Solving Complex Questions: Re(z^2)=? Im(z^2)=?

[~angel~]

Please help with this question. I'm really having trouble.

Let z= r(cosa+isina) be any complex number.

a. Show that lz^2l = lzl^2 (I know how to prove this question, but I just put this down because I'm assuming you'd need it for the following questions)

b. For which (if any) complex numbers is Re(z^2) = (Rez)^2

c. For which (if any) complex numbers is Im(z^2) = (Imz)^2

I tried solving b. and I ended up with an answer involving "cos".

Any help would be greatly appreciated.

 

[StatusX]

Do you mean you got that this would only be true if cos a was a certain value? (let me guess, 1 or -1) ? That's fine, just find out what this means about sin a (if you got what I guessed, sin a must be 0), plug it into z and see what you get. It should be pretty clear why this is the case.

 

[Data]

Try it by converting $z$ back to cartesian coordinates (ie. set $z = a + bi \Longrightarrow z^2 = a^2 - b^2 + 2abi$).

I'm sure if you think about it for a little while you can come up with a reasonable guess for part b without doing any calculation anyways

 

[~angel~]

How did you get $z^2 = a^2 - b^2 + 2abi$? I must be blind or something. Sorry about this. I just don't understand how the negative sign becomes involved and the "i" disappears in b^2.

I asked my tutor and he also said to convert it to cartesian form.

 

[Data]

if $z= a+bi$ then

$$z^2 = (a+bi)(a+bi) = a^2 + 2abi + (bi)^2 = a^2 + 2abi + (-1)b^2 = a^2 - b^2 + 2abi.$$

 

[~angel~]

Thanks I forgot i^2 equals -1. Sorry, I'm completely lost...do you find the real part of that equation for part a.?

 

[Data]

you mean part b, right?

If so, then it's asking you to find out when

$$\left(\mbox{Re} [z]\right)^2 = \mbox{Re} \left[z^2\right]$$

so yes .

 

[~angel~]

Yeah, sorry, i meant part b. I can't seem to get it. The real part of that equation is a^2 - b^2, for Re(z)^2? But for (Re[z])^2, do you find the real z from a+ib (which is a)? I'm assuming I'm doing something wrong. We haven't exactly been taught this stuff.

 

[Data]

Well, you correctly identified

$$\mbox{Re}\left[z^2\right] = a^2 - b^2$$

and

$$\mbox{Re}[z] = a \Longrightarrow \left( \mbox{Re}[z]\right)^2 = a^2$$

so, can you tell me under what conditions

$$a^2 = \left( \mbox{Re}[z]\right)^2 = \mbox{Re}\left[z^2\right] = a^2 - b^2$$

?

 

[~angel~]

When b^2 = 0?

 

[Data]

Right. So if $b^2=0 \Longleftrightarrow b = 0$, then what is $z$?

 

[~angel~]

a? (10 char)

 

[Data]

right... so for what complex numbers $z$ do we have the result part b wants?

remember that $a$ is real!

 

[~angel~]

Is it any real number for a and 0 for b? Now, I'm not completely sure about this.

 

[Data]

Indeed. In other words,

$$\left(\mbox{Re}[z]\right)^2 = \mbox{Re}\left[z^2\right]$$

for $z \in \mathbb{C}$ if and only if $z$ is real, or in other words $z \in \mathbb{R}$ (ie. $z$ has no imaginary part!).

Do you see why this would have been a reasonable guess from the start?

As I'm sure you know, the real part of any real number is just the number itself, and the square of any real number is real also!

 

[~angel~]

Thank you so much

So, for the imaginary part

Im(2abi) = Im(i^2*b^2)

Im (2abi) = I am (-b^2)

So, I have to find out what condidtions I am (2abi) = I am (-b^2)?

So, 2abi = -b^2

How exactly do I go about solving this. Obviously 2ai = -b, but how do you get a real number from this?

 

[Data]

Remember, the "imaginary part" is just the real coefficient of $i$. So if $z = a + bi, \; (a, b \in \mathbb{R})$ then $\mbox{Im}[z]$ is just $b$.

 

[~angel~]

Ok...Im[2ab] = Im[b^2]

2ab = b^2

2a = b

I'm not sure how to get a real number out of this.

 

[Data]

Both sides of that equation are real... and you aren't taking $\mbox{Im}[2ab]$ or $\mbox{Im}\left[b^2\right]$, either.

Here's how it should be presented formally:

Let $z = r\left( \cos \theta + i \sin \theta \right) = a + bi, \ (a, b \in \mathbb{R})$.

Then $z^2 = a^2 - b^2 + 2abi$ so

$$\mbox{Im}\left[z^2\right] = 2ab$$

NOTE: here I take the imaginary part of $z^2$. This function returns to me the real coefficient of $i$ when $z^2$ is expressed in standard form, or in this case, $2ab$. On the other hand, writing it the way you did, really, $\mbox{Im}[2ab] = 0$ since $2ab$ has no imaginary part!

Continuing, we now look at $\mbox{Im}[z]\right = b \Longrightarrow \left(\mbox{Im}[z]\right)^2 = b^2$.

Thus we find that $\mbox{Im}\left[z^2\right] = \left(\mbox{Im}[z]\right)^2$ precisely whenever

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[~angel~]

a + 2ai (do I include the imaginary part)?

 

[Data]

yep, that's fine~

 

[~angel~]

Thank you so much for your help. I really do appreciate it. Sorry for being a pain. I just like to learn rather than just copying the answer and stuff.

 

[Data]

Also, I actually made a little mistake. Can you see where (Hint: what happens if $b=0$?)?

 

[~angel~]

Is this for part b. or c?

If b = 0, then z = a, but since 2a= b, then a = b/2, and therefore z = 0 (this is for part c)

 

[Data]

part c

 

[~angel~]

If b = 0, then z = a, but since 2a= b, then a = b/2, and therefore z = 0 (this is for part c)

 

[~angel~]

but why do you make b = 0?

 

[Data]

Well, that's what you would think from your previous answer. Like I said, I made a mistake, though!

I was looking at the equation

$$2ab = b^2$$

if $b=0$ then both sides are zero automatically, no matter what $a$ is (the other way to think about it is that if $b=0$ then I can't change the equation to $2a = b$ since that would be dividing by zero!). So complex numbers of the form

$$z = a + 2ai$$

OR

$$z = a$$

with $a \in \mathbb{R}$ have the property that

$$\mbox{Im}\left[z^2\right] = \left(\mbox{Im}[z]\right)^2$$