Answered

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Let z = equals 38 (cosine (startfraction pi over 8 endfraction) i sine (startfraction pi over 8 endfraction) ) and w = 2 (cosine (startfraction pi over 16 endfraction) i sine (startfraction pi over 16 endfraction) ) . what is the product of zw?

Sagot :

It sounds like you're saying

[tex]z = \dfrac38 \left(\cos\left(\dfrac\pi8\right) + i \sin\left(\dfrac\pi8\right)\right)[/tex]

[tex]w = 2 \left(\cos\left(\dfrac\pi{16}\right) + i \sin\left(\dfrac\pi{16}\right)\right)[/tex]

The product [tex]zw[/tex] is obtained by multiplying the moduli and adding the arguments. In other words

[tex]z = |z| e^{i\arg(z)} \text{ and } w = |w| e^{i\arg(w)} \implies zw = |z||w| e^{i(\arg(z)+\arg(w))}[/tex]

where [tex]e^{it}=\cos(t)+i\sin(t)[/tex], so that

[tex]zw = \dfrac38\times2 \left(\cos\left(\dfrac\pi8+\dfrac\pi{16}\right) + i \sin\left(\dfrac\pi8 + \dfrac\pi{16}\right)\right) = \boxed{\dfrac34 \left(\cos\left(\dfrac{3\pi}{16}\right) + i \sin\left(\dfrac{3\pi}{16}\right)\right)}[/tex]

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