Proving <mrow class="MJX-TeXAtom-ORD"> | </mrow> cos &#x2061;<!-- ⁡ --> z <mtext>&#

Yasmin Camacho

Yasmin Camacho

Answered question

2022-05-19

Proving | cos z   | 2 + | sin z   | 2 1
My attempt:
For z:=x+iy ,
cos z = cos x cos i y sin x sin i y = cos x cosh y i sin x sinh y sin z = sin x cosh y + i cos x sinh y
So
| cos z   | 2 + | sin z   | 2 = cos 2 x cosh 2 y + sin 2 x sinh 2 y + cosh 2 y sin 2 x + cos 2 x sinh 2 y = cos 2 x ( cosh 2 y + sinh 2 y ) + sin 2 x ( cosh 2 y + sinh 2 y ) = cos 2 y 1.
I'm not sure where I went wrong, and a numerical check on Wolfram Alpha shows that the inequality should be as suggested.

Answer & Explanation

Julien Carrillo

Julien Carrillo

Beginner2022-05-20Added 13 answers

You're almost there:
cos 2 x ( cosh 2 y + sinh 2 y ) + sin 2 x ( cosh 2 y + sinh 2 y ) =
( cos 2 x + sin 2 x ) ( cosh 2 y + sinh 2 y ) =
( cosh 2 y + sinh 2 y ) =
2 sinh 2 y + 1 1
The last step uses the hyperbolic identity
cosh 2 y sinh 2 y = 1
(note the sign difference with the more familiar goniometric one)

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