# 1.4.1 complex number by condition

#### karush

##### Well-known member
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

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#### topsquark

##### Well-known member
MHB Math Helper
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
$$\displaystyle 1 + z = \sqrt{10 + 2z}$$

etc.
$$\displaystyle 1 + z^2 = 10$$

$$\displaystyle 1 + (x + iy)^2 = 10$$

$$\displaystyle 1 + x^2 - y^2 + 2ixy = 10$$

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

(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 $$\displaystyle x = r e^{i \theta }$$ but I don't think this adds anything.

#### karush

##### Well-known member
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
$$\displaystyle 1 + z = \sqrt{10 + 2z}$$

etc.
$$\displaystyle 1 + z^2 = 10$$

$$\displaystyle 1 + (x + iy)^2 = 10$$

$$\displaystyle 1 + x^2 - y^2 + 2ixy = 10$$

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

(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 $$\displaystyle x = r e^{i \theta }$$ but I don't think this adds anything.
ok I see what your basic idea is but how did you get $$\displaystyle 1 + (x + iy)^2 = 10$$ what happened to 2x ?

a+bi=10
a=10
b=0

#### topsquark

##### Well-known member
MHB Math Helper
ok I see what your basic idea is but how did you get $$\displaystyle 1 + (x + iy)^2 = 10$$ what happened to 2x ?
$$\displaystyle 1 + z = \sqrt{10 + 2z}$$

$$\displaystyle (1 + z)^2 = 10 + 2z$$

$$\displaystyle 1 + \cancel{2z} + z^2 = 10 + \cancel{2z}$$

$$\displaystyle 1 + z^2 = 10$$
as before.

-Dan

MHB Math Helper

#### topsquark

##### Well-known member
MHB Math Helper
@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. $$\displaystyle z = re^{i \theta }$$ is probably better in general since you can recall that $$\displaystyle z = r e^{i \theta } = r e^{i ( \theta + 2 \pi k )}$$ which may give you more solutions... such as when solving a cubic equation.

-Dan

#### karush

##### Well-known member
ok
must of been a lot students stumble on that problem

#### topsquark

##### Well-known member
MHB Math Helper
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

##### Well-known member
possible its a typo
came out of a handwritten journal

#### Country Boy

##### Well-known member
MHB Math Helper
The set of complex number includes the real numbers!

3 and -3 are perfectly good complex numbers.

#### topsquark

##### Well-known member
MHB Math Helper
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

#### Country Boy

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MHB Math Helper
Well, I am part of their gang!

#### topsquark

##### Well-known member
MHB Math Helper
Well, I am part of their gang!