Solving a Complex Equation

We solve the equation |z - |z + i|| = (z + 1)i + |z + 1|. The main challenge is to keep the algebra manageable throughout our calculations.
00:00 Considering real and imaginary parts
02:01 Dealing with |y|
04:00 More calculations
06:32 Finishing off

[@kapsel-yg2sk]

Unfortunately, now we have to check the solution

[@MichaelRothwell1]

You are correct in that not all lines in the solution are equivalent. In my solution (please see my comment), all lines are equivalent so there is no need to check the answer found.

[@MichaelRothwell1]

Here is my solution to this nice problem, essentially the same as that in the video. I was rather hoping that the answer would simplify to something nice at the end, but I suppose this would have been asking too much! Intially I struggled to find a geometric interpretation of the equation but rapidly gave up. There is a needed check that √3a-1>0 towards the end of my solution, corresponding to checking that -√3y-1≥0 in the solution in the video, which you missed (note that -√3y-1 must equal a square root expression and not its opposite, which could happen after squaring both sides). Solution: LHS real and ≥0, so RHS real and ≥0. So z+1 is purely imaginary. z+1=±ai where a≥0. If z+1=ai where a≥0, then RHS=-a+a=0. So from LHS, z=ai=|z+i|. So for ai real, a=0, but then z=-1, RHS=0, LHS=1+√2≠0, so no solution. Hence z+1=-ai where a>0, z=-1-ai. So RHS=a+a=2a. Equation becomes |-1-ai-|-1-(a-1)i||=2a |1+√(1+(a-1)²)+ai|=2a (1+√(1+(a-1)²))²+a²=4a² (1+√(1+(a-1)²))²=3a² 1+√(1+(a-1)²)=√3a (as a>0) √(1+(a-1)²)=√3a-1 1+(a-1)²=(√3a-1)² (and √3a-1>0) a²-2a+2=3a²-2√3a+1 2a²-2(√3-1)a-1=0 a=[√3-1+√((√3-1)²+2)]/2 (as a>0. I used the simplified quadratic formula for the solution of ax²+2bx+c=0 which is x=[-b±√(b²-ac)]/a. Note the product of the roots is -1, so only the plus sign will give a positive root) a=[√3-1+√(6-2√3)]/2 Now √36. Also √3-1>0. So 2√3a>√(18-6√3)>√6>2, so √3a-1>0 as required for a valid solution. So z=-1-[√3-1+√(6-2√3)]i/2.

[@aytunch]

Great observations to shorten the solution. Good work Dr

[@coreymonsta7505]

That was a really interesting problem, including the shortcuts/observations used

[@FadkinsDiet]

case analysis on the sign of y: if y is positive the right hand side simplifies to 2y, if y is 0 the rhs is 0, if y is negative the right side is also 0. Makes the rest a little simpler.

[@user-wu8yq1rb9t]

At first: Happy new haircut . And second: Great .... I like it The Beautiful Mind dear Dr. Barker, Thank you so much (I just learned and enjoyed because *i* is one of my favorites parameters). And please *more complex analysis* , thank you.

[@synaestheziac]

Very cool problem. Brilliantly lucid explanation!

[@SpeedyMemes]

wow. nice observations to simplify the problem

[@dr.mikelitoris]

Seoul dynasty better sorry not sorry

[@SpeedyMemes]

@@dr.mikelitoris ??

[@dr.mikelitoris]

@@SpeedyMemes i said what i said

[@prathamrana8916]

Nice 🤘🏻

[@dr.mikelitoris]

Sheesh Dr dripker lookin clean wit da fade 🥶🥶

[@willyh.r.1216]

... an interesting pbm. Thx.

[@Monolith-yb6yl]

You must to check -sqrt(3)y-1 is not negative to understand is your solution valid or not

[@worldnotworld]

How does one come up with these things?

[@DrBarker]

For this equation, I wanted something that looked crazy with nested absolute values of complex numbers, but could be solved without anything too complicated. If it wasn't set up so that we could make the first observation that Re(z) = -1, it could have been much more difficult to solve. Then to actually find a nice equation with reasonably a nice solution, this took some tweaking/trial and error, just running different equations through Wolfram until I was happy. I can't offer any geometric insight into the equation, but maybe we could try to think about the geometric meaning of e.g. |z - |z - w|| to understand what's going on.

[@mathcanbeeasy]

(z+1)*i is real => z+1 = a*i, with a real. => z = -1+a*i. No need for writing z=x+y*i. 🙂

[@karma_kun9833]

Complex numbers ***