Find all solutions $z^2overlinez^3=32$

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up vote
3
down vote

favorite

$z^2overlinez^3=32$

I have bought about the following way:

to simplify it by $z^2overlinez^3=z^2overlinez^2overlinez=(zoverlinez)^2overlinez=|z|^4overlinez$

to replace $z$ or by $z=x+iy$ or $z=re^itheta$ which should I choose? I have tried both with no success

edited Jul 27 at 16:03

greedoid

26.1k93473

asked Jul 27 at 15:58

newhere

742310

add a commentÂ |Â

up vote
3
down vote

favorite

$z^2overlinez^3=32$

I have bought about the following way:

to simplify it by $z^2overlinez^3=z^2overlinez^2overlinez=(zoverlinez)^2overlinez=|z|^4overlinez$

to replace $z$ or by $z=x+iy$ or $z=re^itheta$ which should I choose? I have tried both with no success

edited Jul 27 at 16:03

greedoid

26.1k93473

asked Jul 27 at 15:58

newhere

742310

add a commentÂ |Â

up vote
3
down vote

favorite

$z^2overlinez^3=32$

I have bought about the following way:

to simplify it by $z^2overlinez^3=z^2overlinez^2overlinez=(zoverlinez)^2overlinez=|z|^4overlinez$

to replace $z$ or by $z=x+iy$ or $z=re^itheta$ which should I choose? I have tried both with no success

edited Jul 27 at 16:03

greedoid

26.1k93473

asked Jul 27 at 15:58

newhere

742310

$z^2overlinez^3=32$

I have bought about the following way:

to simplify it by $z^2overlinez^3=z^2overlinez^2overlinez=(zoverlinez)^2overlinez=|z|^4overlinez$

to replace $z$ or by $z=x+iy$ or $z=re^itheta$ which should I choose? I have tried both with no success

edited Jul 27 at 16:03

greedoid

26.1k93473

asked Jul 27 at 15:58

newhere

742310

edited Jul 27 at 16:03

greedoid

26.1k93473

edited Jul 27 at 16:03

greedoid

26.1k93473

edited Jul 27 at 16:03

greedoid

26.1k93473

asked Jul 27 at 15:58

newhere

742310

asked Jul 27 at 15:58

newhere

742310

asked Jul 27 at 15:58

newhere

742310

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
4
down vote

accepted

Yes we have that

$$z^2overlinez^3=z^2overlinez^2overlinez=(zoverlinez)^2overlinez=|z|^4overlinez=32$$

but then $bar z=x$ must be real and since $x^5=32 implies x=2$ the only solution is $z=2$.

edited Jul 27 at 16:13

answered Jul 27 at 16:00

gimusi

64.9k73483

2

I think the observation that $bar z$ must be real is the key to doing this question simply and quickly
â€“Â Mark Bennet
Jul 27 at 16:27

@MarcBennet I agree with you! Thanks
â€“Â gimusi
Jul 27 at 16:29

@gimusi Why did $overlinez$ have to be real? did you conclude it from the powers $|z|^4overlinez=2^5$?
â€“Â newhere
Jul 28 at 18:47

@gimusi rethinking: a product of a real number $|z|^4$ with a real to complex number $overlinez$ which gives a real number then $overlinez$ must be real
â€“Â newhere
Jul 28 at 18:49

@newhere Exactly! Well done.
â€“Â gimusi
Jul 28 at 18:56

add a commentÂ |Â

up vote
1
down vote

If you use $z=re^itheta$ with $r in mathbb R_ge 0$ and $theta in mathbb R$

then $z^2overlinez^3=32$ gives you $r^5 e^-itheta=32e^i2npi$ for integer $n$

so matching coefficients gives $r=2$ and $theta = -2npi$

and thus solutions are of the form $z=2e^-i2npi$ and the only one is $z=2$

answered Jul 27 at 16:13

Henry

92.9k469147

add a commentÂ |Â

up vote
1
down vote

Quite often it useful to see what can you say about an absolute value of a complex number and this is such a problem.

$$z^2overlinez^3=32implies |z^2overlinez^3|=32implies |z|^5 = 32implies |z|=2$$

so $$z^2overlinez^3=32implies |z|^4overlinez=32implies overlinez=2implies z=2$$

edited Jul 27 at 16:18

answered Jul 27 at 16:02

greedoid

26.1k93473

add a commentÂ |Â

up vote
0
down vote

I like using

$z = re^itheta; tag 1$

we have

$bar z = r e^-itheta; tag 2$

we are given that

$z^2 bar z^3 = 32; tag 3$

using (1) and (2) yields

$r^5 e^-itheta = r^2 e^2i theta r^3 e^-3i theta = z^2 bar z^3 = 32, tag 4$

whence

$r^5 = vert r vert^5 = vert r^5 e^-itheta vert = vert 32 vert = 32; tag 5$

it follows that

$r = 2; tag 6$

therefore, by (4),

$32 e^-itheta = 32 Longrightarrow e^-itheta = 1 Longrightarrow e^itheta = 1; tag 7$

therefore (1) becomes

$z = 2. tag 8$

answered Jul 27 at 16:09

Robert Lewis

36.8k22155

add a commentÂ |Â

up vote
0
down vote

We have

$$
z^2bar z^3 = left(zbar zright)^2bar z = |z|^4bar z = 32Rightarrow bar z = alpha + 0(-j) = alpha = z
$$

hence

$$
alpha^5 = 32Rightarrow alpha = 2 = z
$$

answered Jul 27 at 16:25

Cesareo

5,6392412

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
4
down vote

accepted

Yes we have that

$$z^2overlinez^3=z^2overlinez^2overlinez=(zoverlinez)^2overlinez=|z|^4overlinez=32$$

but then $bar z=x$ must be real and since $x^5=32 implies x=2$ the only solution is $z=2$.

edited Jul 27 at 16:13

answered Jul 27 at 16:00

gimusi

64.9k73483

2

I think the observation that $bar z$ must be real is the key to doing this question simply and quickly
â€“Â Mark Bennet
Jul 27 at 16:27

@MarcBennet I agree with you! Thanks
â€“Â gimusi
Jul 27 at 16:29

@gimusi Why did $overlinez$ have to be real? did you conclude it from the powers $|z|^4overlinez=2^5$?
â€“Â newhere
Jul 28 at 18:47

@gimusi rethinking: a product of a real number $|z|^4$ with a real to complex number $overlinez$ which gives a real number then $overlinez$ must be real
â€“Â newhere
Jul 28 at 18:49

@newhere Exactly! Well done.
â€“Â gimusi
Jul 28 at 18:56

add a commentÂ |Â

up vote
4
down vote

accepted

Yes we have that

$$z^2overlinez^3=z^2overlinez^2overlinez=(zoverlinez)^2overlinez=|z|^4overlinez=32$$

but then $bar z=x$ must be real and since $x^5=32 implies x=2$ the only solution is $z=2$.

edited Jul 27 at 16:13

answered Jul 27 at 16:00

gimusi

64.9k73483

2

I think the observation that $bar z$ must be real is the key to doing this question simply and quickly
â€“Â Mark Bennet
Jul 27 at 16:27

@MarcBennet I agree with you! Thanks
â€“Â gimusi
Jul 27 at 16:29

@gimusi Why did $overlinez$ have to be real? did you conclude it from the powers $|z|^4overlinez=2^5$?
â€“Â newhere
Jul 28 at 18:47

@gimusi rethinking: a product of a real number $|z|^4$ with a real to complex number $overlinez$ which gives a real number then $overlinez$ must be real
â€“Â newhere
Jul 28 at 18:49

@newhere Exactly! Well done.
â€“Â gimusi
Jul 28 at 18:56

add a commentÂ |Â

up vote
4
down vote

accepted

Yes we have that

$$z^2overlinez^3=z^2overlinez^2overlinez=(zoverlinez)^2overlinez=|z|^4overlinez=32$$

but then $bar z=x$ must be real and since $x^5=32 implies x=2$ the only solution is $z=2$.

edited Jul 27 at 16:13

answered Jul 27 at 16:00

gimusi

64.9k73483

Yes we have that

$$z^2overlinez^3=z^2overlinez^2overlinez=(zoverlinez)^2overlinez=|z|^4overlinez=32$$

but then $bar z=x$ must be real and since $x^5=32 implies x=2$ the only solution is $z=2$.

edited Jul 27 at 16:13

answered Jul 27 at 16:00

gimusi

64.9k73483

edited Jul 27 at 16:13

answered Jul 27 at 16:00

gimusi

64.9k73483

answered Jul 27 at 16:00

gimusi

64.9k73483

answered Jul 27 at 16:00

gimusi

64.9k73483

2

I think the observation that $bar z$ must be real is the key to doing this question simply and quickly
â€“Â Mark Bennet
Jul 27 at 16:27

@MarcBennet I agree with you! Thanks
â€“Â gimusi
Jul 27 at 16:29

@gimusi Why did $overlinez$ have to be real? did you conclude it from the powers $|z|^4overlinez=2^5$?
â€“Â newhere
Jul 28 at 18:47

@gimusi rethinking: a product of a real number $|z|^4$ with a real to complex number $overlinez$ which gives a real number then $overlinez$ must be real
â€“Â newhere
Jul 28 at 18:49

@newhere Exactly! Well done.
â€“Â gimusi
Jul 28 at 18:56

add a commentÂ |Â

2

I think the observation that $bar z$ must be real is the key to doing this question simply and quickly
â€“Â Mark Bennet
Jul 27 at 16:27

@MarcBennet I agree with you! Thanks
â€“Â gimusi
Jul 27 at 16:29

@gimusi Why did $overlinez$ have to be real? did you conclude it from the powers $|z|^4overlinez=2^5$?
â€“Â newhere
Jul 28 at 18:47

@gimusi rethinking: a product of a real number $|z|^4$ with a real to complex number $overlinez$ which gives a real number then $overlinez$ must be real
â€“Â newhere
Jul 28 at 18:49

@newhere Exactly! Well done.
â€“Â gimusi
Jul 28 at 18:56

I think the observation that $bar z$ must be real is the key to doing this question simply and quickly
â€“Â Mark Bennet
Jul 27 at 16:27

@MarcBennet I agree with you! Thanks
â€“Â gimusi
Jul 27 at 16:29

@gimusi Why did $overlinez$ have to be real? did you conclude it from the powers $|z|^4overlinez=2^5$?
â€“Â newhere
Jul 28 at 18:47

@gimusi rethinking: a product of a real number $|z|^4$ with a real to complex number $overlinez$ which gives a real number then $overlinez$ must be real
â€“Â newhere
Jul 28 at 18:49

@newhere Exactly! Well done.
â€“Â gimusi
Jul 28 at 18:56

add a commentÂ |Â

up vote
1
down vote

If you use $z=re^itheta$ with $r in mathbb R_ge 0$ and $theta in mathbb R$

then $z^2overlinez^3=32$ gives you $r^5 e^-itheta=32e^i2npi$ for integer $n$

so matching coefficients gives $r=2$ and $theta = -2npi$

and thus solutions are of the form $z=2e^-i2npi$ and the only one is $z=2$

answered Jul 27 at 16:13

Henry

92.9k469147

add a commentÂ |Â

up vote
1
down vote

If you use $z=re^itheta$ with $r in mathbb R_ge 0$ and $theta in mathbb R$

then $z^2overlinez^3=32$ gives you $r^5 e^-itheta=32e^i2npi$ for integer $n$

so matching coefficients gives $r=2$ and $theta = -2npi$

and thus solutions are of the form $z=2e^-i2npi$ and the only one is $z=2$

answered Jul 27 at 16:13

Henry

92.9k469147

add a commentÂ |Â

up vote
1
down vote

If you use $z=re^itheta$ with $r in mathbb R_ge 0$ and $theta in mathbb R$

then $z^2overlinez^3=32$ gives you $r^5 e^-itheta=32e^i2npi$ for integer $n$

so matching coefficients gives $r=2$ and $theta = -2npi$

and thus solutions are of the form $z=2e^-i2npi$ and the only one is $z=2$

answered Jul 27 at 16:13

Henry

92.9k469147

If you use $z=re^itheta$ with $r in mathbb R_ge 0$ and $theta in mathbb R$

then $z^2overlinez^3=32$ gives you $r^5 e^-itheta=32e^i2npi$ for integer $n$

so matching coefficients gives $r=2$ and $theta = -2npi$

and thus solutions are of the form $z=2e^-i2npi$ and the only one is $z=2$

answered Jul 27 at 16:13

Henry

92.9k469147

answered Jul 27 at 16:13

Henry

92.9k469147

answered Jul 27 at 16:13

Henry

92.9k469147

answered Jul 27 at 16:13

Henry

92.9k469147

add a commentÂ |Â

up vote
1
down vote

Quite often it useful to see what can you say about an absolute value of a complex number and this is such a problem.

$$z^2overlinez^3=32implies |z^2overlinez^3|=32implies |z|^5 = 32implies |z|=2$$

so $$z^2overlinez^3=32implies |z|^4overlinez=32implies overlinez=2implies z=2$$

edited Jul 27 at 16:18

answered Jul 27 at 16:02

greedoid

26.1k93473

add a commentÂ |Â

up vote
1
down vote

Quite often it useful to see what can you say about an absolute value of a complex number and this is such a problem.

$$z^2overlinez^3=32implies |z^2overlinez^3|=32implies |z|^5 = 32implies |z|=2$$

so $$z^2overlinez^3=32implies |z|^4overlinez=32implies overlinez=2implies z=2$$

edited Jul 27 at 16:18

answered Jul 27 at 16:02

greedoid

26.1k93473

add a commentÂ |Â

up vote
1
down vote

Quite often it useful to see what can you say about an absolute value of a complex number and this is such a problem.

$$z^2overlinez^3=32implies |z^2overlinez^3|=32implies |z|^5 = 32implies |z|=2$$

so $$z^2overlinez^3=32implies |z|^4overlinez=32implies overlinez=2implies z=2$$

edited Jul 27 at 16:18

answered Jul 27 at 16:02

greedoid

26.1k93473

Quite often it useful to see what can you say about an absolute value of a complex number and this is such a problem.

$$z^2overlinez^3=32implies |z^2overlinez^3|=32implies |z|^5 = 32implies |z|=2$$

so $$z^2overlinez^3=32implies |z|^4overlinez=32implies overlinez=2implies z=2$$

edited Jul 27 at 16:18

answered Jul 27 at 16:02

greedoid

26.1k93473

edited Jul 27 at 16:18

answered Jul 27 at 16:02

greedoid

26.1k93473

answered Jul 27 at 16:02

greedoid

26.1k93473

answered Jul 27 at 16:02

greedoid

26.1k93473

add a commentÂ |Â

up vote
0
down vote

I like using

$z = re^itheta; tag 1$

we have

$bar z = r e^-itheta; tag 2$

we are given that

$z^2 bar z^3 = 32; tag 3$

using (1) and (2) yields

$r^5 e^-itheta = r^2 e^2i theta r^3 e^-3i theta = z^2 bar z^3 = 32, tag 4$

whence

$r^5 = vert r vert^5 = vert r^5 e^-itheta vert = vert 32 vert = 32; tag 5$

it follows that

$r = 2; tag 6$

therefore, by (4),

$32 e^-itheta = 32 Longrightarrow e^-itheta = 1 Longrightarrow e^itheta = 1; tag 7$

therefore (1) becomes

$z = 2. tag 8$

answered Jul 27 at 16:09

Robert Lewis

36.8k22155

add a commentÂ |Â

up vote
0
down vote

I like using

$z = re^itheta; tag 1$

we have

$bar z = r e^-itheta; tag 2$

we are given that

$z^2 bar z^3 = 32; tag 3$

using (1) and (2) yields

$r^5 e^-itheta = r^2 e^2i theta r^3 e^-3i theta = z^2 bar z^3 = 32, tag 4$

whence

$r^5 = vert r vert^5 = vert r^5 e^-itheta vert = vert 32 vert = 32; tag 5$

it follows that

$r = 2; tag 6$

therefore, by (4),

$32 e^-itheta = 32 Longrightarrow e^-itheta = 1 Longrightarrow e^itheta = 1; tag 7$

therefore (1) becomes

$z = 2. tag 8$

answered Jul 27 at 16:09

Robert Lewis

36.8k22155

add a commentÂ |Â

up vote
0
down vote

I like using

$z = re^itheta; tag 1$

we have

$bar z = r e^-itheta; tag 2$

we are given that

$z^2 bar z^3 = 32; tag 3$

using (1) and (2) yields

$r^5 e^-itheta = r^2 e^2i theta r^3 e^-3i theta = z^2 bar z^3 = 32, tag 4$

whence

$r^5 = vert r vert^5 = vert r^5 e^-itheta vert = vert 32 vert = 32; tag 5$

it follows that

$r = 2; tag 6$

therefore, by (4),

$32 e^-itheta = 32 Longrightarrow e^-itheta = 1 Longrightarrow e^itheta = 1; tag 7$

therefore (1) becomes

$z = 2. tag 8$

answered Jul 27 at 16:09

Robert Lewis

36.8k22155

I like using

$z = re^itheta; tag 1$

we have

$bar z = r e^-itheta; tag 2$

we are given that

$z^2 bar z^3 = 32; tag 3$

using (1) and (2) yields

$r^5 e^-itheta = r^2 e^2i theta r^3 e^-3i theta = z^2 bar z^3 = 32, tag 4$

whence

$r^5 = vert r vert^5 = vert r^5 e^-itheta vert = vert 32 vert = 32; tag 5$

it follows that

$r = 2; tag 6$

therefore, by (4),

$32 e^-itheta = 32 Longrightarrow e^-itheta = 1 Longrightarrow e^itheta = 1; tag 7$

therefore (1) becomes

$z = 2. tag 8$

answered Jul 27 at 16:09

Robert Lewis

36.8k22155

answered Jul 27 at 16:09

Robert Lewis

36.8k22155

answered Jul 27 at 16:09

Robert Lewis

36.8k22155

answered Jul 27 at 16:09

Robert Lewis

36.8k22155

add a commentÂ |Â

up vote
0
down vote

We have

$$
z^2bar z^3 = left(zbar zright)^2bar z = |z|^4bar z = 32Rightarrow bar z = alpha + 0(-j) = alpha = z
$$

hence

$$
alpha^5 = 32Rightarrow alpha = 2 = z
$$

answered Jul 27 at 16:25

Cesareo

5,6392412

add a commentÂ |Â

up vote
0
down vote

We have

$$
z^2bar z^3 = left(zbar zright)^2bar z = |z|^4bar z = 32Rightarrow bar z = alpha + 0(-j) = alpha = z
$$

hence

$$
alpha^5 = 32Rightarrow alpha = 2 = z
$$

answered Jul 27 at 16:25

Cesareo

5,6392412

add a commentÂ |Â

up vote
0
down vote

We have

$$
z^2bar z^3 = left(zbar zright)^2bar z = |z|^4bar z = 32Rightarrow bar z = alpha + 0(-j) = alpha = z
$$

hence

$$
alpha^5 = 32Rightarrow alpha = 2 = z
$$

answered Jul 27 at 16:25

Cesareo

5,6392412

We have

$$
z^2bar z^3 = left(zbar zright)^2bar z = |z|^4bar z = 32Rightarrow bar z = alpha + 0(-j) = alpha = z
$$

hence

$$
alpha^5 = 32Rightarrow alpha = 2 = z
$$

answered Jul 27 at 16:25

Cesareo

5,6392412

answered Jul 27 at 16:25

Cesareo

5,6392412

answered Jul 27 at 16:25

Cesareo

5,6392412

answered Jul 27 at 16:25

Cesareo

5,6392412

add a commentÂ |Â

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