How do you simplify [2(cos⁡(7π4)+isin⁡(7π4))]÷[22(cos⁡(3π4)+isin⁡(3π4))] and express the result in rectangular form?

Step 1 2(cos⁡(7π4)+isin⁡(7π4))22(cos⁡(3π)+isin⁡(3π4)) =2(cos⁡(7π4)+isin⁡(7π4))(cos⁡(3π4)+isin⁡(3π4)) Step 2 =2ei7π4ei3π4 =2ei(7π4−3π4) =2eiπ =2(cos⁡π+isin⁡π) =2(−1+i⋅0)=−2+i⋅0

Two divide two complex numbers, we divide their moduli and subtract their arguments. So our number will have modulus 222=112=2 and argument of 74π−34π=44π=π So our number will be 2(cos⁡π+isin⁡π)=2(−1+0i)=−2+0i

Step 1 2(cos⁡(7π4)+isin⁡(7π4))22(cos⁡(3π)+isin⁡(3π4)) =2(cos⁡(7π4)+isin⁡(7π4))(cos⁡(3π4)+isin⁡(3π4)) Step 2 =2ei7π4ei3π4 =2ei(7π4−3π4) =2eiπ =2(cos⁡π+isin⁡π) =2(−1+i⋅0)=−2+i⋅0

Two divide two complex numbers, we divide their moduli and subtract their arguments. So our number will have modulus 222=112=2 and argument of 74π−34π=44π=π So our number will be 2(cos⁡π+isin⁡π)=2(−1+0i)=−2+0i

Resolved: "How do you simplify [2(cos⁡(7π4)+isin⁡(7π4))]÷[22(cos⁡(3π4)+isin⁡(3π4))] and express the..."

Jay Mckay

Answered question

2022-02-01

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}))]

and express the result in rectangular form?

Answer & Explanation

fumanchuecf

Beginner2022-02-02Added 21 answers

Step 1

\frac{\sqrt{2} (\cos (\frac{7 π}{4}) + i \sin (\frac{7 π}{4}))}{\frac{\sqrt{2}}{2} (\cos (\frac{3}{π}) + i \sin (\frac{3 π}{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}))}

Step 2

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

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

= 2 e^{i π}

= 2 (\cos π + i \sin π)

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

tainiaadjouctlw

Beginner2022-02-03Added 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} π - \frac{3}{4} π = \frac{4}{4} π = π

So our number will be

2 (\cos π + i \sin π) = 2 (- 1 + 0 i) = - 2 + 0 i

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