henpen
A complex question
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henpen
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\[\left| z-1 \right|+\left| z+1 \right|=4\]
Which turns into the messy
\[\sqrt{(a-1)^2+b^2}+\sqrt{(a+1)^2+b^2}=4\]
Which I have no idea about how to solve graphically.
henpen
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(Where z=a+ib)
henpen
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(And by 'graphically' I mean plotting the possible values of z on the complex plane)
baddinlol
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OK
baddinlol
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| a + bi - 1| + |a + bi + 1| = 4
baddinlol
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| (a -1) + (b)i | + | (a+1) + (b)i | = 4
baddinlol
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Now let me find the modulus which is given by sqrt( re(z)^2 + Im(z)^2)
baddinlol
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\[\sqrt{(a +1)^2 + b^2} + \sqrt{(a-1)^2 + b^2} = 4\]
baddinlol
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Then you should square both sides
baddinlol
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Actually that will be hard
baddinlol
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So move one of the square roots on the left then square both sides
baddinlol
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\[(\sqrt{(a+1)^2 + b^2} )^2= (4 - \sqrt{(a-1)^2 + b^2})^2\]
baddinlol
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Which will give you this
henpen
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There's still a horrible square root sign there, though.
baddinlol
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\[(a+1)^2 + b^2= 16 - 8\sqrt{(a-1)^2 + b^2} + (a-1)^2 + b^2\]
baddinlol
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I will get rid of it soon
baddinlol
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Ok now i am going to cancel some stuff and simplify this equation
baddinlol
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\[a^2 + 2a + 1 + b^2 = 16 - 8\sqrt{(a-1)^2 + b^2} + a^2 - 2a + 1 + b^2\]
baddinlol
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Now i can simplify further
baddinlol
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\[4a -16 = -8\sqrt{(a-1)^2+b^2}\]
henpen
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4a-16=-8sqrt(a^2-2a+1+b^2)
baddinlol
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Now I am going to square both sides again
baddinlol
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\[(4a -16)^2 = 64((a-1)^2 + b^2)\]
baddinlol
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Finally we have no more square roots
baddinlol
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Now i am going to expand
baddinlol
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\[16a^2 -128a + 256 = 64 a^2 -128a + 64 + 64b^2)\]
henpen
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it ends up as 192=48a^2+64b^2 I think
baddinlol
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Now simlify...
baddinlol
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SSo you get
henpen
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That actually agrees with the wolfram link I posted earlier. Thanks very much- I had little idea how to deal with square roots efficiently before!
baddinlol
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Final step: complete the square on a term
baddinlol
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oops
baddinlol
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Sorry yeah that was the answer 192 = 48a^2 +64b^2
baddinlol
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And thats an eclipse.
baddinlol
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No problem. =)