feminizmiki

## Answered question

2021-12-20

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

### Answer & Explanation

Philip Williams

Beginner2021-12-21Added 39 answers

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

Beginner2021-12-22Added 41 answers

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

Skilled2021-12-29Added 437 answers

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: $-\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}+i\left(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\right)=\mathrm{cos}\left(\frac{5\pi }{12}\right)+i\mathrm{sin}\left(\frac{5\pi }{12}\right)$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?