How do you simplify $[sqrt2(cos((7pi)/4)+isin((7pi)/4))]div[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))]$ and express the result in rectangular form?

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Answer 1 · 2018-04-21T01:00:58

$[sqrt2(cos((7pi)/4)+isin((7pi)/4))]div[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))]$

$=[sqrt2(cos((7pi)/4)+isin((7pi)/4))]/[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))]$

$=[2(cos((7pi)/4)+isin((7pi)/4))]/[(cos((3pi)/4)+isin((3pi)/4))]$

$=[2e^(i(7pi)/4)]/[e^(i(3pi)/4)]$

$=2e^(i((7pi)/4-(3pi)/4)]$

$=2e^(ipi)$

$=2(cospi+isinpi)$

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

Answer 2 · 2018-04-21T01:09:35

The answer is $-2+0i$

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

So our number will have modulus

$sqrt2/(sqrt2/2)=1/(1/2)=2$

and argument of

$7/4pi-3/4pi=4/4pi=pi$

So our number will be

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

How do you simplify [sqrt2(cos((7pi)/4)+isin((7pi)/4))]div[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))][sqrt2(cos((7pi)/4)+isin((7pi)/4))]div[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))] and express the result in rectangular form?