Proving the Real Part Summation Property for Complex Numbers

[freezer]

Homework Statement

Prove for complex number z1, z2, ..., zn that:

$\mathbb{R}e\left \{ \sum_{k=1}^{N} z_{k}\right \} = \sum_{k=1}^{N}\mathbb{R}e\left \{ z_{k} \right \}$

Homework Equations

The Attempt at a Solution

Not sure how to setup this problem.

I was thinking:

$\mathbb{R}e\left \{ \sum_{k=1}^{N} {a_{k}} + \sum_{k=1}^{N} {ib_{k}} \right \} = \sum_{k=1}^{N}\mathbb{R}e\left \{ a_{k} + ib_{k}\right \}$

 

[kostas230]

Have you tried induction?

 

[freezer]

Have you tried induction?

I have not, only ever done trig proofs and not exactly sure what to do. With what i found with inductions, i still not sure how to set it up. When i look at this, i don't see a path to get the left to look like the right.

 

[kostas230]

The real part $Re(z)$ of a complex number $z=a+bi$, where $a,b$ are real numbers, can be found by the formula: $Re(z)=(z+\bar{z})/2$, where $\bar{z}=a-bi$.

Next time, study harder before asking for help ;)

 

[freezer]

The real part $Re(z)$ of a complex number $z=a+bi$, where $a,b$ are real numbers, can be found by the formula: $Re(z)=(z+\bar{z})/2$, where $\bar{z}=a-bi$.

Next time, study harder before asking for help ;)

Okay, thanks for the advice.

So we can show that:

$\frac{N}{2}(z + \bar{z}) = \frac{N}{2}(z + \bar{z})$

 

[kostas230]

Ummm, no. If $z=\sum_{i=1}^N z_i$ then:

$$Re(z)=\frac{\sum_{i=1}^N z_i+\bar{z}_i}{2}=\sum_{i=1}^N \frac{z_i+\bar{z}_i}{2}=\sum_{i=1}^N Re(z_i)$$

I don't get why it's that hard.

 

[Tobias Funke]

Ummm, no. If $z=\sum_{i=1}^N z_i$ then:

$$Re(z)=\frac{\sum_{i=1}^N z_i+\bar{z}_i}{2}=\sum_{i=1}^N \frac{z_i+\bar{z}_i}{2}=\sum_{i=1}^N Re(z_i)$$

I don't get why it's that hard.

Maybe using sigma notation and induction just makes it look hard to someone who doesn't have much experience with proofs. I think the supposedly horrifying lack of rigor in

$$Re[(a_1+b_1i)+(a_2+b_2i)+\dots+(a_N+b_Ni)]=\dots=Re(a_1+b_1i)+Re(a_2+b_2i)+\dots+Re(a_N+b_Ni)$$

is worth it if it makes the proof easier to read and understand. freezer, can you fill in the middle steps?

 

[freezer]

Maybe using sigma notation and induction just makes it look hard to someone who doesn't have much experience with proofs. I think the supposedly horrifying lack of rigor in

$$Re[(a_1+b_1i)+(a_2+b_2i)+\dots+(a_N+b_Ni)]=\dots=Re(a_1+b_1i)+Re(a_2+b_2i)+\dots+Re(a_N+b_Ni)$$

is worth it if it makes the proof easier to read and understand. freezer, can you fill in the middle steps?

I suppose my inadequacy in proofs is apparent, as horrifying as that may be. Having been out of academics for over 20 years, I spend a lot more time on re-learning things much that of the homework expects you to have recent experience with. So please accept my apology for being a bit slow.

As for the intermediate steps in the series, I can see that distributing the Re gets from one series to the next but what else is to be shown in between? Once the Re is distributed, we can work toward the right hand side.

 

[Tobias Funke]

I suppose my inadequacy in proofs is apparent, as horrifying as that may be. Having been out of academics for over 20 years, I spend a lot more time on re-learning things much that of the homework expects you to have recent experience with. So please accept my apology for being a bit slow.

As for the intermediate steps in the series, I can see that distributing the Re gets from one series to the next but what else is to be shown in between? Once the Re is distributed, we can work toward the right hand side.

I wasn't thinking so much about "distributing" the Re on the left side (the point of the proof is to show that you can actually do this, so it would be circular reasoning to use it in the proof), more like adding the numbers first and then taking the real part. I'm almost going to give away the whole proof here, but it follows so quickly from basic definitions that it's hard not to:

$$(a_1+b_1i)+(a_2+b_2i)=(a_1+a_2)+(b_1+b_2)i.$$

Now just extend this to more than two numbers and you're well on your way. I was being a little sarcastic with the "horrifying lack of rigor" remark, because most people don't think anything with ... is an acceptable proof, and induction is required. That's technically true, I suppose, and you should try to understand kostas230's proof as well, but I think the ellipses give the idea of the proof better than induction in this case.