Show that: 1^° |sin(x+iy)|=|sin x+sin iy|; 2^°…

Show that: $\ 1^\circ\ |\sin(x+iy)|=|\sin x+\sin iy|;\quad 2^\circ\ |\sinh(x+iy)|=|\sinh x+\sinh iy|$.

Key evaluations

$$ \sin(iy)=\frac{e^{i\cdot iy}-e^{-i\cdot iy}}{2i}=\frac{e^{-y}-e^{y}}{2i}=-\frac1i\sinh y=i\sinh y $$

$$ \sinh(iy)=\frac{e^{iy}-e^{-iy}}2=\frac{(\cos y+i\sin y)-(\cos y-i\sin y)}2=i\sin y $$

Part $1^\circ$

Left side, using the sine split formula: $\ |\sin(x+iy)|=\sqrt{\sin^2x+\sinh^2y}$ (proved earlier).

Right side: $\ \sin x+\sin(iy)=\sin x+i\sinh y$, so $\ |\sin x+\sin iy|=\sqrt{\sin^2x+\sinh^2y}$.

$$ \boxed{|\sin(x+iy)|=|\sin x+\sin iy|=\sqrt{\sin^2x+\sinh^2y}} $$

Part $2^\circ$

Left side: from $\sinh(x+iy)=\sinh x\cos y+i\cosh x\sin y$ and $\cosh^2x=1+\sinh^2x$:

$$ |\sinh(x+iy)|^2=\sinh^2x\cos^2y+\cosh^2x\sin^2y=\sinh^2x+\sin^2y $$

Right side: $\ \sinh x+\sinh(iy)=\sinh x+i\sin y$, so $\ |\sinh x+\sinh iy|=\sqrt{\sinh^2x+\sin^2y}$.

$$ \boxed{|\sinh(x+iy)|=|\sinh x+\sinh iy|=\sqrt{\sinh^2x+\sin^2y}} $$

Why this works

Both identities hold because in each case the "cross term" that would normally appear when expanding a modulus of a sum vanishes: the two summands ($\sin x$ real, $\sin(iy)=i\sinh y$ purely imaginary — likewise for $\sinh$) sit on perpendicular axes, so $|a+ib|=\sqrt{a^2+b^2}$ applies directly and coincides with the modulus of the full complex value.