Calculate the argument of $ (-2{sqrt 3} - 2i)^5$
$begingroup$
So I'm supposed to calcualte the argument and the answer is supposed to be $frac{11}{6}π$.
Instead of that I'm getting $frac{5}{6}π$ because
$frac{-2}{-2{sqrt 3}} = frac{1}{{sqrt 3}} $ the argument of that being $frac{π}{{6}}$ and when I multiply it by $5,$ I get $frac{5}{6}π$
where is my mistake?
complex-numbers
edited Jan 13 at 23:24
user376343
3,9834829
asked Jan 13 at 23:22
ythhtrgythhtrg
383
$endgroup$
add a comment |
$begingroup$
So I'm supposed to calcualte the argument and the answer is supposed to be $frac{11}{6}π$.
Instead of that I'm getting $frac{5}{6}π$ because
$frac{-2}{-2{sqrt 3}} = frac{1}{{sqrt 3}} $ the argument of that being $frac{π}{{6}}$ and when I multiply it by $5,$ I get $frac{5}{6}π$
where is my mistake?
complex-numbers
edited Jan 13 at 23:24
user376343
3,9834829
asked Jan 13 at 23:22
ythhtrgythhtrg
383
$endgroup$
$begingroup$
Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
$endgroup$
– D.B.
Jan 13 at 23:28
$begingroup$
More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
$endgroup$
– obscurans
Jan 13 at 23:38
add a comment |
$begingroup$
So I'm supposed to calcualte the argument and the answer is supposed to be $frac{11}{6}π$.
Instead of that I'm getting $frac{5}{6}π$ because
$frac{-2}{-2{sqrt 3}} = frac{1}{{sqrt 3}} $ the argument of that being $frac{π}{{6}}$ and when I multiply it by $5,$ I get $frac{5}{6}π$
where is my mistake?
complex-numbers
edited Jan 13 at 23:24
user376343
3,9834829
asked Jan 13 at 23:22
ythhtrgythhtrg
383
$endgroup$
So I'm supposed to calcualte the argument and the answer is supposed to be $frac{11}{6}π$.
Instead of that I'm getting $frac{5}{6}π$ because
$frac{-2}{-2{sqrt 3}} = frac{1}{{sqrt 3}} $ the argument of that being $frac{π}{{6}}$ and when I multiply it by $5,$ I get $frac{5}{6}π$
where is my mistake?
complex-numbers
complex-numbers
edited Jan 13 at 23:24
user376343
3,9834829
asked Jan 13 at 23:22
ythhtrgythhtrg
383
edited Jan 13 at 23:24
user376343
3,9834829
asked Jan 13 at 23:22
ythhtrgythhtrg
383
edited Jan 13 at 23:24
user376343
3,9834829
edited Jan 13 at 23:24
user376343
3,9834829
edited Jan 13 at 23:24
user376343
3,9834829
3,9834829
asked Jan 13 at 23:22
ythhtrgythhtrg
383
asked Jan 13 at 23:22
ythhtrgythhtrg
383
asked Jan 13 at 23:22
ythhtrgythhtrg
383
383
$begingroup$
Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
$endgroup$
– D.B.
Jan 13 at 23:28
$begingroup$
More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
$endgroup$
– obscurans
Jan 13 at 23:38
add a comment |
$begingroup$
Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
$endgroup$
– D.B.
Jan 13 at 23:28
$begingroup$
More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
$endgroup$
– obscurans
Jan 13 at 23:38
$begingroup$
Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
$endgroup$
– D.B.
Jan 13 at 23:28
$begingroup$
Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
$endgroup$
– D.B.
Jan 13 at 23:28
$begingroup$
More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
$endgroup$
– obscurans
Jan 13 at 23:38
$begingroup$
More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
$endgroup$
– obscurans
Jan 13 at 23:38
add a comment |
2 Answers
2
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$begingroup$
$$arg((-2-2sqrt{3})^5) = arg((-1)^5)+arg((2+2sqrt{3})^5)$$
$$=arg(-1)+arg((2+2sqrt{3})^5)$$
$$=pi+5arg(2+2sqrt{3}) = pi+5pi/6 = 11pi/6.$$
answered Jan 13 at 23:34
D.B.D.B.
1,61029
$endgroup$
add a comment |
$begingroup$
$$(-2sqrt3-2i)^5=left(4left(-frac{sqrt3}{2}-frac{1}{2}iright)right)^5=$$
$$=1024(cos(210^{circ}cdot5)+isin(210^{circ}cdot5))=$$
$$=1024(cos330^{circ}+isin330^{circ}),$$ which gives the answer:
$$330^{circ}.$$
answered Jan 13 at 23:30
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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$begingroup$
$$arg((-2-2sqrt{3})^5) = arg((-1)^5)+arg((2+2sqrt{3})^5)$$
$$=arg(-1)+arg((2+2sqrt{3})^5)$$
$$=pi+5arg(2+2sqrt{3}) = pi+5pi/6 = 11pi/6.$$
answered Jan 13 at 23:34
D.B.D.B.
1,61029
$endgroup$
add a comment |
$begingroup$
$$arg((-2-2sqrt{3})^5) = arg((-1)^5)+arg((2+2sqrt{3})^5)$$
$$=arg(-1)+arg((2+2sqrt{3})^5)$$
$$=pi+5arg(2+2sqrt{3}) = pi+5pi/6 = 11pi/6.$$
answered Jan 13 at 23:34
D.B.D.B.
1,61029
$endgroup$
add a comment |
$begingroup$
$$arg((-2-2sqrt{3})^5) = arg((-1)^5)+arg((2+2sqrt{3})^5)$$
$$=arg(-1)+arg((2+2sqrt{3})^5)$$
$$=pi+5arg(2+2sqrt{3}) = pi+5pi/6 = 11pi/6.$$
answered Jan 13 at 23:34
D.B.D.B.
1,61029
$endgroup$
$$arg((-2-2sqrt{3})^5) = arg((-1)^5)+arg((2+2sqrt{3})^5)$$
$$=arg(-1)+arg((2+2sqrt{3})^5)$$
$$=pi+5arg(2+2sqrt{3}) = pi+5pi/6 = 11pi/6.$$
answered Jan 13 at 23:34
D.B.D.B.
1,61029
edited Jan 13 at 23:44
answered Jan 13 at 23:34
D.B.D.B.
1,61029
answered Jan 13 at 23:34
D.B.D.B.
1,61029
answered Jan 13 at 23:34
D.B.D.B.
1,61029
1,61029
add a comment |
add a comment |
$begingroup$
$$(-2sqrt3-2i)^5=left(4left(-frac{sqrt3}{2}-frac{1}{2}iright)right)^5=$$
$$=1024(cos(210^{circ}cdot5)+isin(210^{circ}cdot5))=$$
$$=1024(cos330^{circ}+isin330^{circ}),$$ which gives the answer:
$$330^{circ}.$$
answered Jan 13 at 23:30
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
$$(-2sqrt3-2i)^5=left(4left(-frac{sqrt3}{2}-frac{1}{2}iright)right)^5=$$
$$=1024(cos(210^{circ}cdot5)+isin(210^{circ}cdot5))=$$
$$=1024(cos330^{circ}+isin330^{circ}),$$ which gives the answer:
$$330^{circ}.$$
answered Jan 13 at 23:30
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
$$(-2sqrt3-2i)^5=left(4left(-frac{sqrt3}{2}-frac{1}{2}iright)right)^5=$$
$$=1024(cos(210^{circ}cdot5)+isin(210^{circ}cdot5))=$$
$$=1024(cos330^{circ}+isin330^{circ}),$$ which gives the answer:
$$330^{circ}.$$
answered Jan 13 at 23:30
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
$$(-2sqrt3-2i)^5=left(4left(-frac{sqrt3}{2}-frac{1}{2}iright)right)^5=$$
$$=1024(cos(210^{circ}cdot5)+isin(210^{circ}cdot5))=$$
$$=1024(cos330^{circ}+isin330^{circ}),$$ which gives the answer:
$$330^{circ}.$$
answered Jan 13 at 23:30
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 13 at 23:30
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 13 at 23:30
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 13 at 23:30
Michael RozenbergMichael Rozenberg
111k1897201
111k1897201
add a comment |
add a comment |
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$begingroup$
Notice that you are in the third quadrant. If you were in the first quadrant, the answer would be $5pi/6$. But because you are in the third quadrant, just add $pi$: $5pi/6+pi = 11pi/6$.
$endgroup$
– D.B.
Jan 13 at 23:28
$begingroup$
More directly, the argument of $-2sqrt{3}-2i$ is not $pi/6$, it's actually $7pi/6$, and you can multiply that by 5 mod $2pi$
$endgroup$
– obscurans
Jan 13 at 23:38