Mixing
Simplify the complex expression…
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
edited Nov 22 '18 at 4:44
naveen dankal
4,52821348
asked Nov 21 '18 at 10:33
Kayla Martin
636
|
show 2 more comments
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
edited Nov 22 '18 at 4:44
naveen dankal
4,52821348
asked Nov 21 '18 at 10:33
Kayla Martin
636
What is $operatorname{mod}$ in this context?
– José Carlos Santos
Nov 21 '18 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 '18 at 10:48
Do you mean $lvert e^{itheta}-1rvert$?
– José Carlos Santos
Nov 21 '18 at 10:51
Yes, exactly. Same for the +1 term.
– Kayla Martin
Nov 21 '18 at 10:52
Then why don't you just type|e^{itheta}-1|?
– José Carlos Santos
Nov 21 '18 at 10:54
|
show 2 more comments
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
edited Nov 22 '18 at 4:44
naveen dankal
4,52821348
asked Nov 21 '18 at 10:33
Kayla Martin
636
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
edited Nov 22 '18 at 4:44
naveen dankal
4,52821348
asked Nov 21 '18 at 10:33
Kayla Martin
636
edited Nov 22 '18 at 4:44
naveen dankal
4,52821348
asked Nov 21 '18 at 10:33
Kayla Martin
636
edited Nov 22 '18 at 4:44
naveen dankal
4,52821348
edited Nov 22 '18 at 4:44
naveen dankal
4,52821348
edited Nov 22 '18 at 4:44
naveen dankal
4,52821348
4,52821348
asked Nov 21 '18 at 10:33
Kayla Martin
636
asked Nov 21 '18 at 10:33
Kayla Martin
636
asked Nov 21 '18 at 10:33
Kayla Martin
636
636
What is $operatorname{mod}$ in this context?
– José Carlos Santos
Nov 21 '18 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 '18 at 10:48
Do you mean $lvert e^{itheta}-1rvert$?
– José Carlos Santos
Nov 21 '18 at 10:51
Yes, exactly. Same for the +1 term.
– Kayla Martin
Nov 21 '18 at 10:52
Then why don't you just type|e^{itheta}-1|?
– José Carlos Santos
Nov 21 '18 at 10:54
|
show 2 more comments
What is $operatorname{mod}$ in this context?
– José Carlos Santos
Nov 21 '18 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 '18 at 10:48
Do you mean $lvert e^{itheta}-1rvert$?
– José Carlos Santos
Nov 21 '18 at 10:51
Yes, exactly. Same for the +1 term.
– Kayla Martin
Nov 21 '18 at 10:52
Then why don't you just type|e^{itheta}-1|?
– José Carlos Santos
Nov 21 '18 at 10:54
What is $operatorname{mod}$ in this context?
– José Carlos Santos
Nov 21 '18 at 10:34
What is $operatorname{mod}$ in this context?
– José Carlos Santos
Nov 21 '18 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 '18 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 '18 at 10:48
Do you mean $lvert e^{itheta}-1rvert$?
– José Carlos Santos
Nov 21 '18 at 10:51
Do you mean $lvert e^{itheta}-1rvert$?
– José Carlos Santos
Nov 21 '18 at 10:51
Yes, exactly. Same for the +1 term.
– Kayla Martin
Nov 21 '18 at 10:52
Yes, exactly. Same for the +1 term.
– Kayla Martin
Nov 21 '18 at 10:52
Then why don't you just type
|e^{itheta}-1|?– José Carlos Santos
Nov 21 '18 at 10:54
Then why don't you just type
|e^{itheta}-1|?– José Carlos Santos
Nov 21 '18 at 10:54
|
show 2 more comments
1 Answer
1
active
oldest
votes
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 $theta$.
answered Nov 21 '18 at 11:41
José Carlos Santos
151k22123224
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 '18 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 '18 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 '18 at 12:14
I cannot help you here, since I don't understand where your doubt is.
– José Carlos Santos
Nov 21 '18 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 '18 at 12:39
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
votes
active
oldest
votes
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 $theta$.
answered Nov 21 '18 at 11:41
José Carlos Santos
151k22123224
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 '18 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 '18 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 '18 at 12:14
I cannot help you here, since I don't understand where your doubt is.
– José Carlos Santos
Nov 21 '18 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 '18 at 12:39
|
show 2 more comments
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 $theta$.
answered Nov 21 '18 at 11:41
José Carlos Santos
151k22123224
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 '18 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 '18 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 '18 at 12:14
I cannot help you here, since I don't understand where your doubt is.
– José Carlos Santos
Nov 21 '18 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 '18 at 12:39
|
show 2 more comments
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 $theta$.
answered Nov 21 '18 at 11:41
José Carlos Santos
151k22123224
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 $theta$.
answered Nov 21 '18 at 11:41
José Carlos Santos
151k22123224
edited Nov 21 '18 at 12:35
answered Nov 21 '18 at 11:41
José Carlos Santos
151k22123224
answered Nov 21 '18 at 11:41
José Carlos Santos
151k22123224
answered Nov 21 '18 at 11:41
José Carlos Santos
151k22123224
151k22123224
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 '18 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 '18 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 '18 at 12:14
I cannot help you here, since I don't understand where your doubt is.
– José Carlos Santos
Nov 21 '18 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 '18 at 12:39
|
show 2 more comments
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 '18 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 '18 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 '18 at 12:14
I cannot help you here, since I don't understand where your doubt is.
– José Carlos Santos
Nov 21 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 12:14
I cannot help you here, since I don't understand where your doubt is.
– José Carlos Santos
Nov 21 '18 at 12:35
I cannot help you here, since I don't understand where your doubt is.
– José Carlos Santos
Nov 21 '18 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 '18 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 '18 at 12:39
|
show 2 more comments
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What is $operatorname{mod}$ in this context?
– José Carlos Santos
Nov 21 '18 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 '18 at 10:48
Do you mean $lvert e^{itheta}-1rvert$?
– José Carlos Santos
Nov 21 '18 at 10:51
Yes, exactly. Same for the +1 term.
– Kayla Martin
Nov 21 '18 at 10:52
Then why don't you just type
|e^{itheta}-1|?– José Carlos Santos
Nov 21 '18 at 10:54