Solving (a+ib)^2 = (c+id)^2: Understanding the Problem and Finding Solutions

[squenshl]

Homework Statement

How do I show that if (a+ib)² = (c+id)² then (a+ib) = $$\pm$$(c+id)

Homework Equations

The Attempt at a Solution

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

 

[gabbagabbahey]

The Attempt at a Solution

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

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)

 

[squenshl]

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

 

[gabbagabbahey]

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

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=___?

 

[jgens]

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.

 

[gabbagabbahey]

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.

That works fine, provided you've already proven that

[tex]z_1z_2=0\implies z_1=0\;\;\;\text{or}\;\;\;z_2=0[/itex]<br /> <br /> for complex numbers.[/tex]

 

[squenshl]

a² = -d² & a² = c²
a = $$\pm$$id & a = $$\pm$$c

 

[gabbagabbahey]

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 $b=\frac{cd}{a}$?

 

[squenshl]

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)

 

[gabbagabbahey]

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

 

[squenshl]

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

 

[gabbagabbahey]

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.

 

[squenshl]

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

 

[gabbagabbahey]

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

 

[squenshl]

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