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Find the number of complex numbers $z$ such that $|z - 3 - 5i| = 2$ and $|z - 6 - 6i| = 4.$

 Dec 6, 2020
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Remember that |a + bi| = a^2 + b^2.

 

So from equations, (z - 3)^2 + (z - 5)^2 = 4 and (z - 6)^2 + (z - 6)^2 = 4.

(x - 3)^2 = 4, (y - 5)^2 = 4, (x - 6)^2 = 4, (y - 6)^2 = 4

Two possible values of x, two possible value of y.  Therefore, four solutions for z.

 Dec 6, 2020

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