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Show that there are complex numbers z satisfying |z-a| + |z+a| = 2|c| if and only...

Show that there are complex numbers z satisfying |z-a| + |z+a| = 2|c| if and only if |a| \leq |c|.

If possible, without using triangle inequality. I have seen it done with setting z =

ia\sqrt{\frac{|c|^2}{|a|^2} - 1}

but I do not understand where that number comes from. I have also seen it done with setting z = x + iy and found that confusing.

I would appreciate any help!

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がf 1 62- 2

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