How do you simplify $[\sqrt{2} (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}))] \div [\frac{\sqrt{2}}{2} (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))]$ $[sqrt2(cos((7pi)/4)+isin((7pi)/4))]div[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))]$ and express the result in rectangular form?

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How do you simplify $[\sqrt{2} (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}))] \div [\frac{\sqrt{2}}{2} (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))]$ $[sqrt2(cos((7pi)/4)+isin((7pi)/4))]div[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))]$ and express the result in rectangular form?

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Answer 1 · 2018-04-21T01:00:58

$[\sqrt{2} (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}))] \div [\frac{\sqrt{2}}{2} (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))]$ $[sqrt2(cos((7pi)/4)+isin((7pi)/4))]div[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))]$

$= \frac{\sqrt{2} (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}))}{\frac{\sqrt{2}}{2} (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))}$ $=[sqrt2(cos((7pi)/4)+isin((7pi)/4))]/[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))]$

$= \frac{2 (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}))}{(cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))}$ $=[2(cos((7pi)/4)+isin((7pi)/4))]/[(cos((3pi)/4)+isin((3pi)/4))]$

$= \frac{2 e^{i \frac{7 π}{4}}}{e^{i \frac{3 π}{4}}}$ $=[2e^(i(7pi)/4)]/[e^(i(3pi)/4)]$

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

$= 2 e^{i π}$ $=2e^(ipi)$

$= 2 (cos π + i sin π)$ $=2(cospi+isinpi)$

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

Answer 2 · 2018-04-21T01:09:35

The answer is $- 2 + 0 i$ $-2+0i$

Explanation:

Two divide two complex numbers, we divide their moduli and subtract their arguments.

So our number will have modulus

$\frac{\sqrt{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{\frac{1}{2}} = 2$ $sqrt2/(sqrt2/2)=1/(1/2)=2$

and argument of

$\frac{7}{4} π - \frac{3}{4} π = \frac{4}{4} π = π$ $7/4pi-3/4pi=4/4pi=pi$

So our number will be

$2 (cos π + i sin π) = 2 (- 1 + 0 i) = - 2 + 0 i$ $2(cospi+isinpi)=2(-1+0i)=-2+0i$

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How do you simplify $[\sqrt{2} (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}))] \div [\frac{\sqrt{2}}{2} (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))]$ $[sqrt2(cos((7pi)/4)+isin((7pi)/4))]div[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))]$ and express the result in rectangular form?

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Explanation:

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How do you simplify $[\sqrt{2} (cos (\frac{7 π}{4}) + i sin (\frac{7 π}{4}))] \div [\frac{\sqrt{2}}{2} (cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4}))]$ $[sqrt2(cos((7pi)/4)+isin((7pi)/4))]div[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))]$ and express the result in rectangular form?