MHB 1.4.1 complex number by condition

[karush]

1.4.1 Miliani HS
Find all complex numbers x which satisfy the given condition
$\begin{array}{rl}
1+x&=\sqrt{10+2x} \\
(1+x)^2&=10+2x\\
1+2x+x^2&=10+2x\\
x^2-9&=0\\
(x-3)(x+3)&=0
\end{array}$
ok looks these are not complex numbers unless we go back the the radical

 

[topsquark]

1.4.1 Miliani HS
Find all complex numbers x which satisfy the given condition
$\begin{array}{rl}
1+x&=\sqrt{10+2x} \\
(1+x)^2&=10+2x\\
1+2x+x^2&=10+2x\\
x^2-9&=0\\
(x-3)(x+3)&=0
\end{array}$
ok looks these are not complex numbers unless we go back the the radical

I realize what you are saying but real numbers are complex numbers!

Let's re-arrange the variables in a more standard way. Let z = x + iy.

Then
[math]1 + z = \sqrt{10 + 2z}[/math]

etc.
[math]1 + z^2 = 10[/math]

[math]1 + (x + iy)^2 = 10[/math]

[math]1 + x^2 - y^2 + 2ixy = 10[/math]

Equating real and imaginary parts on both sides gives:
[math]\begin{cases} 1 + x^2 - y^2 = 10 \\ 2xy = 0 \end{cases}[/math]

(Hint: If x = 0 can we have a real value for y?)

So what values of x and y can we have and thus what values of z?

-Dan

Addendum: You can also let [math]x = r e^{i \theta }[/math] but I don't think this adds anything.

 

[karush]

I realize what you are saying but real numbers are complex numbers!

Let's re-arrange the variables in a more standard way. Let z = x + iy.

Then
[math]1 + z = \sqrt{10 + 2z}[/math]

etc.
[math]1 + z^2 = 10[/math]

[math]1 + (x + iy)^2 = 10[/math]

[math]1 + x^2 - y^2 + 2ixy = 10[/math]

Equating real and imaginary parts on both sides gives:
[math]\begin{cases} 1 + x^2 - y^2 = 10 \\ 2xy = 0 \end{cases}[/math]

(Hint: If x = 0 can we have a real value for y?)

So what values of x and y can we have and thus what values of z?

-Dan

Addendum: You can also let [math]x = r e^{i \theta }[/math] but I don't think this adds anything.

ok I see what your basic idea is but how did you get [math]1 + (x + iy)^2 = 10[/math] what happened to 2x ?

 

[maxkor]

a+bi=10
a=10
b=0

 

[topsquark]

ok I see what your basic idea is but how did you get [math]1 + (x + iy)^2 = 10[/math] what happened to 2x ?

Nothing changed from your original.
[math]1 + z = \sqrt{10 + 2z}[/math]

[math](1 + z)^2 = 10 + 2z[/math]

[math]1 + \cancel{2z} + z^2 = 10 + \cancel{2z}[/math]

[math]1 + z^2 = 10[/math]
as before.

-Dan

 

[topsquark]

a+bi=10
a=10
b=0

Where did you get a + bi = 10 from?

-Dan

 

[topsquark]

@karush: What I'm trying to say is that the only possible solution (complex or real) is x = 3. It doesn't matter how you solve it. The z = x + iy method is probably best (to my mind) as it reminds you to think in terms of complex numbers instead of reals. [math]z = re^{i \theta }[/math] is probably better in general since you can recall that [math]z = r e^{i \theta } = r e^{i ( \theta + 2 \pi k )}[/math] which may give you more solutions... such as when solving a cubic equation.

-Dan

 

[karush]

ok
must of been a lot students stumble on that problem:(

 

[topsquark]

ok
must of been a lot students stumble on that problem:(

My guess is that there is a typo in the original problem. Most problems will not ask for complex solutions if there are only real solutions. You are right to be suspicious.

-Dan

 

[karush]

possible its a typo
came out of a handwritten journal

 

[HOI]

The set of complex number includes the real numbers!

3 and -3 are perfectly good complex numbers.

 

[topsquark]

The set of complex number includes the real numbers!

3 and -3 are perfectly good complex numbers.

No, these are evil complex numbers. I know these two personally.

-Dan

 

[HOI]

Well, I am part of their gang!

 

[topsquark]

Well, I am part of their gang!

Actually, it's a "triad." Hahahahahaha!

-Dan