### Solution Found!

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

**Chapter 17, Problem 42**

(choose chapter or problem)

**QUESTION:**

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

### Questions & Answers

**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}\)