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(Where z=a+ib)

(And by 'graphically' I mean plotting the possible values of z on the complex plane)

OK

| a + bi - 1| + |a + bi + 1| = 4

| (a -1) + (b)i | + | (a+1) + (b)i | = 4

Now let me find the modulus which is given by sqrt( re(z)^2 + Im(z)^2)

\[\sqrt{(a +1)^2 + b^2} + \sqrt{(a-1)^2 + b^2} = 4\]

Then you should square both sides

Actually that will be hard

So move one of the square roots on the left then square both sides

\[(\sqrt{(a+1)^2 + b^2} )^2= (4 - \sqrt{(a-1)^2 + b^2})^2\]

Which will give you this

There's still a horrible square root sign there, though.

\[(a+1)^2 + b^2= 16 - 8\sqrt{(a-1)^2 + b^2} + (a-1)^2 + b^2\]

I will get rid of it soon

Ok now i am going to cancel some stuff and simplify this equation

\[a^2 + 2a + 1 + b^2 = 16 - 8\sqrt{(a-1)^2 + b^2} + a^2 - 2a + 1 + b^2\]

Now i can simplify further

\[4a -16 = -8\sqrt{(a-1)^2+b^2}\]

4a-16=-8sqrt(a^2-2a+1+b^2)

Now I am going to square both sides again

\[(4a -16)^2 = 64((a-1)^2 + b^2)\]

Finally we have no more square roots

Now i am going to expand

\[16a^2 -128a + 256 = 64 a^2 -128a + 64 + 64b^2)\]

it ends up as 192=48a^2+64b^2 I think

Now simlify...

SSo you get

Final step: complete the square on a term

oops

Sorry yeah that was the answer 192 = 48a^2 +64b^2

And thats an eclipse.

No problem. =)