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Find the range of values of $|z|$ and $\arg (z)$ for $$|z-4-4i| = 2 \sqrt{2}.$$

I'm aware that you can solve this geometrically by drawing a circle on the argand diagram and finding out the information from there. I was curious if there is an algebraic method for solving the problem of finding the range of values for the modulus and argument (i.e. without needing to consider it in the argand plane)?

Thanks

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Hint. Note that the equation is solved by $$z=4+4i+2 \sqrt{2}(\cos(t)+i\sin(t))=4+2 \sqrt{2}\cos(t) +i\left(4+2 \sqrt{2}\sin(t)\right)$$ with $t\in [0,2\pi)$. Hence $$|z|^2=(4+2 \sqrt{2}\cos(t))^2 +(4+2 \sqrt{2}\sin(t))^2=40+16\sqrt{2}(\cos(t)+\sin(t))\\ =40+32\sin(t+\pi/4).$$ Hence $$40-32\leq |z|^2\leq 40+32.$$ Can you take it from here?

As regards the argument consider the ratio $$\tan(\arg(z))=\frac{\mbox{Im}(z)}{\mbox{Re}(z)}=\frac{4+2 \sqrt{2}\sin(t)}{4+2 \sqrt{2}\cos(t)}.$$ Finally you should find that the minimum argument is $\pi/12$. What is the maximum argument?

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  • $\begingroup$ Do we find then when $40+ 32 \sin (t+ \pi /3) = 2$? How do we do the argument? $\endgroup$ – frog1944 Aug 14 '17 at 9:23
  • $\begingroup$ @frog1944 Note that $\sin(t+\pi/4)=\sin(t)\cos(\pi/4)+\cos(t)\sin(\pi/4)=(\sin(t)+\cos(t))/\sqrt{2}$. As regards the argument see my edited answer. $\endgroup$ – Robert Z Aug 14 '17 at 9:39
  • $\begingroup$ Ok, why is the equation solved by $z=4+4i+2 \sqrt{2}(\cos(t)+i\sin(t))=4+2 \sqrt{2}\cos(t) +i\left(4+2 \sqrt{2}\sin(t)\right)$? $\endgroup$ – frog1944 Aug 14 '17 at 9:55
  • $\begingroup$ Because $|(z-4-4i) /(2 \sqrt{2})|=1$ and any complex number of modulus $1$ is given by $e^{it}=\cos(t)+i\sin(t)$. $\endgroup$ – Robert Z Aug 14 '17 at 10:01
  • $\begingroup$ Oh, that makes sense. Understand most of it, how did you find that the minimum argument is $\pi / 12$? I understand that that is the minimum value of the function, but without graphing it, how can you determine it? $\endgroup$ – frog1944 Aug 14 '17 at 10:32
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By $|z+w|\geq|z|-|w|$ we have $$2 \sqrt{2}=|z-(4+4i)| \geq |z|-|4+4i|=|z|-4 \sqrt{2}\color{red}{\implies}|z|\leq6 \sqrt{2}$$ $$2 \sqrt{2}=|z-(4+4i)| \geq |4+4i|-|z|=4 \sqrt{2}-|z|\color{red}{\implies}|z|\geq2 \sqrt{2}$$

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  • $\begingroup$ Cool! How do you do the argument part? $\endgroup$ – frog1944 Aug 14 '17 at 9:23
  • $\begingroup$ Unfortunately this way works only where the center of circle is of the form $z_0\pm iz_0$ that is lies on $y=\pm x$, in this case $\arg z=n\dfrac{\pi}{4}$. $\endgroup$ – Nosrati Aug 14 '17 at 10:27

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