Resolution exercises on complex numbers
How do I solve this equation in the field of complex numbers?:
$$|z|^2 - z|z| + z = 0 $$
My solutions are:
$$z_1 = 0$$
$$z_2 = -1$$
complex-numbers
asked Nov 21 '18 at 7:35
Vincenzo Iannucci
343
add a comment |
How do I solve this equation in the field of complex numbers?:
$$|z|^2 - z|z| + z = 0 $$
My solutions are:
$$z_1 = 0$$
$$z_2 = -1$$
complex-numbers
asked Nov 21 '18 at 7:35
Vincenzo Iannucci
343
1
$-1$ is not a solution as you would get $1+1-1not =0$
– Henry
Nov 21 '18 at 11:20
add a comment |
How do I solve this equation in the field of complex numbers?:
$$|z|^2 - z|z| + z = 0 $$
My solutions are:
$$z_1 = 0$$
$$z_2 = -1$$
complex-numbers
asked Nov 21 '18 at 7:35
Vincenzo Iannucci
343
How do I solve this equation in the field of complex numbers?:
$$|z|^2 - z|z| + z = 0 $$
My solutions are:
$$z_1 = 0$$
$$z_2 = -1$$
complex-numbers
complex-numbers
asked Nov 21 '18 at 7:35
Vincenzo Iannucci
343
asked Nov 21 '18 at 7:35
Vincenzo Iannucci
343
edited Nov 21 '18 at 7:41
asked Nov 21 '18 at 7:35
Vincenzo Iannucci
343
asked Nov 21 '18 at 7:35
Vincenzo Iannucci
343
asked Nov 21 '18 at 7:35
Vincenzo Iannucci
343
343
1
$-1$ is not a solution as you would get $1+1-1not =0$
– Henry
Nov 21 '18 at 11:20
add a comment |
1
$-1$ is not a solution as you would get $1+1-1not =0$
– Henry
Nov 21 '18 at 11:20
1
1
$-1$ is not a solution as you would get $1+1-1not =0$
– Henry
Nov 21 '18 at 11:20
$-1$ is not a solution as you would get $1+1-1not =0$
– Henry
Nov 21 '18 at 11:20
add a comment |
5 Answers
5
active
oldest
votes
$$z=dfrac{|z|^2}{|z|-1}$$ which is real
If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable
If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$
answered Nov 21 '18 at 7:58
lab bhattacharjee
223k15156274
Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
– ruakh
Nov 21 '18 at 23:29
add a comment |
Observe that $z=0$ is a solution of the equation
$$|z|^2 - z|z| + z = 0.$$
( $z=-1$ is not a solution !)
Therefore let $z ne 0$ be a further solution of this equation. We get , since $|z|^2=z overline{z}:$
$overline{z}-|z|+1=0$. This gives $overline{z}=|z|-1 in mathbb R$.Hence
$z=|z|-1 in mathbb R$.
If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.
answered Nov 21 '18 at 7:42
Fred
44.2k1845
add a comment |
WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real
$$r(r-r(cos t+isin t)+(cos t+isin t))=0$$
If $rne0,$ equating the real & the imaginary parts
$r-rcos t+cos t=0=(1-r)sin t$
Case $#1:$
If $r=1,1=0$ which is untenable
If $sin t=0,$
Case $#2A:cos t=1,r=0$ which is untenable
Case $#2A:cos t=-1,r+r-1=0iff r=?$
answered Nov 21 '18 at 7:42
lab bhattacharjee
223k15156274
add a comment |
You may also proceed as follows:
- Rewrite the equation to
$$|z|^2 - z|z| + z = 0 Leftrightarrow boxed{|z|^2 = z(|z|-1)}$$
- Therefore, $color{blue}{z}$ must be $color{blue}{mbox{real}}$ and so we have $color{blue}{|z|^2 = z^2}$.
- Noting the solution $boxed{z = 0}$ we get
$$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$
answered Nov 21 '18 at 8:29
trancelocation
9,1051521
add a comment |
Let $r = e^{itheta}$.
We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.
Factorising, we get $r = 0$, giving $z = 0$ as one solution or:
$r - re^{itheta} + e^{itheta} = 0$
$e^{itheta} = frac{r}{r-1}$
From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.
$frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.
Therefore the two solutions are $z = 0$ and $z = -frac 12$.
answered Nov 21 '18 at 8:58
Deepak
16.8k11436
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$$z=dfrac{|z|^2}{|z|-1}$$ which is real
If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable
If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$
answered Nov 21 '18 at 7:58
lab bhattacharjee
223k15156274
Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
– ruakh
Nov 21 '18 at 23:29
add a comment |
$$z=dfrac{|z|^2}{|z|-1}$$ which is real
If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable
If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$
answered Nov 21 '18 at 7:58
lab bhattacharjee
223k15156274
Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
– ruakh
Nov 21 '18 at 23:29
add a comment |
$$z=dfrac{|z|^2}{|z|-1}$$ which is real
If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable
If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$
answered Nov 21 '18 at 7:58
lab bhattacharjee
223k15156274
$$z=dfrac{|z|^2}{|z|-1}$$ which is real
If $z>0,|z|=+z$ $$0=z^2-z^2+ziff z=0$$ which is untenable
If $zle0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$
answered Nov 21 '18 at 7:58
lab bhattacharjee
223k15156274
edited Nov 21 '18 at 13:51
answered Nov 21 '18 at 7:58
lab bhattacharjee
223k15156274
answered Nov 21 '18 at 7:58
lab bhattacharjee
223k15156274
answered Nov 21 '18 at 7:58
lab bhattacharjee
223k15156274
223k15156274
Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
– ruakh
Nov 21 '18 at 23:29
add a comment |
Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
– ruakh
Nov 21 '18 at 23:29
Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
– ruakh
Nov 21 '18 at 23:29
Your first step only works when $|z| ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.)
– ruakh
Nov 21 '18 at 23:29
add a comment |
Observe that $z=0$ is a solution of the equation
$$|z|^2 - z|z| + z = 0.$$
( $z=-1$ is not a solution !)
Therefore let $z ne 0$ be a further solution of this equation. We get , since $|z|^2=z overline{z}:$
$overline{z}-|z|+1=0$. This gives $overline{z}=|z|-1 in mathbb R$.Hence
$z=|z|-1 in mathbb R$.
If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.
answered Nov 21 '18 at 7:42
Fred
44.2k1845
add a comment |
Observe that $z=0$ is a solution of the equation
$$|z|^2 - z|z| + z = 0.$$
( $z=-1$ is not a solution !)
Therefore let $z ne 0$ be a further solution of this equation. We get , since $|z|^2=z overline{z}:$
$overline{z}-|z|+1=0$. This gives $overline{z}=|z|-1 in mathbb R$.Hence
$z=|z|-1 in mathbb R$.
If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.
answered Nov 21 '18 at 7:42
Fred
44.2k1845
add a comment |
Observe that $z=0$ is a solution of the equation
$$|z|^2 - z|z| + z = 0.$$
( $z=-1$ is not a solution !)
Therefore let $z ne 0$ be a further solution of this equation. We get , since $|z|^2=z overline{z}:$
$overline{z}-|z|+1=0$. This gives $overline{z}=|z|-1 in mathbb R$.Hence
$z=|z|-1 in mathbb R$.
If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.
answered Nov 21 '18 at 7:42
Fred
44.2k1845
Observe that $z=0$ is a solution of the equation
$$|z|^2 - z|z| + z = 0.$$
( $z=-1$ is not a solution !)
Therefore let $z ne 0$ be a further solution of this equation. We get , since $|z|^2=z overline{z}:$
$overline{z}-|z|+1=0$. This gives $overline{z}=|z|-1 in mathbb R$.Hence
$z=|z|-1 in mathbb R$.
If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.
answered Nov 21 '18 at 7:42
Fred
44.2k1845
answered Nov 21 '18 at 7:42
Fred
44.2k1845
answered Nov 21 '18 at 7:42
Fred
44.2k1845
answered Nov 21 '18 at 7:42
Fred
44.2k1845
44.2k1845
add a comment |
add a comment |
WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real
$$r(r-r(cos t+isin t)+(cos t+isin t))=0$$
If $rne0,$ equating the real & the imaginary parts
$r-rcos t+cos t=0=(1-r)sin t$
Case $#1:$
If $r=1,1=0$ which is untenable
If $sin t=0,$
Case $#2A:cos t=1,r=0$ which is untenable
Case $#2A:cos t=-1,r+r-1=0iff r=?$
answered Nov 21 '18 at 7:42
lab bhattacharjee
223k15156274
add a comment |
WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real
$$r(r-r(cos t+isin t)+(cos t+isin t))=0$$
If $rne0,$ equating the real & the imaginary parts
$r-rcos t+cos t=0=(1-r)sin t$
Case $#1:$
If $r=1,1=0$ which is untenable
If $sin t=0,$
Case $#2A:cos t=1,r=0$ which is untenable
Case $#2A:cos t=-1,r+r-1=0iff r=?$
answered Nov 21 '18 at 7:42
lab bhattacharjee
223k15156274
add a comment |
WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real
$$r(r-r(cos t+isin t)+(cos t+isin t))=0$$
If $rne0,$ equating the real & the imaginary parts
$r-rcos t+cos t=0=(1-r)sin t$
Case $#1:$
If $r=1,1=0$ which is untenable
If $sin t=0,$
Case $#2A:cos t=1,r=0$ which is untenable
Case $#2A:cos t=-1,r+r-1=0iff r=?$
answered Nov 21 '18 at 7:42
lab bhattacharjee
223k15156274
WLOG $z=r(cos t+isin t)$ where $r>0,t$ are real
$$r(r-r(cos t+isin t)+(cos t+isin t))=0$$
If $rne0,$ equating the real & the imaginary parts
$r-rcos t+cos t=0=(1-r)sin t$
Case $#1:$
If $r=1,1=0$ which is untenable
If $sin t=0,$
Case $#2A:cos t=1,r=0$ which is untenable
Case $#2A:cos t=-1,r+r-1=0iff r=?$
answered Nov 21 '18 at 7:42
lab bhattacharjee
223k15156274
answered Nov 21 '18 at 7:42
lab bhattacharjee
223k15156274
answered Nov 21 '18 at 7:42
lab bhattacharjee
223k15156274
answered Nov 21 '18 at 7:42
lab bhattacharjee
223k15156274
223k15156274
add a comment |
add a comment |
You may also proceed as follows:
- Rewrite the equation to
$$|z|^2 - z|z| + z = 0 Leftrightarrow boxed{|z|^2 = z(|z|-1)}$$
- Therefore, $color{blue}{z}$ must be $color{blue}{mbox{real}}$ and so we have $color{blue}{|z|^2 = z^2}$.
- Noting the solution $boxed{z = 0}$ we get
$$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$
answered Nov 21 '18 at 8:29
trancelocation
9,1051521
add a comment |
You may also proceed as follows:
- Rewrite the equation to
$$|z|^2 - z|z| + z = 0 Leftrightarrow boxed{|z|^2 = z(|z|-1)}$$
- Therefore, $color{blue}{z}$ must be $color{blue}{mbox{real}}$ and so we have $color{blue}{|z|^2 = z^2}$.
- Noting the solution $boxed{z = 0}$ we get
$$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$
answered Nov 21 '18 at 8:29
trancelocation
9,1051521
add a comment |
You may also proceed as follows:
- Rewrite the equation to
$$|z|^2 - z|z| + z = 0 Leftrightarrow boxed{|z|^2 = z(|z|-1)}$$
- Therefore, $color{blue}{z}$ must be $color{blue}{mbox{real}}$ and so we have $color{blue}{|z|^2 = z^2}$.
- Noting the solution $boxed{z = 0}$ we get
$$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$
answered Nov 21 '18 at 8:29
trancelocation
9,1051521
You may also proceed as follows:
- Rewrite the equation to
$$|z|^2 - z|z| + z = 0 Leftrightarrow boxed{|z|^2 = z(|z|-1)}$$
- Therefore, $color{blue}{z}$ must be $color{blue}{mbox{real}}$ and so we have $color{blue}{|z|^2 = z^2}$.
- Noting the solution $boxed{z = 0}$ we get
$$|z|^2 = z(|z|-1) stackrel{z in mathbb{R}, z neq 0}{Leftrightarrow}z = |z|-1 Rightarrow boxed{z = -frac{1}{2}}$$
answered Nov 21 '18 at 8:29
trancelocation
9,1051521
edited Nov 21 '18 at 15:38
answered Nov 21 '18 at 8:29
trancelocation
9,1051521
answered Nov 21 '18 at 8:29
trancelocation
9,1051521
answered Nov 21 '18 at 8:29
trancelocation
9,1051521
9,1051521
add a comment |
add a comment |
Let $r = e^{itheta}$.
We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.
Factorising, we get $r = 0$, giving $z = 0$ as one solution or:
$r - re^{itheta} + e^{itheta} = 0$
$e^{itheta} = frac{r}{r-1}$
From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.
$frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.
Therefore the two solutions are $z = 0$ and $z = -frac 12$.
answered Nov 21 '18 at 8:58
Deepak
16.8k11436
add a comment |
Let $r = e^{itheta}$.
We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.
Factorising, we get $r = 0$, giving $z = 0$ as one solution or:
$r - re^{itheta} + e^{itheta} = 0$
$e^{itheta} = frac{r}{r-1}$
From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.
$frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.
Therefore the two solutions are $z = 0$ and $z = -frac 12$.
answered Nov 21 '18 at 8:58
Deepak
16.8k11436
add a comment |
Let $r = e^{itheta}$.
We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.
Factorising, we get $r = 0$, giving $z = 0$ as one solution or:
$r - re^{itheta} + e^{itheta} = 0$
$e^{itheta} = frac{r}{r-1}$
From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.
$frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.
Therefore the two solutions are $z = 0$ and $z = -frac 12$.
answered Nov 21 '18 at 8:58
Deepak
16.8k11436
Let $r = e^{itheta}$.
We get $r^2 - re^{itheta}cdot r + re^{itheta} = 0$.
Factorising, we get $r = 0$, giving $z = 0$ as one solution or:
$r - re^{itheta} + e^{itheta} = 0$
$e^{itheta} = frac{r}{r-1}$
From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{itheta} in mathbb{R}$. Hence $e^{itheta} = pm 1$.
$frac{r}{r-1} = 1$ gives no solution, but $frac{r}{r-1} = -1 implies r = frac 12$. Since $e^{itheta} = -1$ that gives $z = -frac 12$.
Therefore the two solutions are $z = 0$ and $z = -frac 12$.
answered Nov 21 '18 at 8:58
Deepak
16.8k11436
answered Nov 21 '18 at 8:58
Deepak
16.8k11436
answered Nov 21 '18 at 8:58
Deepak
16.8k11436
answered Nov 21 '18 at 8:58
Deepak
16.8k11436
16.8k11436
add a comment |
add a comment |
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1
$-1$ is not a solution as you would get $1+1-1not =0$
– Henry
Nov 21 '18 at 11:20