Complex Numbers Help 2

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Scrutinize

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So I'm stuck on this question, not really sure what it means.

z2 = z (with a bar on top) so conjugate I think


The answers are like 0,1 and -1/2 +( root(3)/2 )i

How do you get that from that simple z statement??

I tried doing (a+bi)2 = (a-bi) but I just got a ton of different expressions that I couldn't simplify. Any help on what this means would be greatly appreciated!]

(PS: I'm new to complex numbers so apologies if this is a dumb question, just couldn't find an answer on google)

Thank you very much!
 
Scrutinize said:
So I'm stuck on this question, not really sure what it means.
z2 = z (with a bar on top) so conjugate I think
The answers are like 0,1 and -1/2 +( root(3)/2 )i
How do you get that from that simple z statement??
Click to expand...
\(\displaystyle z^2=\overline{~z~}\)
\(\displaystyle a^2-b^2+2abi=a-bi \Rightarrow~a^2-b^2=a~\&~2ab=-b \)

Can you continue?
 
pka said:
\(\displaystyle z^2=\overline{~z~}\)
\(\displaystyle a^2-b^2+2abi=a-bi \Rightarrow~a^2-b^2=a~\&~2ab=-b \)

Can you continue?
Click to expand...
How did you get from


\(\displaystyle a^2-b^2+2abi=a-bi \ \)

To that?

\(\displaystyle a^2-b^2=a~\&~2ab=-b \)


Also I don't really know what to do from there either?


Sorry I just don't really understand this topic too much I guess.
 
Scrutinize said:
How did you get from
\(\displaystyle a^2-b^2+2abi=a-bi \ \)
To that? \(\displaystyle a^2-b^2=a~\&~2ab=-b \)
Click to expand...
If each of \(\displaystyle z~\&~w\) is a complex number & \(\displaystyle z=w\) then the real & imaginary parts are equal.
That is \(\displaystyle \Re(z)=\Re(w)~\&~\Im(z)=\Im(w)\)
 
pka said:
If each of \(\displaystyle z~\&~w\) is a complex number & \(\displaystyle z=w\) then the real & imaginary parts are equal.
That is \(\displaystyle \Re(z)=\Re(w)~\&~\Im(z)=\Im(w)\)
Click to expand...
Not gonna lie, I'm still confused and that kind of confused me more. Sorry.

Like I get what you're saying kind of but I also don't.
 
The z = z bar is not a general truth. It is a special case, and you are to find the complex numbers for which it is true.

The definition of z bar, or the conjugate of z, is

[MATH]z = a + bi \iff \bar z = a - bi.[/MATH]
[MATH]\text {THEREFORE, } z = a + bi \text { and }\bar z = z^2\implies[/MATH]
[MATH]a - bi = (a + bi)^2 = a^2 - b^2 + 2abi.[/MATH] One of the identities I showed you yesterday.

[MATH]a - bi = = a^2 - b^2 + 2abi \implies a = a^2 - b^2 \text { and } -\ b = 2ab.[/MATH]
[MATH]b= 0 \implies a = a^2 - 0^2 = a^2 \implies a^2 - a = 0 \implies[/MATH]
[MATH]a^2 - a = 0 \implies a(a - 1) = 0 \implies a = 0 \text { or } a = 1.[/MATH]
So two answers are (1) z = 0 + 0i, and (2) z = 1 + 0i. These are the same as the real numbers 0 and 1. looks good because z and its conjugate are the same whenever b = 0, and only 1 and 0 in the real numbers equal their squares.

[MATH]b \ne 0 \text { and } -\ b = 2ab \implies -\ 1 = 2a \implies a =\ - \dfrac{1}{2}.[/MATH]
[MATH]a = -\ \dfrac{1}{2} \text { and } a = a^2 - b^2 \implies[/MATH]
[MATH]b^2 = a - a^2 = -\ \dfrac{1}{2} - \left ( -\ \dfrac{1}{2} \right )^2 = -\ \dfrac{2}{4} - \dfrac{1}{4} = (-\ 1) * \dfrac{3}{2^2} \implies[/MATH]
[MATH]b = \pm \dfrac{i\sqrt{3}}{2}.[/MATH]
So we get two more answers: (3) -(1/2) + (i * sqrt(3))/2, and (4) -(1/2) - (i * sqrt(3))/2. Let's check those out.

To do so, we need another identity.

[MATH]z = a - bi \implies z^2 = a^2 - 2abi + (-\ bi)^2 = a^2 + b^2(i^2) - 2abi = a^2 - b^2 - 2abi.[/MATH]
[MATH]z = -\ \dfrac{1}{2} + \dfrac{i\sqrt{3}}{2} \implies \bar z = -\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \text { and}[/MATH]
[MATH]z^2 = \left (-\ \dfrac{1}{2} \right)^2 - \left ( \dfrac{\sqrt{3}}{2} \right )^2 + 2 * \left ( -\ \dfrac{1}{2} \right ) * \dfrac{i \sqrt{3}}{2} \implies [/MATH]
[MATH]z^2 = \dfrac{1}{4} - \dfrac{3}{4} - \dfrac{i \sqrt{3}}{2} = -\ \dfrac{1}{2} - \dfrac{i \sqrt{3}}{2} = \bar z. [/MATH]

[MATH]z = -\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \implies \bar z = -\ \dfrac{1}{2} + \dfrac{i\sqrt{3}}{2} \text { and}[/MATH]
[MATH]z^2 = \left (-\ \dfrac{1}{2} \right)^2 - \left (-\ \dfrac{\sqrt{3}}{2} \right )^2 - 2 * \left ( -\ \dfrac{1}{2} \right ) * \dfrac{i \sqrt{3}}{2} \implies [/MATH]
[MATH]z^2 = \dfrac{1}{4} - \dfrac{3}{4} + \dfrac{i \sqrt{3}}{2} = -\ \dfrac{1}{2} + \dfrac{i \sqrt{3}}{2} = \bar z. [/MATH]
There is nothing deep here. Remember the identities I told you yesterday and watch out for the signs.
 
Last edited:
If you are to have this type of question involving complex numbers , then you need the basic ideas.
if \(\displaystyle C=\cos(0.6\pi)+i\sin(0.6\pi)\) then the real part of \(\displaystyle C,~\Re(C)= \cos(0.6\pi)\) and the imaginary part of \(\displaystyle C,~\Im(C)=\sin(0.6\pi)\) (note no \(\displaystyle \bf i)\)
Now working with complex equations real parts equal real parts and imaginary parts equal imaginary parts.
Read the above carefully twice. Then here is what we did.
Given that \(\displaystyle z^2=\overline{\,z\,}\) we know that \(\displaystyle \Re(z^2)=\Re(\overline{\,z\,})~\&~\Im(z^2)=\Im(\overline{\,z\,})\) or \(\displaystyle a^2-b^2=a~\&~2ab=-b\).
Because both of \(\displaystyle \bf a~\&~b\) are real numbers, these are ordinary algebra problems.
 
JeffM said:
The z = z bar is not a general truth. It is a special case, and you are to find the complex numbers for which it is true.

The definition of z bar, or the conjugate of z, is

[MATH]z = a + bi \iff \bar z = a - bi.[/MATH]
[MATH]\text {THEREFORE, } z = a + bi \text { and }\bar z = z^2\implies[/MATH]
[MATH]a - bi = (a + bi)^2 = a^2 - b^2 + 2abi.[/MATH] One of the identities I showed you yesterday.

[MATH]a - bi = = a^2 - b^2 + 2abi \implies a = a^2 - b^2 \text { and } -\ b = 2ab.[/MATH]
[MATH]b= 0 \implies a = a^2 - 0^2 = a^2 \implies a^2 - a = 0 \implies[/MATH]
[MATH]a^2 - a = 0 \implies a(a - 1) = 0 \implies a = 0 \text { or } a = 1.[/MATH]
So two answers are (1) z = 0 + 0i, and (2) z = 1 + 0i. These are the same as the real numbers 0 and 1. looks good because z and its conjugate are the same whenever b = 0, and only 1 and 0 in the real numbers equal their squares.

[MATH]b \ne 0 \text { and } -\ b = 2ab \implies -\ 1 = 2a \implies a =\ - \dfrac{1}{2}.[/MATH]
[MATH]a = -\ \dfrac{1}{2} \text { and } a = a^2 - b^2 \implies[/MATH]
[MATH]b^2 = a - a^2 = -\ \dfrac{1}{2} - \left ( -\ \dfrac{1}{2} \right )^2 = -\ \dfrac{2}{4} - \dfrac{1}{4} = (-\ 1) * \dfrac{3}{2^2} \implies[/MATH]
[MATH]b = \pm \dfrac{i\sqrt{3}}{2}.[/MATH]
So we get two more answers: (3) -(1/2) + (i * sqrt(3))/2, and (4) -(1/2) - (i * sqrt(3))/2. Let's check those out.

To do so, we need another identity.

[MATH]z = a - bi \implies z^2 = a^2 - 2abi + (-\ bi)^2 = a^2 + b^2(i^2) - 2abi = a^2 - b^2 - 2abi.[/MATH]
[MATH]z = -\ \dfrac{1}{2} + \dfrac{i\sqrt{3}}{2} \implies \bar z = -\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \text { and}[/MATH]
[MATH]z^2 = \left (-\ \dfrac{1}{2} \right)^2 - \left ( \dfrac{\sqrt{3}}{2} \right )^2 + 2 * \left ( -\ \dfrac{1}{2} \right ) * \dfrac{i \sqrt{3}}{2} \implies [/MATH]
[MATH]z^2 = \dfrac{1}{4} - \dfrac{3}{4} - \dfrac{i \sqrt{3}}{2} = -\ \dfrac{1}{2} - \dfrac{i \sqrt{3}}{2} = \bar z. [/MATH]

[MATH]z = -\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \implies \bar z = -\ \dfrac{1}{2} + \dfrac{i\sqrt{3}}{2} \text { and}[/MATH]
[MATH]z^2 = \left (-\ \dfrac{1}{2} \right)^2 - \left (-\ \dfrac{\sqrt{3}}{2} \right )^2 - 2 * \left ( -\ \dfrac{1}{2} \right ) * \dfrac{i \sqrt{3}}{2} \implies [/MATH]
[MATH]z^2 = \dfrac{1}{4} - \dfrac{3}{4} + \dfrac{i \sqrt{3}}{2} = -\ \dfrac{1}{2} + \dfrac{i \sqrt{3}}{2} = \bar z. [/MATH]
There is nothing deep here. Remember the identities I told you yesterday and watch out for the signs.
Click to expand...
You write:

[MATH]b = \pm \dfrac{i\sqrt{3}}{2}.[/MATH]
'b' cannot be an imaginary number. There is a small typo that led to this impossible situation. You wrote:

[MATH]....... a = a^2 - b^2 \implies[/MATH]
[MATH]b^2 = a - a^2 .......[/MATH]
That should be:

[MATH]....... a = a^2 - b^2 \implies[/MATH]
[MATH]b^2 = a^2 - a.......[/MATH]
and so on......
 
Subhotosh Khan said:
You write:

[MATH]b = \pm \dfrac{i\sqrt{3}}{2}.[/MATH]
'b' cannot be an imaginary number. There is a small typo that led to this impossible situation. You wrote:

[MATH]....... a = a^2 - b^2 \implies[/MATH]
[MATH]b^2 = a - a^2 .......[/MATH]
That should be:

[MATH]....... a = a^2 - b^2 \implies[/MATH]
[MATH]b^2 = a^2 - a.......[/MATH]
and so on......
Click to expand...
And I said to be careful of signs.

1 hour in the corner
 
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