feminizmiki

2021-12-20

Find the product of the complex numbers ${z}_{1}=\mathrm{cos}\frac{\pi }{6}+i\mathrm{sin}\frac{\pi }{6}$ and ${z}_{2}=\mathrm{cos}\frac{\pi }{4}+i\mathrm{sin}\frac{\pi }{4}$

Philip Williams

Step 1
The given two complex numbers are:
${z}_{1}=\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}\left(\frac{\pi }{6}\right)$
${z}_{2}=\mathrm{cos}\left(\frac{\pi }{4}+i\mathrm{sin}\left(\frac{\pi }{4}\right)$
we have to find the product of the given two complex numbers, we have to write the answer in polar form.
Step 2
Therefore,
${z}_{1}{z}_{2}=\left(\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}\left(\frac{\pi }{6}\right)\right)\left(\mathrm{cos}\left(\frac{\pi }{4}\right)+i\mathrm{sin}\left(\frac{\pi }{4}\right)\right)$
$=\mathrm{cos}\left(\frac{\pi }{6}\right)\mathrm{cos}\left(\frac{\pi }{4}\right)+i\mathrm{cos}\left(\frac{\pi }{6}\right)\mathrm{sin}\left(\frac{\pi }{4}\right)$
$+i\mathrm{sin}\left(\frac{\pi }{6}\right)\mathrm{cos}\left(\frac{\pi }{4}\right)+{i}^{2}\mathrm{sin}\left(\frac{\pi }{6}\right)\mathrm{sin}\left(\frac{\pi }{4}\right)$
$=\mathrm{cos}\left(\frac{\pi }{6}\right)\mathrm{cos}\left(\frac{\pi }{4}\right)+{i}^{2}\mathrm{sin}\left(\frac{\pi }{6}\right)\mathrm{sin}\left(\frac{\pi }{4}\right)$
$+i\left(\mathrm{cos}\left(\frac{\pi }{6}\right)\mathrm{sin}\left(\frac{\pi }{4}\right)+\mathrm{sin}\left(\frac{\pi }{6}\right)\mathrm{cos}\left(\frac{\pi }{4}\right)\right)$
$=\mathrm{cos}\left(\frac{\pi }{6}\right)\mathrm{cos}\left(\frac{\pi }{4}\right)+\left(-1\right)\mathrm{sin}\left(\frac{\pi }{6}\right)\mathrm{sin}\left(\frac{\pi }{4}\right)$
$+i\left(\mathrm{cos}\left(\frac{\pi }{6}\right)\mathrm{sin}\left(\frac{\pi }{4}\right)+\mathrm{sin}\left(\frac{\pi }{6}\right)\mathrm{cos}\left(\frac{\pi }{4}\right)\right)$ (because ${i}^{2}=-1$)

Donald Cheek

Step 1
The equation complex numbers:
${z}_{1}×{z}_{2}=\left(\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}\left(\frac{\pi }{6}\right)\right)×\left(\mathrm{cos}\left(\frac{\pi }{4}\right)+i\mathrm{sin}\left(\frac{\pi }{4}\right)\right):$
$=\frac{\sqrt{2}\left(-1+\sqrt{3}\right)}{4}+i\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{4}$
Answer: $\frac{\sqrt{2}\left(-1+\sqrt{3}\right)}{4}+i\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{4}$

RizerMix

Step 1
Your input: simplify and calculate different forms of
$\left(\mathrm{cos}\left(\frac{\pi }{4}\right)+i\mathrm{sin}\left(\frac{\pi }{4}\right)\right)\left(\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}\left(\frac{\pi }{6}\right)\right)$
Rewrite:
$=\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}i}{2}\right)\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)$
Use FOIL to multiply (for steps, see foil calculator), don't forget that ${i}^{2}=-1$
$\left(\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}i}{2}\right)\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)\right)=\left(-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+i\left(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\right)\right)$
Simplify:
$\left(-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+i\left(\frac{\sqrt{2}}{4}+\frac{6}{4}\right)\right)=\left(-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+\frac{i\left(\sqrt{2}+\sqrt{6}\right)}{4}\right)$
Hence: $=-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+\frac{i\left(\sqrt{2}+\sqrt{6}\right)}{4}$
Step 2
Polar form
For a complex number a+bi, polar form given by $r\left(\mathrm{cos}\left(\theta \right)+i\mathrm{sin}\left(\theta \right)\right)$, where $r=\sqrt{{a}^{2}+{b}^{2}}$ and $\theta =a\mathrm{tan}\left(\frac{b}{a}\right)$
We have that
Thus, $r=\sqrt{\left(-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}{\right)}^{2}+\left(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}{\right)}^{2}}=1$
Also, $\theta =a\mathrm{tan}\left(\frac{\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}}{-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}}\right)=\frac{5\pi }{12}$
Therefore: $-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+i\left(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\right)=\mathrm{cos}\left(\frac{5\pi }{12}\right)+i\mathrm{sin}\left(\frac{5\pi }{12}\right)$

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