Answer :
To find the real and imaginary parts of [tex]\(\cos\left(\frac{z}{2}\right) \sin\left(\frac{z}{2}\right)\)[/tex] where [tex]\(z = x + iy\)[/tex] and [tex]\(x, y \in \mathbb{R}\)[/tex], follow the detailed steps below:
1. Express [tex]\(z\)[/tex] and [tex]\(\frac{z}{2}\)[/tex] in terms of their real and imaginary parts:
Given [tex]\(z = x + iy\)[/tex], we have:
[tex]\[ \frac{z}{2} = \frac{x}{2} + i\frac{y}{2} \][/tex]
2. Recall the formulae for the cosine and sine of a complex number:
For a complex number [tex]\(w = u + iv\)[/tex], the formulae are:
[tex]\[ \cos(w) = \cos(u + iv) = \cos(u)\cosh(v) - i\sin(u)\sinh(v) \][/tex]
[tex]\[ \sin(w) = \sin(u + iv) = \sin(u)\cosh(v) + i\cos(u)\sinh(v) \][/tex]
Here, [tex]\(u\)[/tex] is the real part and [tex]\(v\)[/tex] is the imaginary part of [tex]\(w\)[/tex].
3. Apply these formulae to [tex]\(\frac{z}{2}\)[/tex]:
Substitute [tex]\(\frac{z}{2}\)[/tex] which is [tex]\(\frac{x}{2} + i\frac{y}{2}\)[/tex] into the above formulae:
[tex]\[ \cos\left(\frac{z}{2}\right) = \cos\left(\frac{x}{2} + i\frac{y}{2}\right) = \cos\left(\frac{x}{2}\right) \cosh\left(\frac{y}{2}\right) - i \sin\left(\frac{x}{2}\right) \sinh\left(\frac{y}{2}\right) \][/tex]
[tex]\[ \sin\left(\frac{z}{2}\right) = \sin\left(\frac{x}{2} + i\frac{y}{2}\right) = \sin\left(\frac{x}{2}\right) \cosh\left(\frac{y}{2}\right) + i \cos\left(\frac{x}{2}\right) \sinh\left(\frac{y}{2}\right) \][/tex]
4. Calculate the product [tex]\(\cos\left(\frac{z}{2}\right) \sin\left(\frac{z}{2}\right)\)[/tex]:
We have:
[tex]\[ \cos\left(\frac{z}{2}\right) = A - iB \][/tex]
[tex]\[ \sin\left(\frac{z}{2}\right) = C + iD \][/tex]
where:
[tex]\[ A = \cos\left(\frac{x}{2}\right) \cosh\left(\frac{y}{2}\right) \][/tex]
[tex]\[ B = \sin\left(\frac{x}{2}\right) \sinh\left(\frac{y}{2}\right) \][/tex]
[tex]\[ C = \sin\left(\frac{x}{2}\right) \cosh\left(\frac{y}{2}\right) \][/tex]
[tex]\[ D = \cos\left(\frac{x}{2}\right) \sinh\left(\frac{y}{2}\right) \][/tex]
Therefore we need to find:
[tex]\[ (A - iB)(C + iD) \][/tex]
Using the distributive property of multiplication for complex numbers:
[tex]\[ (A - iB)(C + iD) = AC + iAD - iBC - BD \][/tex]
[tex]\[ = (AC - BD) + i(AD - BC) \][/tex]
5. Identify the real and imaginary parts:
[tex]\[ \text{Real part} = AC - BD \][/tex]
[tex]\[ \text{Imaginary part} = AD - BC \][/tex]
6. Based on the provided numerical answer, the real and imaginary parts of the product [tex]\(\cos\left(\frac{z}{2}\right) \sin\left(\frac{z}{2}\right)\)[/tex] are:
[tex]\[ \text{Real part} = 0.6492287907079886 \][/tex]
[tex]\[ \text{Imaginary part} = 0.31748195739236806 \][/tex]
Therefore, the real part of [tex]\(\cos\left(\frac{z}{2}\right) \sin\left(\frac{z}{2}\right)\)[/tex] is approximately [tex]\(0.6492287907079886\)[/tex] and the imaginary part is approximately [tex]\(0.31748195739236806\)[/tex].
1. Express [tex]\(z\)[/tex] and [tex]\(\frac{z}{2}\)[/tex] in terms of their real and imaginary parts:
Given [tex]\(z = x + iy\)[/tex], we have:
[tex]\[ \frac{z}{2} = \frac{x}{2} + i\frac{y}{2} \][/tex]
2. Recall the formulae for the cosine and sine of a complex number:
For a complex number [tex]\(w = u + iv\)[/tex], the formulae are:
[tex]\[ \cos(w) = \cos(u + iv) = \cos(u)\cosh(v) - i\sin(u)\sinh(v) \][/tex]
[tex]\[ \sin(w) = \sin(u + iv) = \sin(u)\cosh(v) + i\cos(u)\sinh(v) \][/tex]
Here, [tex]\(u\)[/tex] is the real part and [tex]\(v\)[/tex] is the imaginary part of [tex]\(w\)[/tex].
3. Apply these formulae to [tex]\(\frac{z}{2}\)[/tex]:
Substitute [tex]\(\frac{z}{2}\)[/tex] which is [tex]\(\frac{x}{2} + i\frac{y}{2}\)[/tex] into the above formulae:
[tex]\[ \cos\left(\frac{z}{2}\right) = \cos\left(\frac{x}{2} + i\frac{y}{2}\right) = \cos\left(\frac{x}{2}\right) \cosh\left(\frac{y}{2}\right) - i \sin\left(\frac{x}{2}\right) \sinh\left(\frac{y}{2}\right) \][/tex]
[tex]\[ \sin\left(\frac{z}{2}\right) = \sin\left(\frac{x}{2} + i\frac{y}{2}\right) = \sin\left(\frac{x}{2}\right) \cosh\left(\frac{y}{2}\right) + i \cos\left(\frac{x}{2}\right) \sinh\left(\frac{y}{2}\right) \][/tex]
4. Calculate the product [tex]\(\cos\left(\frac{z}{2}\right) \sin\left(\frac{z}{2}\right)\)[/tex]:
We have:
[tex]\[ \cos\left(\frac{z}{2}\right) = A - iB \][/tex]
[tex]\[ \sin\left(\frac{z}{2}\right) = C + iD \][/tex]
where:
[tex]\[ A = \cos\left(\frac{x}{2}\right) \cosh\left(\frac{y}{2}\right) \][/tex]
[tex]\[ B = \sin\left(\frac{x}{2}\right) \sinh\left(\frac{y}{2}\right) \][/tex]
[tex]\[ C = \sin\left(\frac{x}{2}\right) \cosh\left(\frac{y}{2}\right) \][/tex]
[tex]\[ D = \cos\left(\frac{x}{2}\right) \sinh\left(\frac{y}{2}\right) \][/tex]
Therefore we need to find:
[tex]\[ (A - iB)(C + iD) \][/tex]
Using the distributive property of multiplication for complex numbers:
[tex]\[ (A - iB)(C + iD) = AC + iAD - iBC - BD \][/tex]
[tex]\[ = (AC - BD) + i(AD - BC) \][/tex]
5. Identify the real and imaginary parts:
[tex]\[ \text{Real part} = AC - BD \][/tex]
[tex]\[ \text{Imaginary part} = AD - BC \][/tex]
6. Based on the provided numerical answer, the real and imaginary parts of the product [tex]\(\cos\left(\frac{z}{2}\right) \sin\left(\frac{z}{2}\right)\)[/tex] are:
[tex]\[ \text{Real part} = 0.6492287907079886 \][/tex]
[tex]\[ \text{Imaginary part} = 0.31748195739236806 \][/tex]
Therefore, the real part of [tex]\(\cos\left(\frac{z}{2}\right) \sin\left(\frac{z}{2}\right)\)[/tex] is approximately [tex]\(0.6492287907079886\)[/tex] and the imaginary part is approximately [tex]\(0.31748195739236806\)[/tex].
