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

[DottZakapa]

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

 

[fresh_42]

You could multiply it as $(z\cdot \bar{z})\cdot z.$ But you can also guess the solution.

 

[DottZakapa]

$a^3+ia^2b+b^2a+ib^3=8i$
\begin{cases}
a^3+b^2a=0 \\
ia^2b+ib^3=8i
\end{cases}

 

[fresh_42]

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.

 

[DottZakapa]

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

 

[fresh_42]

mmm ok so? sorry I'm not picking

You have a product which multiplies to zero, so what does that mean for the factors?

 

[DottZakapa]

no solution?

 

[Fred Wright]

Another way is to expand $(x+iy)^3$ in a binomial expansion and compare real and imaginary parts in the equation.

 

[DottZakapa]

no solution?

a and b are 0

 

[Mark44]

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?

 

[DottZakapa]

at the moment nothing comes up on my mind sorry, completely blank

 

[fresh_42]

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 $a^2+b^2 \neq 0$?

 

[DottZakapa]

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

 

[fresh_42]

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 $z^2\bar{z}=(a+ib)^2(a-ib)=\ldots = 8i$?

 

[DaveE]

Have you learned the polar form Ae^iΘ 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.

 

[DottZakapa]

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

 

[fresh_42]

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 $a$?

 

[DottZakapa]

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 $a$?

i don't know, sorry

 

[fresh_42]

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.

 

[DottZakapa]

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

 

[fresh_42]

Yes. And now put $a=0$ into the second equation which we have:

\begin{cases}
ia^2b+ib^3=8i
\end{cases}

 

[DottZakapa]

Yes. And now put $a=0$ into the second equation which we have:

b= -2i?

 

[fresh_42]

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 $b$.

 

[DottZakapa]

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 $Z=2i$

 

[mathwonk]

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.

 

[Elbraido]

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.

$a^2+b^2\neq 0$ isn't this assuming that $a$ and $b$ are purely real which isn't implied in the question?

 

[fresh_42]

$a^2+b^2\neq 0$ isn't this assuming that $a$ and $b$ are purely real which isn't implied in the question?

$a,b$ are real, since $z$ is written as $z=a+ib$ before the calculation started:

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.

Now if $a=b=0$ then $z=0$ which doesn't solve the equation $z^2\bar{z}=8i .$

 

[anuttarasammyak]

(Z/2)^2\ \overline{Z/2}=i
So we know |Z/2|=1 so
Z/2=i

 

[mathwonk]

after reading many of your comments, maybe the easiest way to begin, without being too clever, is to take absolute value of both sides, getting |Z|^3 = 8, so |Z|=2. Then Z^2.Zbar = Z.|Z|^2 = 8i, gives the answer quickly. This is of course the idea behind the previous post's solution as well, but maybe less clever. I.e. as a general rule, taking absolute values may simplify a complex number situation, and give useful partial information.

 

[jbunniii]

Polar coordinates make this very simple. Let $z = re^{i\theta}$. Then $z^2 \overline{z} = (r^2 e^{i2\theta})(re^{-i\theta}) = r^3 e^{i\theta} = 8i$, from which it's obvious what $r$ and $\theta$ must be.