Prove $|\cos^2(z)| + |\sin^2(z)| > 1$ for complex numbers $z$ with nonzero imaginary part

Question

Prove $$|\cos^2(z)| + |\sin^2(z)| > 1$$ for $\operatorname{Im}(z) \ne 0$

I know from using the triangle inequality, $|x+y| \leq |x| + |y|$, that $|\cos^2(z)| + |\sin^2(z)| \geq 1$ but I don't know how to carry on to show that it is strictly greater than.

Answer 1

Hint: Recall from geometry that$|z_0-z_2|=|z_0-z_1|+|z_1-z_2|$ means that $z_1$ is on the line between $z_0$ and $z_2$.

We know that $\cos^2(z)=1-\sin^2(z)$, so $$ \begin{align} |\sin^2(z)|+|\cos^2(z)| &=|\sin^2(z)-0|+|1-\sin^2(z)|\\ &=1 \end{align} $$ means that $\sin^2(z)$ is on the line between $0$ and $1$.

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Hint 2: Note that $$ \begin{align} \sin(x+iy)&=\sin(x)\cosh(y)+i\cos(x)\sinh(y)\\ |\sin(x+iy)|^2&=\sin^2(x)\cosh^2(y)+\cos^2(x)\sinh^2(y) \end{align} $$ and $$ \begin{align} \cos(x+iy)&=\cos(x)\cosh(y)-i\sin(x)\sinh(y)\\ |\cos(x+iy)|^2&=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y) \end{align} $$ Then recall that $\sin^2(x)+\cos^2(x)=1$ and $\cosh^2(y)+\sinh^2(y)=\cosh(2y)$.

Answer 2

for $z = x + iy$ with $y< 0$(see the comment) this can be always arranged because either $z$ or its conjugate $\bar z$ has negative imaginary part. we have $\cos(z) + i\sin(z) = e^{iz} = e^{-y}e^{ix}$ now take the absolute value of both sides and squaring it gives $$|\cos z|^2 + |\sin z|^2 = e^{-y} > 1.$$

edit: the above answer is wrong. see the comment by robjohn and my response.

Answer 3

Since $$ \sin(x+iy)=\sin x \cosh y +i\cos x \sinh y \\ \cos(x+iy)=\cos x \cosh y -i\sin x \sinh y, $$ you have $$ \left|\sin(x+iy)\right|^2\ge \sin^2 x \cosh^2 y \\ \left|\cos(x+iy)\right|^2\ge \cos^2 x \cosh^2 y, $$ and $$ \left|\sin(x+iy)\right|^2 + \left|\cos(x+iy)\right|^2\ge (\cos^2 x + \sin^2 x)\cosh^2 y=\cosh^2 y. $$ You can conclude that $$ \left|\sin z\right|^2 + \left|\cos z\right|^2\ge \cosh^2 \left({\text{Im }}z\right). $$ Since $\cosh x > 1$ for all $x\neq 0$, the desired result follows.