Mixing
Complex number (finding range of values)
$begingroup$
Let $z_1 = 2 e^{ipi/6}$ and $z_2 = re^{itheta}$, where $r>0$ and $0letheta<2pi$. Find the range of values of $r$ and $theta$ for which $z_1z_2$ is:
a) a real number greater than $5$
b) a purely imaginary number with modulus less than $1$
This question was in my math textbook and I can't figure it out. There are no worked example to show the method in the textbook. Any hint on how I should approach it. I converted $z_1$ from polar to rectangular form and got $z_1 = sqrt 3 + i$
I think $theta$ has to be greater than $pi$ and less than $2pi$ because that will make value of $sintheta$ negative which would give the multiplication of $z_1z_2$ the form of $(a+b)(a-b)$ removing the imaginary part and leaving only real number. But I don't know how to get the number to be greater than $5$.
Answer in the textbook is
a) $r>5/2$ and $theta=11pi/6$
b) $r<1/2$ and $r > 0$ and $theta=pi/3$ or $theta=4pi/3$
complex-numbers polar-coordinates
edited Feb 1 at 11:20
Dylan
14.2k31127
asked Jan 29 at 20:15
AkhilAkhil
1084
$endgroup$
add a comment |
$begingroup$
Let $z_1 = 2 e^{ipi/6}$ and $z_2 = re^{itheta}$, where $r>0$ and $0letheta<2pi$. Find the range of values of $r$ and $theta$ for which $z_1z_2$ is:
a) a real number greater than $5$
b) a purely imaginary number with modulus less than $1$
This question was in my math textbook and I can't figure it out. There are no worked example to show the method in the textbook. Any hint on how I should approach it. I converted $z_1$ from polar to rectangular form and got $z_1 = sqrt 3 + i$
I think $theta$ has to be greater than $pi$ and less than $2pi$ because that will make value of $sintheta$ negative which would give the multiplication of $z_1z_2$ the form of $(a+b)(a-b)$ removing the imaginary part and leaving only real number. But I don't know how to get the number to be greater than $5$.
Answer in the textbook is
a) $r>5/2$ and $theta=11pi/6$
b) $r<1/2$ and $r > 0$ and $theta=pi/3$ or $theta=4pi/3$
complex-numbers polar-coordinates
edited Feb 1 at 11:20
Dylan
14.2k31127
asked Jan 29 at 20:15
AkhilAkhil
1084
$endgroup$
1
$begingroup$
Hint: cis$theta_1 times $cis$theta_2 = $cis$(theta_1+theta_2)$.
$endgroup$
– Math Lover
Jan 29 at 20:20
$begingroup$
What is 'cis' here?
$endgroup$
– Berci
Jan 29 at 21:59
$begingroup$
$cis(theta)=cos(theta)+icdot sin(theta)$, where $i^2=-1$
$endgroup$
– J. W. Tanner
Jan 29 at 22:36
add a comment |
$begingroup$
Let $z_1 = 2 e^{ipi/6}$ and $z_2 = re^{itheta}$, where $r>0$ and $0letheta<2pi$. Find the range of values of $r$ and $theta$ for which $z_1z_2$ is:
a) a real number greater than $5$
b) a purely imaginary number with modulus less than $1$
This question was in my math textbook and I can't figure it out. There are no worked example to show the method in the textbook. Any hint on how I should approach it. I converted $z_1$ from polar to rectangular form and got $z_1 = sqrt 3 + i$
I think $theta$ has to be greater than $pi$ and less than $2pi$ because that will make value of $sintheta$ negative which would give the multiplication of $z_1z_2$ the form of $(a+b)(a-b)$ removing the imaginary part and leaving only real number. But I don't know how to get the number to be greater than $5$.
Answer in the textbook is
a) $r>5/2$ and $theta=11pi/6$
b) $r<1/2$ and $r > 0$ and $theta=pi/3$ or $theta=4pi/3$
complex-numbers polar-coordinates
edited Feb 1 at 11:20
Dylan
14.2k31127
asked Jan 29 at 20:15
AkhilAkhil
1084
$endgroup$
Let $z_1 = 2 e^{ipi/6}$ and $z_2 = re^{itheta}$, where $r>0$ and $0letheta<2pi$. Find the range of values of $r$ and $theta$ for which $z_1z_2$ is:
a) a real number greater than $5$
b) a purely imaginary number with modulus less than $1$
This question was in my math textbook and I can't figure it out. There are no worked example to show the method in the textbook. Any hint on how I should approach it. I converted $z_1$ from polar to rectangular form and got $z_1 = sqrt 3 + i$
I think $theta$ has to be greater than $pi$ and less than $2pi$ because that will make value of $sintheta$ negative which would give the multiplication of $z_1z_2$ the form of $(a+b)(a-b)$ removing the imaginary part and leaving only real number. But I don't know how to get the number to be greater than $5$.
Answer in the textbook is
a) $r>5/2$ and $theta=11pi/6$
b) $r<1/2$ and $r > 0$ and $theta=pi/3$ or $theta=4pi/3$
complex-numbers polar-coordinates
complex-numbers polar-coordinates
edited Feb 1 at 11:20
Dylan
14.2k31127
asked Jan 29 at 20:15
AkhilAkhil
1084
edited Feb 1 at 11:20
Dylan
14.2k31127
asked Jan 29 at 20:15
AkhilAkhil
1084
edited Feb 1 at 11:20
Dylan
14.2k31127
edited Feb 1 at 11:20
Dylan
14.2k31127
edited Feb 1 at 11:20
Dylan
14.2k31127
14.2k31127
asked Jan 29 at 20:15
AkhilAkhil
1084
asked Jan 29 at 20:15
AkhilAkhil
1084
asked Jan 29 at 20:15
AkhilAkhil
1084
1084
1
$begingroup$
Hint: cis$theta_1 times $cis$theta_2 = $cis$(theta_1+theta_2)$.
$endgroup$
– Math Lover
Jan 29 at 20:20
$begingroup$
What is 'cis' here?
$endgroup$
– Berci
Jan 29 at 21:59
$begingroup$
$cis(theta)=cos(theta)+icdot sin(theta)$, where $i^2=-1$
$endgroup$
– J. W. Tanner
Jan 29 at 22:36
add a comment |
1
$begingroup$
Hint: cis$theta_1 times $cis$theta_2 = $cis$(theta_1+theta_2)$.
$endgroup$
– Math Lover
Jan 29 at 20:20
$begingroup$
What is 'cis' here?
$endgroup$
– Berci
Jan 29 at 21:59
$begingroup$
$cis(theta)=cos(theta)+icdot sin(theta)$, where $i^2=-1$
$endgroup$
– J. W. Tanner
Jan 29 at 22:36
1
1
$begingroup$
Hint: cis$theta_1 times $cis$theta_2 = $cis$(theta_1+theta_2)$.
$endgroup$
– Math Lover
Jan 29 at 20:20
$begingroup$
Hint: cis$theta_1 times $cis$theta_2 = $cis$(theta_1+theta_2)$.
$endgroup$
– Math Lover
Jan 29 at 20:20
$begingroup$
What is 'cis' here?
$endgroup$
– Berci
Jan 29 at 21:59
$begingroup$
What is 'cis' here?
$endgroup$
– Berci
Jan 29 at 21:59
$begingroup$
$cis(theta)=cos(theta)+icdot sin(theta)$, where $i^2=-1$
$endgroup$
– J. W. Tanner
Jan 29 at 22:36
$begingroup$
$cis(theta)=cos(theta)+icdot sin(theta)$, where $i^2=-1$
$endgroup$
– J. W. Tanner
Jan 29 at 22:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $z_1=2e^{ipi/6}$ and $z_2=re^{itheta}$ where anyways $rgt0$ and $0leqthetalt2pi$
Now the product $z_1z_2=2re^{i(theta+ipi/6)}$
The argument of the above complex number is $2r$ and for the first part:
$1.)$$$2r>5$$ which means $$r>frac{5}{2}$$
To make $z_1z_2$ as a real number the argument should be $0$ $or$ $2pi$ which means
$$ theta+pi/6=2pi$$ giving $$theta=11pi/6$$
And for the second Part:
$2.)$ The modulus of $z_1z_2$ is to be less than $1$ so
$$2rlt1$$
Giving $$rltfrac{1}{2}$$ Also the amplitude is a positive quantity, $rgt0$
Combining you get $$0lt rltfrac{1}{2}$$
And to make the number purely imaginary the argument should be either $pi/2$ or $3pi/2$
$$theta+pi/6=pi/2space orspace 3pi/2$$
So the values for $theta$ will be $$theta = pi/3 space or space 4pi/3$$
Hope this helps ….
answered Jan 30 at 16:42
SNEHIL SANYALSNEHIL SANYAL
656110
$endgroup$
add a comment |
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$begingroup$
Let $z_1=2e^{ipi/6}$ and $z_2=re^{itheta}$ where anyways $rgt0$ and $0leqthetalt2pi$
Now the product $z_1z_2=2re^{i(theta+ipi/6)}$
The argument of the above complex number is $2r$ and for the first part:
$1.)$$$2r>5$$ which means $$r>frac{5}{2}$$
To make $z_1z_2$ as a real number the argument should be $0$ $or$ $2pi$ which means
$$ theta+pi/6=2pi$$ giving $$theta=11pi/6$$
And for the second Part:
$2.)$ The modulus of $z_1z_2$ is to be less than $1$ so
$$2rlt1$$
Giving $$rltfrac{1}{2}$$ Also the amplitude is a positive quantity, $rgt0$
Combining you get $$0lt rltfrac{1}{2}$$
And to make the number purely imaginary the argument should be either $pi/2$ or $3pi/2$
$$theta+pi/6=pi/2space orspace 3pi/2$$
So the values for $theta$ will be $$theta = pi/3 space or space 4pi/3$$
Hope this helps ….
answered Jan 30 at 16:42
SNEHIL SANYALSNEHIL SANYAL
656110
$endgroup$
add a comment |
$begingroup$
Let $z_1=2e^{ipi/6}$ and $z_2=re^{itheta}$ where anyways $rgt0$ and $0leqthetalt2pi$
Now the product $z_1z_2=2re^{i(theta+ipi/6)}$
The argument of the above complex number is $2r$ and for the first part:
$1.)$$$2r>5$$ which means $$r>frac{5}{2}$$
To make $z_1z_2$ as a real number the argument should be $0$ $or$ $2pi$ which means
$$ theta+pi/6=2pi$$ giving $$theta=11pi/6$$
And for the second Part:
$2.)$ The modulus of $z_1z_2$ is to be less than $1$ so
$$2rlt1$$
Giving $$rltfrac{1}{2}$$ Also the amplitude is a positive quantity, $rgt0$
Combining you get $$0lt rltfrac{1}{2}$$
And to make the number purely imaginary the argument should be either $pi/2$ or $3pi/2$
$$theta+pi/6=pi/2space orspace 3pi/2$$
So the values for $theta$ will be $$theta = pi/3 space or space 4pi/3$$
Hope this helps ….
answered Jan 30 at 16:42
SNEHIL SANYALSNEHIL SANYAL
656110
$endgroup$
add a comment |
$begingroup$
Let $z_1=2e^{ipi/6}$ and $z_2=re^{itheta}$ where anyways $rgt0$ and $0leqthetalt2pi$
Now the product $z_1z_2=2re^{i(theta+ipi/6)}$
The argument of the above complex number is $2r$ and for the first part:
$1.)$$$2r>5$$ which means $$r>frac{5}{2}$$
To make $z_1z_2$ as a real number the argument should be $0$ $or$ $2pi$ which means
$$ theta+pi/6=2pi$$ giving $$theta=11pi/6$$
And for the second Part:
$2.)$ The modulus of $z_1z_2$ is to be less than $1$ so
$$2rlt1$$
Giving $$rltfrac{1}{2}$$ Also the amplitude is a positive quantity, $rgt0$
Combining you get $$0lt rltfrac{1}{2}$$
And to make the number purely imaginary the argument should be either $pi/2$ or $3pi/2$
$$theta+pi/6=pi/2space orspace 3pi/2$$
So the values for $theta$ will be $$theta = pi/3 space or space 4pi/3$$
Hope this helps ….
answered Jan 30 at 16:42
SNEHIL SANYALSNEHIL SANYAL
656110
$endgroup$
Let $z_1=2e^{ipi/6}$ and $z_2=re^{itheta}$ where anyways $rgt0$ and $0leqthetalt2pi$
Now the product $z_1z_2=2re^{i(theta+ipi/6)}$
The argument of the above complex number is $2r$ and for the first part:
$1.)$$$2r>5$$ which means $$r>frac{5}{2}$$
To make $z_1z_2$ as a real number the argument should be $0$ $or$ $2pi$ which means
$$ theta+pi/6=2pi$$ giving $$theta=11pi/6$$
And for the second Part:
$2.)$ The modulus of $z_1z_2$ is to be less than $1$ so
$$2rlt1$$
Giving $$rltfrac{1}{2}$$ Also the amplitude is a positive quantity, $rgt0$
Combining you get $$0lt rltfrac{1}{2}$$
And to make the number purely imaginary the argument should be either $pi/2$ or $3pi/2$
$$theta+pi/6=pi/2space orspace 3pi/2$$
So the values for $theta$ will be $$theta = pi/3 space or space 4pi/3$$
Hope this helps ….
answered Jan 30 at 16:42
SNEHIL SANYALSNEHIL SANYAL
656110
answered Jan 30 at 16:42
SNEHIL SANYALSNEHIL SANYAL
656110
answered Jan 30 at 16:42
SNEHIL SANYALSNEHIL SANYAL
656110
answered Jan 30 at 16:42
SNEHIL SANYALSNEHIL SANYAL
656110
656110
add a comment |
add a comment |
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1
$begingroup$
Hint: cis$theta_1 times $cis$theta_2 = $cis$(theta_1+theta_2)$.
$endgroup$
– Math Lover
Jan 29 at 20:20
$begingroup$
What is 'cis' here?
$endgroup$
– Berci
Jan 29 at 21:59
$begingroup$
$cis(theta)=cos(theta)+icdot sin(theta)$, where $i^2=-1$
$endgroup$
– J. W. Tanner
Jan 29 at 22:36