Rewriting this complex square root
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for some reason I can to figure out how to rewrite this square root.
I have:
$sqrt{2+i}$
And I need to rewrite it into:
$frac{sqrt{2(sqrt{5} + 2)} + sqrt{-2(sqrt{5} - 2)}}{2}$
Can anybody show me I to do this? I've been trying for an hour now..
complex-numbers
$endgroup$
add a comment |
$begingroup$
for some reason I can to figure out how to rewrite this square root.
I have:
$sqrt{2+i}$
And I need to rewrite it into:
$frac{sqrt{2(sqrt{5} + 2)} + sqrt{-2(sqrt{5} - 2)}}{2}$
Can anybody show me I to do this? I've been trying for an hour now..
complex-numbers
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6
$begingroup$
Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
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– Omnomnomnom
Jun 27 '14 at 19:44
$begingroup$
Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
$endgroup$
– Yves Daoust
Jun 27 '14 at 19:54
1
$begingroup$
@DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 20:18
$begingroup$
Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
$endgroup$
– Dane Bouchie
Jun 27 '14 at 20:22
add a comment |
$begingroup$
for some reason I can to figure out how to rewrite this square root.
I have:
$sqrt{2+i}$
And I need to rewrite it into:
$frac{sqrt{2(sqrt{5} + 2)} + sqrt{-2(sqrt{5} - 2)}}{2}$
Can anybody show me I to do this? I've been trying for an hour now..
complex-numbers
$endgroup$
for some reason I can to figure out how to rewrite this square root.
I have:
$sqrt{2+i}$
And I need to rewrite it into:
$frac{sqrt{2(sqrt{5} + 2)} + sqrt{-2(sqrt{5} - 2)}}{2}$
Can anybody show me I to do this? I've been trying for an hour now..
complex-numbers
complex-numbers
asked Jun 27 '14 at 19:40
user160513user160513
141
141
6
$begingroup$
Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 19:44
$begingroup$
Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
$endgroup$
– Yves Daoust
Jun 27 '14 at 19:54
1
$begingroup$
@DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 20:18
$begingroup$
Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
$endgroup$
– Dane Bouchie
Jun 27 '14 at 20:22
add a comment |
6
$begingroup$
Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 19:44
$begingroup$
Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
$endgroup$
– Yves Daoust
Jun 27 '14 at 19:54
1
$begingroup$
@DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 20:18
$begingroup$
Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
$endgroup$
– Dane Bouchie
Jun 27 '14 at 20:22
6
6
$begingroup$
Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 19:44
$begingroup$
Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 19:44
$begingroup$
Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
$endgroup$
– Yves Daoust
Jun 27 '14 at 19:54
$begingroup$
Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
$endgroup$
– Yves Daoust
Jun 27 '14 at 19:54
1
1
$begingroup$
@DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 20:18
$begingroup$
@DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 20:18
$begingroup$
Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
$endgroup$
– Dane Bouchie
Jun 27 '14 at 20:22
$begingroup$
Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
$endgroup$
– Dane Bouchie
Jun 27 '14 at 20:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
$$(x+yi)^2=2+i$$ so
$$x^2-y^2=2$$
$$2xy=1$$
and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$
So $$x^2+y^2=sqrt{5}$$
(positive since $x$ and $y$ real)
and
$$x^2=frac{sqrt{5}+2}{2}$$
$$y^2=frac{sqrt{5}-2}{2}$$
so
$$x=pmsqrt{frac{sqrt{5}+2}{2}}$$
$$y=pmsqrt{frac{sqrt{5}-2}{2}}$$
Both $x$ and $y$ have same sign since $2xy=1$.
$endgroup$
add a comment |
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1 Answer
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$begingroup$
One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
$$(x+yi)^2=2+i$$ so
$$x^2-y^2=2$$
$$2xy=1$$
and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$
So $$x^2+y^2=sqrt{5}$$
(positive since $x$ and $y$ real)
and
$$x^2=frac{sqrt{5}+2}{2}$$
$$y^2=frac{sqrt{5}-2}{2}$$
so
$$x=pmsqrt{frac{sqrt{5}+2}{2}}$$
$$y=pmsqrt{frac{sqrt{5}-2}{2}}$$
Both $x$ and $y$ have same sign since $2xy=1$.
$endgroup$
add a comment |
$begingroup$
One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
$$(x+yi)^2=2+i$$ so
$$x^2-y^2=2$$
$$2xy=1$$
and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$
So $$x^2+y^2=sqrt{5}$$
(positive since $x$ and $y$ real)
and
$$x^2=frac{sqrt{5}+2}{2}$$
$$y^2=frac{sqrt{5}-2}{2}$$
so
$$x=pmsqrt{frac{sqrt{5}+2}{2}}$$
$$y=pmsqrt{frac{sqrt{5}-2}{2}}$$
Both $x$ and $y$ have same sign since $2xy=1$.
$endgroup$
add a comment |
$begingroup$
One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
$$(x+yi)^2=2+i$$ so
$$x^2-y^2=2$$
$$2xy=1$$
and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$
So $$x^2+y^2=sqrt{5}$$
(positive since $x$ and $y$ real)
and
$$x^2=frac{sqrt{5}+2}{2}$$
$$y^2=frac{sqrt{5}-2}{2}$$
so
$$x=pmsqrt{frac{sqrt{5}+2}{2}}$$
$$y=pmsqrt{frac{sqrt{5}-2}{2}}$$
Both $x$ and $y$ have same sign since $2xy=1$.
$endgroup$
One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
$$(x+yi)^2=2+i$$ so
$$x^2-y^2=2$$
$$2xy=1$$
and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$
So $$x^2+y^2=sqrt{5}$$
(positive since $x$ and $y$ real)
and
$$x^2=frac{sqrt{5}+2}{2}$$
$$y^2=frac{sqrt{5}-2}{2}$$
so
$$x=pmsqrt{frac{sqrt{5}+2}{2}}$$
$$y=pmsqrt{frac{sqrt{5}-2}{2}}$$
Both $x$ and $y$ have same sign since $2xy=1$.
edited Dec 16 '18 at 10:39
Did
248k23225463
248k23225463
answered Jun 27 '14 at 20:50
Rene SchipperusRene Schipperus
32.4k11960
32.4k11960
add a comment |
add a comment |
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6
$begingroup$
Hint: $sqrt{cos theta + i sin theta} = cos(theta/2) + i sin(theta / 2)$. Note $2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right)$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 19:44
$begingroup$
Solve $(x+iy)^2=x^2-y^2+2ixy=2+i$.
$endgroup$
– Yves Daoust
Jun 27 '14 at 19:54
1
$begingroup$
@DaneBouchie Read that again. I say $$ 2 + i = sqrt{5} cdot left(frac{2}{sqrt 5} + i frac{1}{sqrt 5}right) $$
$endgroup$
– Omnomnomnom
Jun 27 '14 at 20:18
$begingroup$
Ah, my bad. I read $.$ as a period for some reason. Even then, lazy reading on my behalf.
$endgroup$
– Dane Bouchie
Jun 27 '14 at 20:22