Complex sin

Algebra Level 2

$\large \Re ( \sin(x+iy)) = \ ?$

Notations:

$\Re(\cdot)$ denotes the real part of a complex number .
$i = \sqrt {-1}$ denotes the imaginary unit .

\sinh(x)\cos(y)

\sinh(x+iy)

\sinh(x)\cosh(y)

\cosh(x+iy)

\sin(x)\cosh(y)

\cos(x+iy)

\sin(x)

2 solutions

ابراهيم فقرا
Dec 9, 2018

The real part is in green .

You should put a backslash "\" before function for example \Re $\Re$ , \sin x $\sin x$ , \sinh x $\sinh x$ , \cos x $\cos x$ and \cosh x $\cosh x$ . Note that the function names $\sin$ , $\cos$ , $\sinh$ , $\cosh$ are not in italic (slanding) but the variable $x$ is in italic.

Chew-Seong Cheong - 2 years, 6 months ago

Chew-Seong Cheong
Dec 10, 2018

$\begin{aligned} \Re (\sin(x+iy)) & = \Re \left(\frac {i\left(e^{-i(x+iy)} - e^{i(x+iy)}\right)}2\right) \\ & = \Re \left(\frac {i\left(e^{-ix+y} - e^{ix-y)}\right)}2\right) \\ & = \Re \left(\frac {i\left(e^y(\cos x - i\sin x) - e^{-y}(\cos x + i\sin x)\right)}2\right) \\ & = \Re \left(\frac {\sin x \left(e^y + e^{-y}\right) + i \cos x \left(e^y - e^{-y}\right)}2 \right) \\ & = \frac {\sin x \left(e^y + e^{-y} \right)}2 \\ & = \boxed{\sin x \cosh y} \end{aligned}$

Complex sin

2 solutions

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