sine, the complex sine is unbounded — its magnitude grows like $\sinh|y|$ as the imaginary part $y$ grows, which is exactly what appears here via the $\cosh(2)$ and $\sinh(2)$ factors.