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Finding the complex square roots of a complex number without a calculator
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The complex number $z$ is given by $z = -1 + (4 sqrt3)i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4sqrt3)i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?
complex-numbers radicals
edited 23 mins ago
José Carlos Santos
127k17102189
asked 41 mins ago
Pegladon
287
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Pegladon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
4
down vote
favorite
The complex number $z$ is given by $z = -1 + (4 sqrt3)i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4sqrt3)i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?
complex-numbers radicals
edited 23 mins ago
José Carlos Santos
127k17102189
asked 41 mins ago
Pegladon
287
New contributor
Pegladon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You could notice that $u=a^2$ and $v=-b^2$ satisfy $begincasesu+v=-1\ uv=-12\ uge 0\ vle0endcases$.
– Saucy O'Path
32 mins ago
Why is $uv=-12?$
– user376343
21 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The complex number $z$ is given by $z = -1 + (4 sqrt3)i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4sqrt3)i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?
complex-numbers radicals
edited 23 mins ago
José Carlos Santos
127k17102189
asked 41 mins ago
Pegladon
287
New contributor
Pegladon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The complex number $z$ is given by $z = -1 + (4 sqrt3)i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4sqrt3)i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?
complex-numbers radicals
complex-numbers radicals
edited 23 mins ago
José Carlos Santos
127k17102189
asked 41 mins ago
Pegladon
287
New contributor
Pegladon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 23 mins ago
José Carlos Santos
127k17102189
asked 41 mins ago
Pegladon
287
New contributor
Pegladon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 23 mins ago
José Carlos Santos
127k17102189
edited 23 mins ago
José Carlos Santos
127k17102189
edited 23 mins ago
José Carlos Santos
127k17102189
127k17102189
asked 41 mins ago
Pegladon
287
New contributor
Pegladon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 41 mins ago
Pegladon
287
asked 41 mins ago
Pegladon
287
287
New contributor
Pegladon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Pegladon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Pegladon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You could notice that $u=a^2$ and $v=-b^2$ satisfy $begincasesu+v=-1\ uv=-12\ uge 0\ vle0endcases$.
– Saucy O'Path
32 mins ago
Why is $uv=-12?$
– user376343
21 mins ago
add a comment |Â
You could notice that $u=a^2$ and $v=-b^2$ satisfy $begincasesu+v=-1\ uv=-12\ uge 0\ vle0endcases$.
– Saucy O'Path
32 mins ago
Why is $uv=-12?$
– user376343
21 mins ago
You could notice that $u=a^2$ and $v=-b^2$ satisfy $begincasesu+v=-1\ uv=-12\ uge 0\ vle0endcases$.
– Saucy O'Path
32 mins ago
You could notice that $u=a^2$ and $v=-b^2$ satisfy $begincasesu+v=-1\ uv=-12\ uge 0\ vle0endcases$.
– Saucy O'Path
32 mins ago
Why is $uv=-12?$
– user376343
21 mins ago
Why is $uv=-12?$
– user376343
21 mins ago
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
3
down vote
accepted
The second equation can be written $ab=2sqrt3$ which gives $b = frac2sqrt3a$. If we substitute back into the first equation we get $a^2 - frac12a^2 = -1 $. Multiplying both sides by $a^2$ gives $a^4 - 12 = - a^2$. This can be written as $a^4 + a^2 - 12 = 0$ which is a quadratic equation solvable for $a^2$.
answered 25 mins ago
sfmiller940
862
Thanks, I managed to get up to $a^4 + a^2 - 12 = 0$ on my own before asking this question, but couldn't figure out how to solve for $a$ without a calculator. Using your suggestion I found $(a^2 + 4)(a^2 - 3) = 0$ therefore $a = 2i text and pm sqrt3$, thanks for the help.
– Pegladon
18 mins ago
add a comment |Â
up vote
1
down vote
Let z^2 = (x + yi)2 = −1+(4√3)i
(x^2 – y^2) + 2xyi = −1+(4√3)i
Compare real parts and imaginary parts,
x^2 – y^2 = -1 (1)
2xy = (4√3) (2)
Now, consider the modulus: |z|2 = |z2|
x2 + y2 = √((-1)^2 + (4√3)^2) = 7 (3)
Solving (1) and (3), we get x^2 = 3 and y2 = 4
x = ±√3 and y = ±2
From (2), x and y are of same sign,
(x = √3 and y = 2) or (x = -√3 or y = -2)
z = ± (√3 + 2i)
answered 24 mins ago
Alselvdor
313
Welcome to MSE. Please see this typesetting reference to learn how to display mathematical expressions.
– Théophile
22 mins ago
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1
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Observe that $|z| = sqrt(-1)^2+(4sqrt3)^2 = sqrt49 = 7$. Therefore the root of $z$ will have length $sqrt 7$, so $a^2+b^2=7$. Combine this with $a^2-b^2=-1$ to get $a$ and $b$.
answered 24 mins ago
Théophile
18k12740
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up vote
1
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Note that$$z=7left(-frac17+frac4sqrt37yright).tag1$$Now, since $left(-frac17right)^2+left(frac4sqrt37right)^2=1$, $(1)$ expresses $z$ as $7bigl(cos(alpha)+sin(alpha)ibigr)$, for some $alpha$. So, a square root of $z$ is $sqrt7left(cosleft(fracalpha2right)+sinleft(fracalpha2right)iright)$. Now, note that if $c=cosleft(fracalpha2right)$ and $s=sinleft(fracalpha2right)$, then $c^2+s^2=1$ and $c^2-s^2=cos(alpha)=-frac17$. This allows you to compute the squae roots of $z$.
answered 24 mins ago
José Carlos Santos
127k17102189
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up vote
0
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One way is to write $z=r^2 e^2theta$ and roots will be $re^theta$ and $re^pi -theta$.
From $z =-1+4sqrt3i$, we obtain $r=7$ and $tan2theta = frac2tantheta1-tan^2theta = -4sqrt3$. Second expression gives you a quadratic equation, $2sqrt3tan^2theta -2sqrt3 + tantheta =0$.
Roots of the above quadratic equation are $tantheta= sqrt3/2,-2/sqrt3$ which form $tantheta$ and $tan(pi-theta)$.
Hence, square roots of $z$ are $(1+sqrt3/2i)frac7sqrt1+3/4 = sqrt7(2+sqrt3i)$ and $(1-2/sqrt3i)frac7sqrt1+4/3 = sqrt7(sqrt3-2i)$.
answered 13 mins ago
zero
14
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That the square roots are $pm(sqrt 3 + 2i)$ can be seen by elementary algebra and trigonometry as follows.
beginalign
& left|-1 + i4sqrt 3right| = sqrt(-1)^2 + (4sqrt 3)^2 = 7. \[10pt]
textTherefore & -1+i4sqrt 3 = 7(cosvarphi + isinvarphi). \[10pt]
textTherefore & pmsqrt-1+i4sqrt 3 = pmsqrt 7 left( cos frac varphi 2 + i sinfracvarphi 2 right).
endalign
Notice that
$$
sin varphi = frac4sqrt 3 7 quad textand quad cosvarphi = frac-1 7
$$
and recall that
beginalign
tanfracvarphi 2 & = fracsinvarphi1+cosvarphi \[12pt]
textso we have tanfracvarphi 2 & = frac4sqrt 37-1 = frac 2 sqrt 3. \[10pt]
textTherefore sinfracvarphi2 & = frac 2 sqrt 7 quad textand quad cosfracvarphi2 = fracsqrt3sqrt7.
endalign
Thus the desired square roots are
$$
pm left( sqrt 3 + 2i right).
$$
answered 12 mins ago
Michael Hardy
207k23189469
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The second equation can be written $ab=2sqrt3$ which gives $b = frac2sqrt3a$. If we substitute back into the first equation we get $a^2 - frac12a^2 = -1 $. Multiplying both sides by $a^2$ gives $a^4 - 12 = - a^2$. This can be written as $a^4 + a^2 - 12 = 0$ which is a quadratic equation solvable for $a^2$.
answered 25 mins ago
sfmiller940
862
Thanks, I managed to get up to $a^4 + a^2 - 12 = 0$ on my own before asking this question, but couldn't figure out how to solve for $a$ without a calculator. Using your suggestion I found $(a^2 + 4)(a^2 - 3) = 0$ therefore $a = 2i text and pm sqrt3$, thanks for the help.
– Pegladon
18 mins ago
add a comment |Â
up vote
3
down vote
accepted
The second equation can be written $ab=2sqrt3$ which gives $b = frac2sqrt3a$. If we substitute back into the first equation we get $a^2 - frac12a^2 = -1 $. Multiplying both sides by $a^2$ gives $a^4 - 12 = - a^2$. This can be written as $a^4 + a^2 - 12 = 0$ which is a quadratic equation solvable for $a^2$.
answered 25 mins ago
sfmiller940
862
Thanks, I managed to get up to $a^4 + a^2 - 12 = 0$ on my own before asking this question, but couldn't figure out how to solve for $a$ without a calculator. Using your suggestion I found $(a^2 + 4)(a^2 - 3) = 0$ therefore $a = 2i text and pm sqrt3$, thanks for the help.
– Pegladon
18 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The second equation can be written $ab=2sqrt3$ which gives $b = frac2sqrt3a$. If we substitute back into the first equation we get $a^2 - frac12a^2 = -1 $. Multiplying both sides by $a^2$ gives $a^4 - 12 = - a^2$. This can be written as $a^4 + a^2 - 12 = 0$ which is a quadratic equation solvable for $a^2$.
answered 25 mins ago
sfmiller940
862
The second equation can be written $ab=2sqrt3$ which gives $b = frac2sqrt3a$. If we substitute back into the first equation we get $a^2 - frac12a^2 = -1 $. Multiplying both sides by $a^2$ gives $a^4 - 12 = - a^2$. This can be written as $a^4 + a^2 - 12 = 0$ which is a quadratic equation solvable for $a^2$.
answered 25 mins ago
sfmiller940
862
answered 25 mins ago
sfmiller940
862
answered 25 mins ago
sfmiller940
862
answered 25 mins ago
sfmiller940
862
862
Thanks, I managed to get up to $a^4 + a^2 - 12 = 0$ on my own before asking this question, but couldn't figure out how to solve for $a$ without a calculator. Using your suggestion I found $(a^2 + 4)(a^2 - 3) = 0$ therefore $a = 2i text and pm sqrt3$, thanks for the help.
– Pegladon
18 mins ago
add a comment |Â
Thanks, I managed to get up to $a^4 + a^2 - 12 = 0$ on my own before asking this question, but couldn't figure out how to solve for $a$ without a calculator. Using your suggestion I found $(a^2 + 4)(a^2 - 3) = 0$ therefore $a = 2i text and pm sqrt3$, thanks for the help.
– Pegladon
18 mins ago
Thanks, I managed to get up to $a^4 + a^2 - 12 = 0$ on my own before asking this question, but couldn't figure out how to solve for $a$ without a calculator. Using your suggestion I found $(a^2 + 4)(a^2 - 3) = 0$ therefore $a = 2i text and pm sqrt3$, thanks for the help.
– Pegladon
18 mins ago
Thanks, I managed to get up to $a^4 + a^2 - 12 = 0$ on my own before asking this question, but couldn't figure out how to solve for $a$ without a calculator. Using your suggestion I found $(a^2 + 4)(a^2 - 3) = 0$ therefore $a = 2i text and pm sqrt3$, thanks for the help.
– Pegladon
18 mins ago
add a comment |Â
up vote
1
down vote
Let z^2 = (x + yi)2 = −1+(4√3)i
(x^2 – y^2) + 2xyi = −1+(4√3)i
Compare real parts and imaginary parts,
x^2 – y^2 = -1 (1)
2xy = (4√3) (2)
Now, consider the modulus: |z|2 = |z2|
x2 + y2 = √((-1)^2 + (4√3)^2) = 7 (3)
Solving (1) and (3), we get x^2 = 3 and y2 = 4
x = ±√3 and y = ±2
From (2), x and y are of same sign,
(x = √3 and y = 2) or (x = -√3 or y = -2)
z = ± (√3 + 2i)
answered 24 mins ago
Alselvdor
313
Welcome to MSE. Please see this typesetting reference to learn how to display mathematical expressions.
– Théophile
22 mins ago
add a comment |Â
up vote
1
down vote
Let z^2 = (x + yi)2 = −1+(4√3)i
(x^2 – y^2) + 2xyi = −1+(4√3)i
Compare real parts and imaginary parts,
x^2 – y^2 = -1 (1)
2xy = (4√3) (2)
Now, consider the modulus: |z|2 = |z2|
x2 + y2 = √((-1)^2 + (4√3)^2) = 7 (3)
Solving (1) and (3), we get x^2 = 3 and y2 = 4
x = ±√3 and y = ±2
From (2), x and y are of same sign,
(x = √3 and y = 2) or (x = -√3 or y = -2)
z = ± (√3 + 2i)
answered 24 mins ago
Alselvdor
313
Welcome to MSE. Please see this typesetting reference to learn how to display mathematical expressions.
– Théophile
22 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let z^2 = (x + yi)2 = −1+(4√3)i
(x^2 – y^2) + 2xyi = −1+(4√3)i
Compare real parts and imaginary parts,
x^2 – y^2 = -1 (1)
2xy = (4√3) (2)
Now, consider the modulus: |z|2 = |z2|
x2 + y2 = √((-1)^2 + (4√3)^2) = 7 (3)
Solving (1) and (3), we get x^2 = 3 and y2 = 4
x = ±√3 and y = ±2
From (2), x and y are of same sign,
(x = √3 and y = 2) or (x = -√3 or y = -2)
z = ± (√3 + 2i)
answered 24 mins ago
Alselvdor
313
Let z^2 = (x + yi)2 = −1+(4√3)i
(x^2 – y^2) + 2xyi = −1+(4√3)i
Compare real parts and imaginary parts,
x^2 – y^2 = -1 (1)
2xy = (4√3) (2)
Now, consider the modulus: |z|2 = |z2|
x2 + y2 = √((-1)^2 + (4√3)^2) = 7 (3)
Solving (1) and (3), we get x^2 = 3 and y2 = 4
x = ±√3 and y = ±2
From (2), x and y are of same sign,
(x = √3 and y = 2) or (x = -√3 or y = -2)
z = ± (√3 + 2i)
answered 24 mins ago
Alselvdor
313
answered 24 mins ago
Alselvdor
313
answered 24 mins ago
Alselvdor
313
answered 24 mins ago
Alselvdor
313
313
Welcome to MSE. Please see this typesetting reference to learn how to display mathematical expressions.
– Théophile
22 mins ago
add a comment |Â
Welcome to MSE. Please see this typesetting reference to learn how to display mathematical expressions.
– Théophile
22 mins ago
Welcome to MSE. Please see this typesetting reference to learn how to display mathematical expressions.
– Théophile
22 mins ago
Welcome to MSE. Please see this typesetting reference to learn how to display mathematical expressions.
– Théophile
22 mins ago
add a comment |Â
up vote
1
down vote
Observe that $|z| = sqrt(-1)^2+(4sqrt3)^2 = sqrt49 = 7$. Therefore the root of $z$ will have length $sqrt 7$, so $a^2+b^2=7$. Combine this with $a^2-b^2=-1$ to get $a$ and $b$.
answered 24 mins ago
Théophile
18k12740
add a comment |Â
up vote
1
down vote
Observe that $|z| = sqrt(-1)^2+(4sqrt3)^2 = sqrt49 = 7$. Therefore the root of $z$ will have length $sqrt 7$, so $a^2+b^2=7$. Combine this with $a^2-b^2=-1$ to get $a$ and $b$.
answered 24 mins ago
Théophile
18k12740
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Observe that $|z| = sqrt(-1)^2+(4sqrt3)^2 = sqrt49 = 7$. Therefore the root of $z$ will have length $sqrt 7$, so $a^2+b^2=7$. Combine this with $a^2-b^2=-1$ to get $a$ and $b$.
answered 24 mins ago
Théophile
18k12740
Observe that $|z| = sqrt(-1)^2+(4sqrt3)^2 = sqrt49 = 7$. Therefore the root of $z$ will have length $sqrt 7$, so $a^2+b^2=7$. Combine this with $a^2-b^2=-1$ to get $a$ and $b$.
answered 24 mins ago
Théophile
18k12740
answered 24 mins ago
Théophile
18k12740
answered 24 mins ago
Théophile
18k12740
answered 24 mins ago
Théophile
18k12740
18k12740
add a comment |Â
add a comment |Â
up vote
1
down vote
Note that$$z=7left(-frac17+frac4sqrt37yright).tag1$$Now, since $left(-frac17right)^2+left(frac4sqrt37right)^2=1$, $(1)$ expresses $z$ as $7bigl(cos(alpha)+sin(alpha)ibigr)$, for some $alpha$. So, a square root of $z$ is $sqrt7left(cosleft(fracalpha2right)+sinleft(fracalpha2right)iright)$. Now, note that if $c=cosleft(fracalpha2right)$ and $s=sinleft(fracalpha2right)$, then $c^2+s^2=1$ and $c^2-s^2=cos(alpha)=-frac17$. This allows you to compute the squae roots of $z$.
answered 24 mins ago
José Carlos Santos
127k17102189
add a comment |Â
up vote
1
down vote
Note that$$z=7left(-frac17+frac4sqrt37yright).tag1$$Now, since $left(-frac17right)^2+left(frac4sqrt37right)^2=1$, $(1)$ expresses $z$ as $7bigl(cos(alpha)+sin(alpha)ibigr)$, for some $alpha$. So, a square root of $z$ is $sqrt7left(cosleft(fracalpha2right)+sinleft(fracalpha2right)iright)$. Now, note that if $c=cosleft(fracalpha2right)$ and $s=sinleft(fracalpha2right)$, then $c^2+s^2=1$ and $c^2-s^2=cos(alpha)=-frac17$. This allows you to compute the squae roots of $z$.
answered 24 mins ago
José Carlos Santos
127k17102189
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that$$z=7left(-frac17+frac4sqrt37yright).tag1$$Now, since $left(-frac17right)^2+left(frac4sqrt37right)^2=1$, $(1)$ expresses $z$ as $7bigl(cos(alpha)+sin(alpha)ibigr)$, for some $alpha$. So, a square root of $z$ is $sqrt7left(cosleft(fracalpha2right)+sinleft(fracalpha2right)iright)$. Now, note that if $c=cosleft(fracalpha2right)$ and $s=sinleft(fracalpha2right)$, then $c^2+s^2=1$ and $c^2-s^2=cos(alpha)=-frac17$. This allows you to compute the squae roots of $z$.
answered 24 mins ago
José Carlos Santos
127k17102189
Note that$$z=7left(-frac17+frac4sqrt37yright).tag1$$Now, since $left(-frac17right)^2+left(frac4sqrt37right)^2=1$, $(1)$ expresses $z$ as $7bigl(cos(alpha)+sin(alpha)ibigr)$, for some $alpha$. So, a square root of $z$ is $sqrt7left(cosleft(fracalpha2right)+sinleft(fracalpha2right)iright)$. Now, note that if $c=cosleft(fracalpha2right)$ and $s=sinleft(fracalpha2right)$, then $c^2+s^2=1$ and $c^2-s^2=cos(alpha)=-frac17$. This allows you to compute the squae roots of $z$.
answered 24 mins ago
José Carlos Santos
127k17102189
answered 24 mins ago
José Carlos Santos
127k17102189
answered 24 mins ago
José Carlos Santos
127k17102189
answered 24 mins ago
José Carlos Santos
127k17102189
127k17102189
add a comment |Â
add a comment |Â
up vote
0
down vote
One way is to write $z=r^2 e^2theta$ and roots will be $re^theta$ and $re^pi -theta$.
From $z =-1+4sqrt3i$, we obtain $r=7$ and $tan2theta = frac2tantheta1-tan^2theta = -4sqrt3$. Second expression gives you a quadratic equation, $2sqrt3tan^2theta -2sqrt3 + tantheta =0$.
Roots of the above quadratic equation are $tantheta= sqrt3/2,-2/sqrt3$ which form $tantheta$ and $tan(pi-theta)$.
Hence, square roots of $z$ are $(1+sqrt3/2i)frac7sqrt1+3/4 = sqrt7(2+sqrt3i)$ and $(1-2/sqrt3i)frac7sqrt1+4/3 = sqrt7(sqrt3-2i)$.
answered 13 mins ago
zero
14
add a comment |Â
up vote
0
down vote
One way is to write $z=r^2 e^2theta$ and roots will be $re^theta$ and $re^pi -theta$.
From $z =-1+4sqrt3i$, we obtain $r=7$ and $tan2theta = frac2tantheta1-tan^2theta = -4sqrt3$. Second expression gives you a quadratic equation, $2sqrt3tan^2theta -2sqrt3 + tantheta =0$.
Roots of the above quadratic equation are $tantheta= sqrt3/2,-2/sqrt3$ which form $tantheta$ and $tan(pi-theta)$.
Hence, square roots of $z$ are $(1+sqrt3/2i)frac7sqrt1+3/4 = sqrt7(2+sqrt3i)$ and $(1-2/sqrt3i)frac7sqrt1+4/3 = sqrt7(sqrt3-2i)$.
answered 13 mins ago
zero
14
add a comment |Â
up vote
0
down vote
up vote
0
down vote
One way is to write $z=r^2 e^2theta$ and roots will be $re^theta$ and $re^pi -theta$.
From $z =-1+4sqrt3i$, we obtain $r=7$ and $tan2theta = frac2tantheta1-tan^2theta = -4sqrt3$. Second expression gives you a quadratic equation, $2sqrt3tan^2theta -2sqrt3 + tantheta =0$.
Roots of the above quadratic equation are $tantheta= sqrt3/2,-2/sqrt3$ which form $tantheta$ and $tan(pi-theta)$.
Hence, square roots of $z$ are $(1+sqrt3/2i)frac7sqrt1+3/4 = sqrt7(2+sqrt3i)$ and $(1-2/sqrt3i)frac7sqrt1+4/3 = sqrt7(sqrt3-2i)$.
answered 13 mins ago
zero
14
One way is to write $z=r^2 e^2theta$ and roots will be $re^theta$ and $re^pi -theta$.
From $z =-1+4sqrt3i$, we obtain $r=7$ and $tan2theta = frac2tantheta1-tan^2theta = -4sqrt3$. Second expression gives you a quadratic equation, $2sqrt3tan^2theta -2sqrt3 + tantheta =0$.
Roots of the above quadratic equation are $tantheta= sqrt3/2,-2/sqrt3$ which form $tantheta$ and $tan(pi-theta)$.
Hence, square roots of $z$ are $(1+sqrt3/2i)frac7sqrt1+3/4 = sqrt7(2+sqrt3i)$ and $(1-2/sqrt3i)frac7sqrt1+4/3 = sqrt7(sqrt3-2i)$.
answered 13 mins ago
zero
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answered 13 mins ago
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answered 13 mins ago
zero
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answered 13 mins ago
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That the square roots are $pm(sqrt 3 + 2i)$ can be seen by elementary algebra and trigonometry as follows.
beginalign
& left|-1 + i4sqrt 3right| = sqrt(-1)^2 + (4sqrt 3)^2 = 7. \[10pt]
textTherefore & -1+i4sqrt 3 = 7(cosvarphi + isinvarphi). \[10pt]
textTherefore & pmsqrt-1+i4sqrt 3 = pmsqrt 7 left( cos frac varphi 2 + i sinfracvarphi 2 right).
endalign
Notice that
$$
sin varphi = frac4sqrt 3 7 quad textand quad cosvarphi = frac-1 7
$$
and recall that
beginalign
tanfracvarphi 2 & = fracsinvarphi1+cosvarphi \[12pt]
textso we have tanfracvarphi 2 & = frac4sqrt 37-1 = frac 2 sqrt 3. \[10pt]
textTherefore sinfracvarphi2 & = frac 2 sqrt 7 quad textand quad cosfracvarphi2 = fracsqrt3sqrt7.
endalign
Thus the desired square roots are
$$
pm left( sqrt 3 + 2i right).
$$
answered 12 mins ago
Michael Hardy
207k23189469
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up vote
0
down vote
That the square roots are $pm(sqrt 3 + 2i)$ can be seen by elementary algebra and trigonometry as follows.
beginalign
& left|-1 + i4sqrt 3right| = sqrt(-1)^2 + (4sqrt 3)^2 = 7. \[10pt]
textTherefore & -1+i4sqrt 3 = 7(cosvarphi + isinvarphi). \[10pt]
textTherefore & pmsqrt-1+i4sqrt 3 = pmsqrt 7 left( cos frac varphi 2 + i sinfracvarphi 2 right).
endalign
Notice that
$$
sin varphi = frac4sqrt 3 7 quad textand quad cosvarphi = frac-1 7
$$
and recall that
beginalign
tanfracvarphi 2 & = fracsinvarphi1+cosvarphi \[12pt]
textso we have tanfracvarphi 2 & = frac4sqrt 37-1 = frac 2 sqrt 3. \[10pt]
textTherefore sinfracvarphi2 & = frac 2 sqrt 7 quad textand quad cosfracvarphi2 = fracsqrt3sqrt7.
endalign
Thus the desired square roots are
$$
pm left( sqrt 3 + 2i right).
$$
answered 12 mins ago
Michael Hardy
207k23189469
add a comment |Â
up vote
0
down vote
up vote
0
down vote
That the square roots are $pm(sqrt 3 + 2i)$ can be seen by elementary algebra and trigonometry as follows.
beginalign
& left|-1 + i4sqrt 3right| = sqrt(-1)^2 + (4sqrt 3)^2 = 7. \[10pt]
textTherefore & -1+i4sqrt 3 = 7(cosvarphi + isinvarphi). \[10pt]
textTherefore & pmsqrt-1+i4sqrt 3 = pmsqrt 7 left( cos frac varphi 2 + i sinfracvarphi 2 right).
endalign
Notice that
$$
sin varphi = frac4sqrt 3 7 quad textand quad cosvarphi = frac-1 7
$$
and recall that
beginalign
tanfracvarphi 2 & = fracsinvarphi1+cosvarphi \[12pt]
textso we have tanfracvarphi 2 & = frac4sqrt 37-1 = frac 2 sqrt 3. \[10pt]
textTherefore sinfracvarphi2 & = frac 2 sqrt 7 quad textand quad cosfracvarphi2 = fracsqrt3sqrt7.
endalign
Thus the desired square roots are
$$
pm left( sqrt 3 + 2i right).
$$
answered 12 mins ago
Michael Hardy
207k23189469
That the square roots are $pm(sqrt 3 + 2i)$ can be seen by elementary algebra and trigonometry as follows.
beginalign
& left|-1 + i4sqrt 3right| = sqrt(-1)^2 + (4sqrt 3)^2 = 7. \[10pt]
textTherefore & -1+i4sqrt 3 = 7(cosvarphi + isinvarphi). \[10pt]
textTherefore & pmsqrt-1+i4sqrt 3 = pmsqrt 7 left( cos frac varphi 2 + i sinfracvarphi 2 right).
endalign
Notice that
$$
sin varphi = frac4sqrt 3 7 quad textand quad cosvarphi = frac-1 7
$$
and recall that
beginalign
tanfracvarphi 2 & = fracsinvarphi1+cosvarphi \[12pt]
textso we have tanfracvarphi 2 & = frac4sqrt 37-1 = frac 2 sqrt 3. \[10pt]
textTherefore sinfracvarphi2 & = frac 2 sqrt 7 quad textand quad cosfracvarphi2 = fracsqrt3sqrt7.
endalign
Thus the desired square roots are
$$
pm left( sqrt 3 + 2i right).
$$
answered 12 mins ago
Michael Hardy
207k23189469
edited 3 mins ago
answered 12 mins ago
Michael Hardy
207k23189469
answered 12 mins ago
Michael Hardy
207k23189469
answered 12 mins ago
Michael Hardy
207k23189469
207k23189469
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You could notice that $u=a^2$ and $v=-b^2$ satisfy $begincasesu+v=-1\ uv=-12\ uge 0\ vle0endcases$.
– Saucy O'Path
32 mins ago
Why is $uv=-12?$
– user376343
21 mins ago