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#