Simplify the complex expression…











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I am asked to simplify the complex expression $$frac{1}{2}(|{e^{i{theta}}-1}^2|+|{e^{i{theta}}+1}|)$$



I have gotten to $$frac{1}{2}((2-2costheta)+(2+2costheta))$$
1. Do I expand to get $$frac{1}{2}(4)$$ OR
2. Do I factor out the 2 and get $$frac{1}{2}(2(1-costheta)+2(1+costheta))$$



Is this answer complete?
Additional question: How would the method change if the argument was negative? i.e. if it was $$frac{1}{2}(|{e^{-i{theta}}-1}^2|+|{e^{-i{theta}}+1}|)$$






calculus complex-analysis analysis complex-numbers polar-coordinates







share|cite|improve this question
























  • What is $operatorname{mod}$ in this context?
    – José Carlos Santos
    Nov 21 at 10:34










  • It is meant to be Modulus, like the absolute value... I wasnt sure how else to do it. It should be around both terms.
    – Kayla Martin
    Nov 21 at 10:48












  • Do you mean $lvert e^{itheta}-1rvert$?
    – José Carlos Santos
    Nov 21 at 10:51










  • Yes, exactly. Same for the +1 term.
    – Kayla Martin
    Nov 21 at 10:52










  • Then why don't you just type |e^{itheta}-1|?
    – José Carlos Santos
    Nov 21 at 10:54















up vote
0
down vote

favorite












I am asked to simplify the complex expression $$frac{1}{2}(|{e^{i{theta}}-1}^2|+|{e^{i{theta}}+1}|)$$



I have gotten to $$frac{1}{2}((2-2costheta)+(2+2costheta))$$
1. Do I expand to get $$frac{1}{2}(4)$$ OR
2. Do I factor out the 2 and get $$frac{1}{2}(2(1-costheta)+2(1+costheta))$$



Is this answer complete?
Additional question: How would the method change if the argument was negative? i.e. if it was $$frac{1}{2}(|{e^{-i{theta}}-1}^2|+|{e^{-i{theta}}+1}|)$$






calculus complex-analysis analysis complex-numbers polar-coordinates







share|cite|improve this question
























  • What is $operatorname{mod}$ in this context?
    – José Carlos Santos
    Nov 21 at 10:34










  • It is meant to be Modulus, like the absolute value... I wasnt sure how else to do it. It should be around both terms.
    – Kayla Martin
    Nov 21 at 10:48












  • Do you mean $lvert e^{itheta}-1rvert$?
    – José Carlos Santos
    Nov 21 at 10:51










  • Yes, exactly. Same for the +1 term.
    – Kayla Martin
    Nov 21 at 10:52










  • Then why don't you just type |e^{itheta}-1|?
    – José Carlos Santos
    Nov 21 at 10:54













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am asked to simplify the complex expression $$frac{1}{2}(|{e^{i{theta}}-1}^2|+|{e^{i{theta}}+1}|)$$



I have gotten to $$frac{1}{2}((2-2costheta)+(2+2costheta))$$
1. Do I expand to get $$frac{1}{2}(4)$$ OR
2. Do I factor out the 2 and get $$frac{1}{2}(2(1-costheta)+2(1+costheta))$$



Is this answer complete?
Additional question: How would the method change if the argument was negative? i.e. if it was $$frac{1}{2}(|{e^{-i{theta}}-1}^2|+|{e^{-i{theta}}+1}|)$$






calculus complex-analysis analysis complex-numbers polar-coordinates







share|cite|improve this question















I am asked to simplify the complex expression $$frac{1}{2}(|{e^{i{theta}}-1}^2|+|{e^{i{theta}}+1}|)$$



I have gotten to $$frac{1}{2}((2-2costheta)+(2+2costheta))$$
1. Do I expand to get $$frac{1}{2}(4)$$ OR
2. Do I factor out the 2 and get $$frac{1}{2}(2(1-costheta)+2(1+costheta))$$



Is this answer complete?
Additional question: How would the method change if the argument was negative? i.e. if it was $$frac{1}{2}(|{e^{-i{theta}}-1}^2|+|{e^{-i{theta}}+1}|)$$






calculus complex-analysis analysis complex-numbers polar-coordinates




calculus complex-analysis analysis complex-numbers polar-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 4:44









naveen dankal

4,53321348




4,53321348










asked Nov 21 at 10:33









Kayla Martin

636




636












  • What is $operatorname{mod}$ in this context?
    – José Carlos Santos
    Nov 21 at 10:34










  • It is meant to be Modulus, like the absolute value... I wasnt sure how else to do it. It should be around both terms.
    – Kayla Martin
    Nov 21 at 10:48












  • Do you mean $lvert e^{itheta}-1rvert$?
    – José Carlos Santos
    Nov 21 at 10:51










  • Yes, exactly. Same for the +1 term.
    – Kayla Martin
    Nov 21 at 10:52










  • Then why don't you just type |e^{itheta}-1|?
    – José Carlos Santos
    Nov 21 at 10:54


















  • What is $operatorname{mod}$ in this context?
    – José Carlos Santos
    Nov 21 at 10:34










  • It is meant to be Modulus, like the absolute value... I wasnt sure how else to do it. It should be around both terms.
    – Kayla Martin
    Nov 21 at 10:48












  • Do you mean $lvert e^{itheta}-1rvert$?
    – José Carlos Santos
    Nov 21 at 10:51










  • Yes, exactly. Same for the +1 term.
    – Kayla Martin
    Nov 21 at 10:52










  • Then why don't you just type |e^{itheta}-1|?
    – José Carlos Santos
    Nov 21 at 10:54
















What is $operatorname{mod}$ in this context?
– José Carlos Santos
Nov 21 at 10:34




What is $operatorname{mod}$ in this context?
– José Carlos Santos
Nov 21 at 10:34












It is meant to be Modulus, like the absolute value... I wasnt sure how else to do it. It should be around both terms.
– Kayla Martin
Nov 21 at 10:48






It is meant to be Modulus, like the absolute value... I wasnt sure how else to do it. It should be around both terms.
– Kayla Martin
Nov 21 at 10:48














Do you mean $lvert e^{itheta}-1rvert$?
– José Carlos Santos
Nov 21 at 10:51




Do you mean $lvert e^{itheta}-1rvert$?
– José Carlos Santos
Nov 21 at 10:51












Yes, exactly. Same for the +1 term.
– Kayla Martin
Nov 21 at 10:52




Yes, exactly. Same for the +1 term.
– Kayla Martin
Nov 21 at 10:52












Then why don't you just type |e^{itheta}-1|?
– José Carlos Santos
Nov 21 at 10:54




Then why don't you just type |e^{itheta}-1|?
– José Carlos Santos
Nov 21 at 10:54










1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










The complet answer is $2$. That follows from both of your approaches. That is$$(forallthetainmathbb R):lvert e^{itheta}-1rvert^2+lvert e^{-itheta}-1rvert^2=2.$$Note that this is for all real numbers $theta$. Therefore, it makes no sense to ask what happens if we deal with $-theta$ instead of $theta$.






share|cite|improve this answer























  • I have a different question that is exactly the same as this but the argument is negative (like I added toward the bottom of the question) so that is why I was asking. Are you saying that the answer will be the same for both?
    – Kayla Martin
    Nov 21 at 12:04












  • If you prove something for an arbitrary $thetainmathbb R$, you shall have proved it both for positive and for negative real numbers.
    – José Carlos Santos
    Nov 21 at 12:10










  • Is this because of the modulus in this question? I don't quite get what you mean sorry. I am doing past exam papers so I have no answers to refer to either.
    – Kayla Martin
    Nov 21 at 12:14










  • I cannot help you here, since I don't understand where your doubt is.
    – José Carlos Santos
    Nov 21 at 12:35










  • Is it a different answer for a negative argument or is it the same if everything else in the question is the same?
    – Kayla Martin
    Nov 21 at 12:39











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1 Answer
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1 Answer
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active

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up vote
0
down vote



accepted










The complet answer is $2$. That follows from both of your approaches. That is$$(forallthetainmathbb R):lvert e^{itheta}-1rvert^2+lvert e^{-itheta}-1rvert^2=2.$$Note that this is for all real numbers $theta$. Therefore, it makes no sense to ask what happens if we deal with $-theta$ instead of $theta$.






share|cite|improve this answer























  • I have a different question that is exactly the same as this but the argument is negative (like I added toward the bottom of the question) so that is why I was asking. Are you saying that the answer will be the same for both?
    – Kayla Martin
    Nov 21 at 12:04












  • If you prove something for an arbitrary $thetainmathbb R$, you shall have proved it both for positive and for negative real numbers.
    – José Carlos Santos
    Nov 21 at 12:10










  • Is this because of the modulus in this question? I don't quite get what you mean sorry. I am doing past exam papers so I have no answers to refer to either.
    – Kayla Martin
    Nov 21 at 12:14










  • I cannot help you here, since I don't understand where your doubt is.
    – José Carlos Santos
    Nov 21 at 12:35










  • Is it a different answer for a negative argument or is it the same if everything else in the question is the same?
    – Kayla Martin
    Nov 21 at 12:39















up vote
0
down vote



accepted










The complet answer is $2$. That follows from both of your approaches. That is$$(forallthetainmathbb R):lvert e^{itheta}-1rvert^2+lvert e^{-itheta}-1rvert^2=2.$$Note that this is for all real numbers $theta$. Therefore, it makes no sense to ask what happens if we deal with $-theta$ instead of $theta$.






share|cite|improve this answer























  • I have a different question that is exactly the same as this but the argument is negative (like I added toward the bottom of the question) so that is why I was asking. Are you saying that the answer will be the same for both?
    – Kayla Martin
    Nov 21 at 12:04












  • If you prove something for an arbitrary $thetainmathbb R$, you shall have proved it both for positive and for negative real numbers.
    – José Carlos Santos
    Nov 21 at 12:10










  • Is this because of the modulus in this question? I don't quite get what you mean sorry. I am doing past exam papers so I have no answers to refer to either.
    – Kayla Martin
    Nov 21 at 12:14










  • I cannot help you here, since I don't understand where your doubt is.
    – José Carlos Santos
    Nov 21 at 12:35










  • Is it a different answer for a negative argument or is it the same if everything else in the question is the same?
    – Kayla Martin
    Nov 21 at 12:39













up vote
0
down vote



accepted







up vote
0
down vote



accepted






The complet answer is $2$. That follows from both of your approaches. That is$$(forallthetainmathbb R):lvert e^{itheta}-1rvert^2+lvert e^{-itheta}-1rvert^2=2.$$Note that this is for all real numbers $theta$. Therefore, it makes no sense to ask what happens if we deal with $-theta$ instead of $theta$.






share|cite|improve this answer














The complet answer is $2$. That follows from both of your approaches. That is$$(forallthetainmathbb R):lvert e^{itheta}-1rvert^2+lvert e^{-itheta}-1rvert^2=2.$$Note that this is for all real numbers $theta$. Therefore, it makes no sense to ask what happens if we deal with $-theta$ instead of $theta$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 12:35

























answered Nov 21 at 11:41









José Carlos Santos

147k22117218




147k22117218












  • I have a different question that is exactly the same as this but the argument is negative (like I added toward the bottom of the question) so that is why I was asking. Are you saying that the answer will be the same for both?
    – Kayla Martin
    Nov 21 at 12:04












  • If you prove something for an arbitrary $thetainmathbb R$, you shall have proved it both for positive and for negative real numbers.
    – José Carlos Santos
    Nov 21 at 12:10










  • Is this because of the modulus in this question? I don't quite get what you mean sorry. I am doing past exam papers so I have no answers to refer to either.
    – Kayla Martin
    Nov 21 at 12:14










  • I cannot help you here, since I don't understand where your doubt is.
    – José Carlos Santos
    Nov 21 at 12:35










  • Is it a different answer for a negative argument or is it the same if everything else in the question is the same?
    – Kayla Martin
    Nov 21 at 12:39


















  • I have a different question that is exactly the same as this but the argument is negative (like I added toward the bottom of the question) so that is why I was asking. Are you saying that the answer will be the same for both?
    – Kayla Martin
    Nov 21 at 12:04












  • If you prove something for an arbitrary $thetainmathbb R$, you shall have proved it both for positive and for negative real numbers.
    – José Carlos Santos
    Nov 21 at 12:10










  • Is this because of the modulus in this question? I don't quite get what you mean sorry. I am doing past exam papers so I have no answers to refer to either.
    – Kayla Martin
    Nov 21 at 12:14










  • I cannot help you here, since I don't understand where your doubt is.
    – José Carlos Santos
    Nov 21 at 12:35










  • Is it a different answer for a negative argument or is it the same if everything else in the question is the same?
    – Kayla Martin
    Nov 21 at 12:39
















I have a different question that is exactly the same as this but the argument is negative (like I added toward the bottom of the question) so that is why I was asking. Are you saying that the answer will be the same for both?
– Kayla Martin
Nov 21 at 12:04






I have a different question that is exactly the same as this but the argument is negative (like I added toward the bottom of the question) so that is why I was asking. Are you saying that the answer will be the same for both?
– Kayla Martin
Nov 21 at 12:04














If you prove something for an arbitrary $thetainmathbb R$, you shall have proved it both for positive and for negative real numbers.
– José Carlos Santos
Nov 21 at 12:10




If you prove something for an arbitrary $thetainmathbb R$, you shall have proved it both for positive and for negative real numbers.
– José Carlos Santos
Nov 21 at 12:10












Is this because of the modulus in this question? I don't quite get what you mean sorry. I am doing past exam papers so I have no answers to refer to either.
– Kayla Martin
Nov 21 at 12:14




Is this because of the modulus in this question? I don't quite get what you mean sorry. I am doing past exam papers so I have no answers to refer to either.
– Kayla Martin
Nov 21 at 12:14












I cannot help you here, since I don't understand where your doubt is.
– José Carlos Santos
Nov 21 at 12:35




I cannot help you here, since I don't understand where your doubt is.
– José Carlos Santos
Nov 21 at 12:35












Is it a different answer for a negative argument or is it the same if everything else in the question is the same?
– Kayla Martin
Nov 21 at 12:39




Is it a different answer for a negative argument or is it the same if everything else in the question is the same?
– Kayla Martin
Nov 21 at 12:39


















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