Jay Mckay

2022-02-01

How do you simplify $\left[\sqrt{2}\left(\mathrm{cos}\left(\frac{7\pi }{4}\right)+i\mathrm{sin}\left(\frac{7\pi }{4}\right)\right)\right]÷\left[\frac{\sqrt{2}}{2}\left(\mathrm{cos}\left(\frac{3\pi }{4}\right)+i\mathrm{sin}\left(\frac{3\pi }{4}\right)\right)\right]$
and express the result in rectangular form?

fumanchuecf

Expert

Step 1
$\frac{\sqrt{2}\left(\mathrm{cos}\left(\frac{7\pi }{4}\right)+i\mathrm{sin}\left(\frac{7\pi }{4}\right)\right)}{\frac{\sqrt{2}}{2}\left(\mathrm{cos}\left(\frac{3}{\pi }\right)+i\mathrm{sin}\left(\frac{3\pi }{4}\right)\right)}$
$=\frac{2\left(\mathrm{cos}\left(\frac{7\pi }{4}\right)+i\mathrm{sin}\left(\frac{7\pi }{4}\right)\right)}{\left(\mathrm{cos}\left(\frac{3\pi }{4}\right)+i\mathrm{sin}\left(\frac{3\pi }{4}\right)\right)}$
Step 2
$=\frac{2{e}^{i\frac{7\pi }{4}}}{{e}^{i\frac{3\pi }{4}}}$
$=2{e}^{i\left(\frac{7\pi }{4}-\frac{3\pi }{4}\right)}$
$=2{e}^{i\pi }$
$=2\left(\mathrm{cos}\pi +i\mathrm{sin}\pi \right)$
$=2\left(-1+i\cdot 0\right)=-2+i\cdot 0$

Expert

Two divide two complex numbers, we divide their moduli and subtract their arguments.
So our number will have modulus
$\frac{\sqrt{2}}{\frac{\sqrt{2}}{2}}=\frac{1}{\frac{1}{2}}=2$
and argument of
$\frac{7}{4}\pi -\frac{3}{4}\pi =\frac{4}{4}\pi =\pi$
So our number will be
$2\left(\mathrm{cos}\pi +i\mathrm{sin}\pi \right)=2\left(-1+0i\right)=-2+0i$

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