Mixing
Solve $z^2 + 4|z| + 4 = 0$ in $mathbb{C}$
$begingroup$
How can I solve this equation:
$z^2 + 4|z| + 4 = 0$?
If I rewrite $z$ as $a+ib$, I get 2 real solutions. But they are wrong.
Any help?
My attempt. By letting $z=a+ib$, we have that
$$(a+ib)^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
$$ a^2 + 2iab - b^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
$$a^2 -b^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
Then $2ab = 0 $.
So, for $a = 0 implies z=2$ and $b = 0implies z=-2$
complex-numbers
edited Jan 27 at 14:53
Robert Z
101k1070143
asked Jan 27 at 13:42
th3gr3ymanth3gr3yman
113
$endgroup$
add a comment |
$begingroup$
How can I solve this equation:
$z^2 + 4|z| + 4 = 0$?
If I rewrite $z$ as $a+ib$, I get 2 real solutions. But they are wrong.
Any help?
My attempt. By letting $z=a+ib$, we have that
$$(a+ib)^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
$$ a^2 + 2iab - b^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
$$a^2 -b^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
Then $2ab = 0 $.
So, for $a = 0 implies z=2$ and $b = 0implies z=-2$
complex-numbers
edited Jan 27 at 14:53
Robert Z
101k1070143
asked Jan 27 at 13:42
th3gr3ymanth3gr3yman
113
$endgroup$
3
$begingroup$
If you could share your attempted solution, perhaps someone could tell you where you are wrong, in stead of writing up a full solution which may or may not help you find your mistake.
$endgroup$
– Servaes
Jan 27 at 13:44
$begingroup$
Tell us more about your attempt
$endgroup$
– Robert Z
Jan 27 at 13:47
add a comment |
$begingroup$
How can I solve this equation:
$z^2 + 4|z| + 4 = 0$?
If I rewrite $z$ as $a+ib$, I get 2 real solutions. But they are wrong.
Any help?
My attempt. By letting $z=a+ib$, we have that
$$(a+ib)^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
$$ a^2 + 2iab - b^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
$$a^2 -b^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
Then $2ab = 0 $.
So, for $a = 0 implies z=2$ and $b = 0implies z=-2$
complex-numbers
edited Jan 27 at 14:53
Robert Z
101k1070143
asked Jan 27 at 13:42
th3gr3ymanth3gr3yman
113
$endgroup$
How can I solve this equation:
$z^2 + 4|z| + 4 = 0$?
If I rewrite $z$ as $a+ib$, I get 2 real solutions. But they are wrong.
Any help?
My attempt. By letting $z=a+ib$, we have that
$$(a+ib)^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
$$ a^2 + 2iab - b^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
$$a^2 -b^2 + 4 sqrt{a^2+b^2} + 4 = 0 $$
Then $2ab = 0 $.
So, for $a = 0 implies z=2$ and $b = 0implies z=-2$
complex-numbers
complex-numbers
edited Jan 27 at 14:53
Robert Z
101k1070143
asked Jan 27 at 13:42
th3gr3ymanth3gr3yman
113
edited Jan 27 at 14:53
Robert Z
101k1070143
asked Jan 27 at 13:42
th3gr3ymanth3gr3yman
113
edited Jan 27 at 14:53
Robert Z
101k1070143
edited Jan 27 at 14:53
Robert Z
101k1070143
edited Jan 27 at 14:53
Robert Z
101k1070143
101k1070143
asked Jan 27 at 13:42
th3gr3ymanth3gr3yman
113
asked Jan 27 at 13:42
th3gr3ymanth3gr3yman
113
asked Jan 27 at 13:42
th3gr3ymanth3gr3yman
113
113
3
$begingroup$
If you could share your attempted solution, perhaps someone could tell you where you are wrong, in stead of writing up a full solution which may or may not help you find your mistake.
$endgroup$
– Servaes
Jan 27 at 13:44
$begingroup$
Tell us more about your attempt
$endgroup$
– Robert Z
Jan 27 at 13:47
add a comment |
3
$begingroup$
If you could share your attempted solution, perhaps someone could tell you where you are wrong, in stead of writing up a full solution which may or may not help you find your mistake.
$endgroup$
– Servaes
Jan 27 at 13:44
$begingroup$
Tell us more about your attempt
$endgroup$
– Robert Z
Jan 27 at 13:47
3
3
$begingroup$
If you could share your attempted solution, perhaps someone could tell you where you are wrong, in stead of writing up a full solution which may or may not help you find your mistake.
$endgroup$
– Servaes
Jan 27 at 13:44
$begingroup$
If you could share your attempted solution, perhaps someone could tell you where you are wrong, in stead of writing up a full solution which may or may not help you find your mistake.
$endgroup$
– Servaes
Jan 27 at 13:44
$begingroup$
Tell us more about your attempt
$endgroup$
– Robert Z
Jan 27 at 13:47
$begingroup$
Tell us more about your attempt
$endgroup$
– Robert Z
Jan 27 at 13:47
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Hint: $z^2 + 4|z| + 4 = 0$ iff $z^2 =-4 |z| - 4 le 0$ and so $z=bi$ for some $b$.
answered Jan 27 at 13:54
lhflhf
167k11172403
$endgroup$
add a comment |
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$begingroup$
Hint: $z^2 + 4|z| + 4 = 0$ iff $z^2 =-4 |z| - 4 le 0$ and so $z=bi$ for some $b$.
answered Jan 27 at 13:54
lhflhf
167k11172403
$endgroup$
add a comment |
$begingroup$
Hint: $z^2 + 4|z| + 4 = 0$ iff $z^2 =-4 |z| - 4 le 0$ and so $z=bi$ for some $b$.
answered Jan 27 at 13:54
lhflhf
167k11172403
$endgroup$
add a comment |
$begingroup$
Hint: $z^2 + 4|z| + 4 = 0$ iff $z^2 =-4 |z| - 4 le 0$ and so $z=bi$ for some $b$.
answered Jan 27 at 13:54
lhflhf
167k11172403
$endgroup$
Hint: $z^2 + 4|z| + 4 = 0$ iff $z^2 =-4 |z| - 4 le 0$ and so $z=bi$ for some $b$.
answered Jan 27 at 13:54
lhflhf
167k11172403
answered Jan 27 at 13:54
lhflhf
167k11172403
answered Jan 27 at 13:54
lhflhf
167k11172403
answered Jan 27 at 13:54
lhflhf
167k11172403
167k11172403
add a comment |
add a comment |
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$begingroup$
If you could share your attempted solution, perhaps someone could tell you where you are wrong, in stead of writing up a full solution which may or may not help you find your mistake.
$endgroup$
– Servaes
Jan 27 at 13:44
$begingroup$
Tell us more about your attempt
$endgroup$
– Robert Z
Jan 27 at 13:47