Complex numbers: Solve ##Z^2\bar{Z}=8i##

  • Thread starter DottZakapa
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  • #1
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Homework Statement:

Solve ##Z^2\bar{Z}=8i##

Relevant Equations:

Complex numbers
Solve ##Z^2\bar{Z}=8i##

i am confused on how to proceed

i have tried to substitute ##z=a+ib## solve the conjugate and the square, then separate the real from the imaginary and put all in a system, but becomes too complicated
 

Answers and Replies

  • #2
13,560
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You could multiply it as ##(z\cdot \bar{z})\cdot z.## But you can also guess the solution.
 
  • #3
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##a^3+ia^2b+b^2a+ib^3=8i##
\begin{cases}
a^3+b^2a=0 \\
ia^2b+ib^3=8i
\end{cases}
 
  • #4
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It is easier if you write ##a^3+b^2a=a\cdot (a^2+b^2)=0##, and ##a^2+b^2## cannot be zero.
 
  • #5
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It is easier if you write ##a^3+b^2a=a\cdot (a^2+b^2)=0##, and ##a^2+b^2## cannot be zero.
mmm ok so? sorry i'm not picking :oldconfused:
 
  • #6
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mmm ok so? sorry i'm not picking :oldconfused:
You have a product which multiplies to zero, so what does that mean for the factors?
 
  • #7
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no solution?
 
  • #8
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Another way is to expand ##(x+iy)^3## in a binomial expansion and compare real and imaginary parts in the equation.
 
  • #10
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a and b are 0
No
if you write ##a^3+b^2a=a\cdot (a^2+b^2)=0##, and ##a^2+b^2## cannot be zero.
You have a product which multiplies to zero, so what does that mean for the factors?
no solution?
No. To simplify what fresh_42 said, if ##a\cdot b = 0##, and b cannot be zero, what can you say about a?
 
  • #11
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at the moment nothing comes up on my mind sorry, completely blank
 
  • #12
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at the moment nothing comes up on my mind sorry, completely blank
To get ##2\cdot x=0## you must have? And then, why is ##a^2+b^2 \neq 0##?
 
  • #13
210
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x must be 0
To get ##2\cdot x=0## you must have?
x is equal to 0
And then, why is ##a^2+b^2 \neq 0##?
this i can't answer
 
  • #14
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x must be 0

x is equal to 0

this i can't answer
Yes, ##x=0##. So if ##a(a^2+b^2)=0## we must have either ##a=0## or ##a^2+b^2=0##. But squares of real numbers ##a,b## are never negative, so their sum isn't either. And there is only one way that ##a^2+b^2=0## can hold. Now does such a solution work for ##z^2\bar{z}=(a+ib)^2(a-ib)=\ldots = 8i ##?
 
  • #15
DaveE
Gold Member
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Have you learned the polar form Ae and how to multiply and conjugation with it? It's easier to see that way.

BTW, if this isn't familiar to you yet, then ignore it. You can solve it either way.
 
  • #16
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Yes, ##x=0##. So if ##a(a^2+b^2)=0## we must have either ##a=0## or ##a^2+b^2=0##. But squares of real numbers ##a,b## are never negative, so their sum isn't either. And there is only one way that ##a^2+b^2=0## can hold. Now does such a solution work for ##z^2\bar{z}=(a+ib)^2(a-ib)=\ldots = 8i ##?
so al so b is equal to 0
 
  • #17
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so al so b is equal to 0
Yes, from ##a^2+b^2=0## we get ##a=b=0##. But then ##z=0## and this cannot be. Now that ##a^2+b^2\neq 0##, what do we get for ##a##?
 
  • #18
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Yes, from ##a^2+b^2=0## we get ##a=b=0##. But then ##z=0## and this cannot be. Now that ##a^2+b^2\neq 0##, what do we get for ##a##?
:cry: i don't know, sorry
 
  • #19
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We have ##a\cdot (a^2+b^2)=0## and ##a^2+b^2\neq 0.## That is the same as ##x\cdot 2= 0##. Just as if ##a^2+b^2## was ##2## and ##a## was ##x##. It isn't, but it is the same situation.
 
  • #20
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We have ##a\cdot (a^2+b^2)=0## and ##a^2+b^2\neq 0.## That is the same as ##x\cdot 2= 0##. Just as if ##a^2+b^2## was ##2## and ##a## was ##x##. It isn't, but it is the same situation.
a is 0
 
  • #22
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:ok:Yes. And now put ##a=0## into the second equation which we have:
b= -2i?
 
  • #23
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b= -2i?
No. You have written ##ib^3=8i## Now we divide both sides by ## i ## and have ##b^3 = 8.## No minus sign around. And ##b## is real! We set ##z=a+ib=0+ib=ib## and only need ##b##.
 
  • #24
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No. You have written ##ib^3=8i## Now we divide both sides by ## i ## and have ##b^3 = 8.## No minus sign around. And ##b## is real! We set ##z=a+ib=0+ib=ib## and only need ##b##.
yes sorry, was late for me and my brain very tired 😅 .##b=2## and ##Z=0+i2## therefore ##Z=2i##
 
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  • #25
mathwonk
Science Advisor
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since z.zbar = |z|^2, the answer is a real scalar, namely |z|^2, times z, and from the form of the answer, 8i, z is obviously 2i.

Ahh, but fresh said this long ago, in post #2.
 

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