Show for all complex numbers z on the circle x 2 y 2 4 that Zz 6 8iZ 12

Chapter 17, Problem 42

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QUESTION:

Show for all complex numbers z on the circle \(x^{2}+y^{2}=4\) that \(|z+6+8 i| \leq 12\).

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QUESTION:

Show for all complex numbers z on the circle \(x^{2}+y^{2}=4\) that \(|z+6+8 i| \leq 12\).

ANSWER:

Step 1 of 2

Show for all complex numbers z on the circle \(x^{2}+y^{2}=4\) that \(|z+6+8 i| \leq 12\)

Let \(z=x+i y\) then we know that \(|z|=\sqrt{x^{2}+y^{2}}\)

\(\Rightarrow|z|^{2}=x^{2}+y^{2}\)

We have that \(x^{2}+y^{2}=4\) (by hypothesis)

\(\begin{array}{l}
\Rightarrow|z|^{2}=4 \\
\Rightarrow|z|=2
\end{array}\)

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