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If $|z_{1}+z_{2}|=|z_{1}|-|z_{2}|=2$ and $|2z_{2}+2i(z_{3}-z_{2})|=|2iz_{3}+(1-2i)z_{2}|=10$

Where $z_{1}=3+4i.$ Then value of $|z_{1}|^2+|z_{2}|^2+|z_{3}|^2$ is

Try: Let $z_{2}=a+ib$ and $z_{3}=c+id$. Then we have $(a+c)^2+(b+d)^2=4.$ and $\sqrt{a^2+b^2}-\sqrt{c^2+d^2}=2$

I have seems that there is an easy way other then substituting $z_{1}=a+ib$ and $z_{2}=c+id.$

Could some help me to solve it , Thanks

Community
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DXT
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2 Answers2

2

HINT

You can directly evaluate $$|z_1|=|3+4i|=5$$ from which you get $$|z_1|-|z_2|=2\Rightarrow\\|z_2|=3$$

So the challenge lies in finding $|z_3|$.

But $$|2z_2+2i(z_3−z_2)|=2|z_2+i(z_3−z_2)|=10\Rightarrow\\|z_2+i(z_3−z_2)|=5\Rightarrow\\|z_2+i(z_3−z_2)|^2=25\Rightarrow\\ (z_2+i(z_3−z_2))(\bar{z_2}-i(\bar{z_3}−\bar{z_2})=25\Rightarrow\\z_2\bar{z_2}-iz_2(\bar{z_3}−\bar{z_2})+i\bar{z_2}(z_3-z_2)+(z_3-z_2)(\bar{z_3}−\bar{z_2})=25\Rightarrow\\|z_2|^2-iz_2\bar{z_3}+iz_2\bar{z_2}+i\bar{z_2}z_3-iz_2\bar{z_2}+z_3\bar{z_3}-\bar{z_2}z_3-z_2\bar{z_3}+z_2\bar{z_2}=25\Rightarrow\\18+|z_3|^2-z_2\bar{z_3}(1+i)+\bar{z_2}z_3(i-1)=25$$

Let $$z_2\bar{z_3}=a$$

So we have $$18+|z_3|^2-z_2\bar{z_3}(1+i)+\bar{z_2}z_3(i-1)=25\Rightarrow\\18+|z_3|^2-a(1+i)-\bar{a}(i-1)=25 (*)$$

Work similarly for $$|2iz_{3}+(1-2i)z_{2}|=10$$ taking advantage of (*) and you will be able to find $a$ and thus calculate $|z_3|$.

MathematicianByMistake
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1

We have $$|z_{1}|=\sqrt{3^2+4^2}=5,\qquad\qquad |z_{2}|=|z_{1}|-2=5-2=3\tag1$$

Now, we see that $$|2z_{2}+2i(z_{3}-z_{2})|=|2iz_{3}+(1-2i)z_{2}|=10$$ is equivalent to $$|\alpha+\beta|=|\alpha-\beta|\tag2$$ and $$|\alpha+\beta|=10\tag3$$ where $$\alpha=2i(z_3-z_2)+\frac 32z_2, \qquad\beta=\frac 12z_2$$

From $(2)$, we have $$(\alpha+\beta)(\bar\alpha+\bar\beta)=(\alpha-\beta)(\bar\alpha-\bar\beta)\iff \alpha\bar\beta+\overline{\alpha\bar\beta}=0\iff \alpha\bar\beta=ki$$ where $k\in\mathbb R$.

So, from $(3)$, we get $$\begin{align}|\alpha+\beta|=10&\iff (\alpha+\beta)(\bar\alpha+\bar\beta)=100 \\\\&\iff \left(\frac{ki}{\bar\beta}+\beta\right)\left(\frac{-ki}{\beta}+\bar\beta\right)=100 \\\\&\iff \frac{k^2}{|\beta|^2}+|\beta|^2=100 \\\\&\iff \frac{4k^2}{9}+\frac 94=100 \\\\&\iff k=\pm\frac 34\sqrt{391}\tag4\end{align}$$

Also, we have $$\begin{align}\alpha\bar\beta=ki& \iff\left(2i(z_3-z_2)+\frac 32z_2\right)\left(\frac 12\overline{z_2}\right)=ki \\\\&\iff i(z_3\overline{z_2}-|z_2|^2)+\frac 34|z_2|^2=ki \\\\&\iff z_3=\frac{9+k+\frac{27}{4}i}{\overline{z_2}}\tag5 \end{align}$$

It follows from $(1)(4)(5)$ that $$|z_1|^2+|z_2|^2+|z_3|^2=5^2+3^2+\frac{(9+k)^2+(\frac{27}{4})^2}{|\overline{z_2}|^2}=\color{red}{\frac{145\pm 3\sqrt{391}}{2}}$$

mathlove
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