Please help with this equation (complex numbers)
- Thread starter palomilu
- Start date
- Joined
- Nov 24, 2012
- Messages
- 3,021
Hello, and welcome to FMH! 
I would write:
[MATH]a+bi=(a^2+2abi-b^2)\sqrt{a^2+b^2}=(a^2-b^2)\sqrt{a^2+b^2}+2ab\sqrt{a^2+b^2}i[/MATH]
Equating like coefficients, we then obtain the system:
[MATH]a=(a^2-b^2)\sqrt{a^2+b^2}[/MATH]
[MATH]b=2ab\sqrt{a^2+b^2}[/MATH]
Can you proceed?
I would write:
[MATH]a+bi=(a^2+2abi-b^2)\sqrt{a^2+b^2}=(a^2-b^2)\sqrt{a^2+b^2}+2ab\sqrt{a^2+b^2}i[/MATH]
Equating like coefficients, we then obtain the system:
[MATH]a=(a^2-b^2)\sqrt{a^2+b^2}[/MATH]
[MATH]b=2ab\sqrt{a^2+b^2}[/MATH]
Can you proceed?
D
Deleted member 4993
Guest
Please post the EXACT problem (verbatim - as it was given to you).
Your statement:
.....b imaginary....
probably is NOT correct!
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,989
Here is a holiday gift: there are but two solutions and both are trivial and obvious.
I am a simple sort of guy.
[MATH]z = z^2 * |z| \implies 1 = z * |z| \text { if } z \ne 0.[/MATH]
But z = 0 satisfies the equation so one solution is [MATH]0 + 0i.[/MATH]
[MATH]z \ne 0 \implies 1 = z * |z| \implies (a + bi) * \sqrt{a^2 + b^2} = 1 \implies b = 0.[/MATH]
[MATH]1 = a * |a| \implies a = 1.[/MATH]
Thus the other solution is [MATH]1 + 0i.[/MATH]
EDIT: If you want, you can work through Mark's simultaneous equations to show that b must be zero and that therefore a must equal 0 or 1 as proof that there are no other solutions, but it never hurts to be a little simple-minded before driving yourself crazy.
Last edited:
Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,598
If b is imaginary, then bi is real making z=a+bi a real number. Then z=1 or z=0.
Where did you run into trouble?