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1.4.1 complex number by condition

karush

Well-known member
Jan 31, 2012
3,062
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
 
Last edited:

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,271
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
Jan 31, 2012
3,062
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 ?
 

maxkor

Member
Jul 22, 2014
62
a+bi=10
a=10
b=0
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,271
ok I see what your basic idea is but how did you get \(\displaystyle 1 + (x + iy)^2 = 10\) what happened to 2x ?
Nothing changed from your original.
\(\displaystyle 1 + z = \sqrt{10 + 2z}\)

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

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

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

-Dan
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,271

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,271
@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
Jan 31, 2012
3,062
ok
must of been a lot students stumble on that problem:(
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,271
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
Jan 31, 2012
3,062
possible its a typo
came out of a handwritten journal