Complex numbers: Prove (Re(2z))^2 - sqrt{5} Im(z^2) + |conjugate of z|^2 is positive

[Qwertyuiop[]]

Hi, I have a question regarding complex numbers. z is a complex number such that z= a+ib , simplify and prove that it's positive.

I will not show all the working but i will write what i got after simplification:
[math]5a^2 + (1-\sqrt{5})b^2[/math]first i need to know if the answer is correct and then i have to prove that it is positive. I know that a^2 and b^2 will always be positive but the expression 1-sqrt(5) is negative so whether the answer is +ve or -ve will depend on the values of a and b we have correct? So can we know if it's going to be +ve or -ve without knowing the actual values of a and b?

 

[Dr.Peterson]

Hi, I have a question regarding complex numbers. z is a complex number such that z= a+ib , simplify and prove that it's positive.
View attachment 36357
I will not show all the working but i will write what i got after simplification:
[math]5a^2 + (1-\sqrt{5})b^2[/math]first i need to know if the answer is correct and then i have to prove that it is positive. I know that a^2 and b^2 will always be positive but the expression 1-sqrt(5) is negative so whether the answer is +ve or -ve will depend on the values of a and b we have correct? So can we know if it's going to be +ve or -ve without knowing the actual values of a and b?

Your result is incorrect. Please show your steps to get it. (I think you found Im(z)^2 instead of Im(z^2). They aren't the same!)

Once you simplify correctly, it will be easy to show it's positive.

 

[Qwertyuiop[]]

Your result is incorrect. Please show your steps to get it. (I think you found Im(z)^2 instead of Im(z^2). They aren't the same!)

Once you simplify correctly, it will be easy to show it's positive.

Oh ok, I took the imaginary part of z which is b and squared it and did the same thing with (Re(2z))^2.
here is the full working:
z=a +ib
for (Re(2z))^2 i took the real part of z which is a.
(2a)^2 = 4a^2

-(sqrt(5)) Im(z^2) : b^2 * -sqrt(5)

and for the complex conjugate of z squared:
I first took the modulus of conjugate of z and then squared it to get a^2 + b^2 then just simplified the expression to get the result above.

Can you then pls explain the difference between lm(z)^2 and lm(z^2)?

 

[Dr.Peterson]

I took the imaginary part of z which is b and squared it

But that is not what it says to do! It says to square z, and find the imaginary part of that.

Do so, and you will see the difference.

Can you then pls explain the difference between lm(z)^2 and lm(z^2)?

As an example, if z = 1 + i, then Im(z)^2 = 1^2 = 1, but Im(z^2) = Im(2i) = 2. They are not the same.

 

[blamocur]

Oh ok, I took the imaginary part of z which is b and squared it and did the same thing with (Re(2z))^2.
here is the full working:
z=a +ib
for (Re(2z))^2 i took the real part of z which is a.
(2a)^2 = 4a^2

-(sqrt(5)) Im(z^2) : b^2 * -sqrt(5)

and for the complex conjugate of z squared:
I first took the modulus of conjugate of z and then squared it to get a^2 + b^2 then just simplified the expression to get the result above.

Can you then pls explain the difference between lm(z)^2 and lm(z^2)?

Now that you know where the mistake was can you work out the correct expression (preferably in [imath]\LaTeX[/imath] format)?

 

[Qwertyuiop[]]

But that is not what it says to do! It says to square z, and find the imaginary part of that.

Do so, and you will see the difference.


As an example, if z = 1 + i, then Im(z)^2 = 1^2 = 1, but Im(z^2) = Im(2i) = 2. They are not the same.

so for Im(z^2) I get:
[math](a+ib)^2 \\ a^2 + 2abi +i^2b^2 \\ a^2-b^2 + 2abi[/math]
so the imaginary part of z^2 is 2ab and the expression becomes [math]4a^2 -2ab\sqrt{5} + a^2 + b^2 \\ 5a^2+b^2 - 2ab\sqrt{5}[/math]Is this correct?

 

[blamocur]

so for Im(z^2) I get:
[math](a+ib)^2 \\ a^2 + 2abi +i^2b^2 \\ a^2-b^2 + 2abi[/math]
so the imaginary part of z^2 is 2ab and the expression becomes [math]4a^2 -2ab\sqrt{5} + a^2 + b^2 \\ 5a^2+b^2 - 2ab\sqrt{5}[/math]Is this correct?

Looks good. Can you now prove that it is non-negative. (It is not always positive since it can be 0).

 

[Steven G]

so for Im(z^2) I get:
[math](a+ib)^2 \\ a^2 + 2abi +i^2b^2 \\ a^2-b^2 + 2abi[/math]
so the imaginary part of z^2 is 2ab and the expression becomes [math]4a^2 -2ab\sqrt{5} + a^2 + b^2 \\ 5a^2+b^2 - 2ab\sqrt{5}[/math]Is this correct?

Looks good to me.

 

[pka]

Hi, I have a question regarding complex numbers. z is a complex number such that z= a+ib , simplify and prove that it's positive.
View attachment 36357

Here are some handy facts: If [imath]z= a+b\bf{i}[/imath] where [imath]\{a,b\}\subset\Re[/imath] then
[imath]|z|=|\overline{\,z\,}|,\;z\cdot\overline{\,z\,}=|z|^2=a^2+b^2,\;\;\&~~.\dfrac{1}{z}=\dfrac{\overline{\,z\,}}{| z|}[/imath]
Now we get [imath]\Re({\overline{\,z\,})}=a\:\&\:\Im({\overline{\,z\,})}=-b[/imath], moreover [imath]\Re(z^2)=a^2-b^2\:\&\:\Im({z^2)}=2ab[/imath]




[imath][/imath][imath][/imath][imath][/imath]

 

[pka]

Here are some handy facts: If [imath]z= a+b\bf{i}[/imath] where [imath]\{a,b\}\subset\Re[/imath] then
[imath]|z|=|\overline{\,z\,}|,\;z\cdot\overline{\,z\,}=|z|^2=a^2+b^2,\;\;\&~~.\dfrac{1}{z}=\dfrac{\overline{\,z\,}}{| z|}[/imath]
Now we get [imath]\Re({\overline{\,z\,})}=a\:\&\:\Im({\overline{\,z\,})}=-b[/imath], moreover [imath]\Re(z^2)=a^2-b^2\:\&\:\Im({z^2)}=2ab[/imath]




[imath][/imath][imath][/imath][imath][/imath]

EDIT CORRECTION: [imath] \dfrac{1}{z}=\dfrac{\overline{\,z\,}}{| z|^2}[/imath]

 

[Qwertyuiop[]]

Looks good. Can you now prove that it is non-negative. (It is not always positive since it can be 0).

The expression forms a perfect square :
[math]5a^2 - 2ab\sqrt{5} + b^2 \\ (\sqrt{5} a - b)^2 \\ \text{and } (\sqrt{5} a - b)^2 \geq 0[/math]