Solving Modulus Equation: Find z=a+bi

  • Thread starter Thread starter BloodyFrozen
  • Start date Start date
  • Tags Tags
    Modulus
Click For Summary
The discussion revolves around solving the modulus equation |z|=|z²+1| for complex numbers z=a+bi. Participants detail their attempts to expand the components and derive the equation, leading to the expression 0=a⁴+2a²b²+b⁴+a²-3b²+1. There is a correction noted regarding the sign in the expansion of z², where it should be -b² instead of +b². The conversation highlights the importance of consistency in mathematical expressions throughout the solution. The thread emphasizes the collaborative effort to clarify and correct the calculations involved in solving the equation.
BloodyFrozen
Messages
353
Reaction score
1

Homework Statement


Find all ##z=a+bi## such that:
|z|=|z^{2}+1|


Homework Equations


The Attempt at a Solution


I expanded the components.
|z|=|z^{2}+1|
z^{2}=a^2-b^2+2abi
\sqrt{a^{2}+b^{2}}=\sqrt{(a^{2}-b^{2}+1)^{2}+(2ab)^{2}}
a^2+b^2=(a^{2}-b^{2}+1)^{2}+(2ab)^{2}
0=a^{4}+2a^{2}b^{2}+b^{4}+a^{2}-3b^{2}+1

I don't see what to do now...
 
Last edited:
Physics news on Phys.org
BloodyFrozen said:

Homework Statement


Find all ##z=a+bi## such that:
|z|=|z^{2}+1|


Homework Equations


The Attempt at a Solution


I expanded the components.
|z|=|z^{2}+1|
z^{2}=a^2+b^2+2abi
That should be -b^2, not +b^2.
 
jbunniii said:
That should be -b^2, not +b^2.

Right. Let me just go fix that.

Edit. Fixed
 
BloodyFrozen said:
Right. Let me just go fix that.

Edit. Fixed

Well, you fixed it in the first line where it appeared, but you still need to fix it everywhere else.
 
jbunniii said:
Well, you fixed it in the first line where it appeared, but you still need to fix it everywhere else.

I fixed then end before, I just forgot to change the middle parts.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 17 ·
Replies
17
Views
2K
Replies
20
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K