# Homework Help: Problem of math Proof

1. May 8, 2010

### squenshl

1. The problem statement, all variables and given/known data
How do I show that if (a+ib)2 = (c+id)2 then (a+ib) = $$\pm$$(c+id)

2. Relevant equations

3. The attempt at a solution
(a+ib)(a+ib) = a2 + 2aib -b2
(c+id)(c+id) = c2 + 2cid - d2
Stuck after this.

2. May 8, 2010

### gabbagabbahey

Re: Proof

Well, if a,b,c and d are all real numbers, then you must have a2-b2=c2-d2 and ab=2cd (for two complex number to be equal, their real parts must be equal and their imaginary parts must be equal)

3. May 8, 2010

### squenshl

Re: Proof

Don't you mean ab = cd. So b = cd/a then (cd/a)2 - b2 = c2 - d2
a4 - c2d2 = a2c2 - a2 - d2
(a2 - c2)(a2 + d2) = 0
Am I on the right track

4. May 8, 2010

### gabbagabbahey

Re: Proof

Yes. Now realize that for (a2 - c2)(a2 + d2) to be zero, either a2= c2 or c2=-d2....but "a", "c" and "d" are real numbers, so a=___?

5. May 8, 2010

### jgens

Re: Proof

If $(a+bi)^2=(c+di)^2$, then clearly $(a+bi)^2 - (c+di)^2 = 0$. Factor this expression and use the zero product property to arrive at the desired result.

6. May 8, 2010

### gabbagabbahey

Re: Proof

That works fine, provided you've already proven that

$$z_1z_2=0\implies z_1=0\;\;\;\text{or}\;\;\;z_2=0[/itex] for complex numbers. 7. May 9, 2010 ### squenshl Re: Proof a2 = -d2 & a2 = c2 a = [tex]\pm$$id & a = $$\pm$$c

Last edited: May 9, 2010
8. May 9, 2010

### gabbagabbahey

Re: Proof

Right, but you can immediately throw away the solution $a=\pm id$ since both $a$and $d$ are supposed to be real numbers....What does the other solution give you for $b$ when you plug it back into $b=\frac{cd}{a}$?

9. May 9, 2010

### squenshl

Re: Proof

b = $$\pm$$d
What do we plug that into?
Since we know what a & b is hence (a+ib) = ($$\pm$$c + i$$\pm$$d) then factorise to get (a+ib) = $$\pm$$(c+id)

Last edited: May 9, 2010
10. May 9, 2010

### gabbagabbahey

Re: Proof

Yeah, basically. If $a=\pm c$ and $b=\pm d$, then $(a+ ib)=\pm(c + i d)$ and you're done.

11. May 9, 2010

### squenshl

Re: Proof

Awesome.
Cheers.
What if I wanted to do it in polar form.

12. May 10, 2010

### gabbagabbahey

Re: Proof

In polar form, you would basically want to show that $(r_1e^{i\phi_1})^2=(r_2e^{i\phi_2})^2$ implies $r_1e^{i\phi_1}=\pm r_2e^{i\phi_2}$ and for that, you would use the fact that $e^{i\phi_2}=e^{i(\phi_2+2\pi n}$ to finds the two unique roots of the equation.

13. May 10, 2010

### squenshl

Re: Proof

What if I wanted to go (a+ib)2 = r12(cos(2$$\vartheta$$) + isin(2$$\vartheta$$))

14. May 10, 2010

### gabbagabbahey

Re: Proof

Then you would just expand (a+ib)^2 again, and compare the real and imaginary parts on both sides of the equation...

15. May 10, 2010

### squenshl

Re: Proof

No thanks. That's nasty.
I'll leave the answer in rectangular form.

Last edited: May 10, 2010