Find $arg(z)$, where $z = -fracsqrt7(1+i)sqrt3+i$
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This is my work so far:
I got the expression into the form $a+bi$ through some tedious algebra. I'll just list the final result here:
$-fracsqrt7(sqrt3+1)4 + fracsqrt7(1-sqrt3)4i$
When I solve for $|z|$ I get $fracsqrt142$, which matches the answer in my textbook. However, when I solve for $arg(z)$ I get $frac11pi12$, while the book lists it as $frac13pi12$. I got my answer by solving for $cos (theta) = fraca$, where $theta = arg(z)$. Nothing seems wrong, so not sure if the answer in the book is a typo or I'm making a mistake.
Edit: I forgot a negative sign, I understand how the answer is derived now. Thanks!
complex-analysis complex-numbers
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up vote
0
down vote
favorite
This is my work so far:
I got the expression into the form $a+bi$ through some tedious algebra. I'll just list the final result here:
$-fracsqrt7(sqrt3+1)4 + fracsqrt7(1-sqrt3)4i$
When I solve for $|z|$ I get $fracsqrt142$, which matches the answer in my textbook. However, when I solve for $arg(z)$ I get $frac11pi12$, while the book lists it as $frac13pi12$. I got my answer by solving for $cos (theta) = fraca$, where $theta = arg(z)$. Nothing seems wrong, so not sure if the answer in the book is a typo or I'm making a mistake.
Edit: I forgot a negative sign, I understand how the answer is derived now. Thanks!
complex-analysis complex-numbers
And I get $pi/12$. I didn't bother with the tedious algebra though.
â Lord Shark the Unknown
Aug 30 at 3:48
I got the expression into the form a+bi
That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$
â dxiv
Aug 30 at 3:50
How'd you come to that answer without the algebra?
â Leorio Paradinight
Aug 30 at 3:50
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is my work so far:
I got the expression into the form $a+bi$ through some tedious algebra. I'll just list the final result here:
$-fracsqrt7(sqrt3+1)4 + fracsqrt7(1-sqrt3)4i$
When I solve for $|z|$ I get $fracsqrt142$, which matches the answer in my textbook. However, when I solve for $arg(z)$ I get $frac11pi12$, while the book lists it as $frac13pi12$. I got my answer by solving for $cos (theta) = fraca$, where $theta = arg(z)$. Nothing seems wrong, so not sure if the answer in the book is a typo or I'm making a mistake.
Edit: I forgot a negative sign, I understand how the answer is derived now. Thanks!
complex-analysis complex-numbers
This is my work so far:
I got the expression into the form $a+bi$ through some tedious algebra. I'll just list the final result here:
$-fracsqrt7(sqrt3+1)4 + fracsqrt7(1-sqrt3)4i$
When I solve for $|z|$ I get $fracsqrt142$, which matches the answer in my textbook. However, when I solve for $arg(z)$ I get $frac11pi12$, while the book lists it as $frac13pi12$. I got my answer by solving for $cos (theta) = fraca$, where $theta = arg(z)$. Nothing seems wrong, so not sure if the answer in the book is a typo or I'm making a mistake.
Edit: I forgot a negative sign, I understand how the answer is derived now. Thanks!
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Aug 30 at 3:56
asked Aug 30 at 3:47
Leorio Paradinight
112
112
And I get $pi/12$. I didn't bother with the tedious algebra though.
â Lord Shark the Unknown
Aug 30 at 3:48
I got the expression into the form a+bi
That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$
â dxiv
Aug 30 at 3:50
How'd you come to that answer without the algebra?
â Leorio Paradinight
Aug 30 at 3:50
add a comment |Â
And I get $pi/12$. I didn't bother with the tedious algebra though.
â Lord Shark the Unknown
Aug 30 at 3:48
I got the expression into the form a+bi
That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$
â dxiv
Aug 30 at 3:50
How'd you come to that answer without the algebra?
â Leorio Paradinight
Aug 30 at 3:50
And I get $pi/12$. I didn't bother with the tedious algebra though.
â Lord Shark the Unknown
Aug 30 at 3:48
And I get $pi/12$. I didn't bother with the tedious algebra though.
â Lord Shark the Unknown
Aug 30 at 3:48
I got the expression into the form a+bi
That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$â dxiv
Aug 30 at 3:50
I got the expression into the form a+bi
That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$â dxiv
Aug 30 at 3:50
How'd you come to that answer without the algebra?
â Leorio Paradinight
Aug 30 at 3:50
How'd you come to that answer without the algebra?
â Leorio Paradinight
Aug 30 at 3:50
add a comment |Â
1 Answer
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$$Arg(z) = Arg(sqrt7(1+i))-Arg(sqrt3+i)=fracpi4-fracpi6=pi/12.$$
It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
â Leorio Paradinight
Aug 30 at 3:54
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$$Arg(z) = Arg(sqrt7(1+i))-Arg(sqrt3+i)=fracpi4-fracpi6=pi/12.$$
It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
â Leorio Paradinight
Aug 30 at 3:54
add a comment |Â
up vote
2
down vote
$$Arg(z) = Arg(sqrt7(1+i))-Arg(sqrt3+i)=fracpi4-fracpi6=pi/12.$$
It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
â Leorio Paradinight
Aug 30 at 3:54
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$Arg(z) = Arg(sqrt7(1+i))-Arg(sqrt3+i)=fracpi4-fracpi6=pi/12.$$
$$Arg(z) = Arg(sqrt7(1+i))-Arg(sqrt3+i)=fracpi4-fracpi6=pi/12.$$
answered Aug 30 at 3:51
Hello_World
3,20321429
3,20321429
It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
â Leorio Paradinight
Aug 30 at 3:54
add a comment |Â
It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
â Leorio Paradinight
Aug 30 at 3:54
It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
â Leorio Paradinight
Aug 30 at 3:54
It seems the book has a typo then, I'm not sure where $frac13pi12$ comes from.
â Leorio Paradinight
Aug 30 at 3:54
add a comment |Â
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And I get $pi/12$. I didn't bother with the tedious algebra though.
â Lord Shark the Unknown
Aug 30 at 3:48
I got the expression into the form a+bi
That's not the best way to go about it in this case. Rather, note that $,arg(1+i)=pi/4,$ and $,arg(sqrt3+i) = ldots,$â dxiv
Aug 30 at 3:50
How'd you come to that answer without the algebra?
â Leorio Paradinight
Aug 30 at 3:50