Jay Mckay

Answered

2022-02-01

How do you simplify $\left[\sqrt{2}(\mathrm{cos}\left(\frac{7\pi}{4}\right)+i\mathrm{sin}\left(\frac{7\pi}{4}\right))\right]\xf7\left[\frac{\sqrt{2}}{2}(\mathrm{cos}\left(\frac{3\pi}{4}\right)+i\mathrm{sin}\left(\frac{3\pi}{4}\right))\right]$

and express the result in rectangular form?

and express the result in rectangular form?

Answer & Explanation

fumanchuecf

Expert

2022-02-02Added 21 answers

Step 1

$\frac{\sqrt{2}(\mathrm{cos}\left(\frac{7\pi}{4}\right)+i\mathrm{sin}\left(\frac{7\pi}{4}\right))}{\frac{\sqrt{2}}{2}(\mathrm{cos}\left(\frac{3}{\pi}\right)+i\mathrm{sin}\left(\frac{3\pi}{4}\right))}$

$=\frac{2(\mathrm{cos}\left(\frac{7\pi}{4}\right)+i\mathrm{sin}\left(\frac{7\pi}{4}\right))}{(\mathrm{cos}\left(\frac{3\pi}{4}\right)+i\mathrm{sin}\left(\frac{3\pi}{4}\right))}$

Step 2

$=\frac{2{e}^{i\frac{7\pi}{4}}}{{e}^{i\frac{3\pi}{4}}}$

$=2{e}^{i(\frac{7\pi}{4}-\frac{3\pi}{4})}$

$=2{e}^{i\pi}$

$=2(\mathrm{cos}\pi +i\mathrm{sin}\pi )$

$=2(-1+i\cdot 0)=-2+i\cdot 0$

Step 2

tainiaadjouctlw

Expert

2022-02-03Added 14 answers

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

$2(\mathrm{cos}\pi +i\mathrm{sin}\pi )=2(-1+0i)=-2+0i$

So our number will have modulus

and argument of

So our number will be

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