200 days streak special 3

Algebra Level 4

Let $z_1$ and $z_2$ be two given complex numbers such that $\dfrac{z_1}{z_2} + \dfrac{z_2}{z_1} = 1$ and $|z_1| = 3$ . Compute $|z_1 - z_2|^2$ .

2 solutions

Sujoy Roy
Jan 10, 2016

Putting $\frac{z_1}{z_2}=t$ , we get $t+\frac{1}{t}=1$ or, $t^2-t+1=0$ or, $t=\frac{1+\sqrt{3}i}{2}$ or, $|t|=1$ or, $|z_1|=|z_2|=3$ .

Now $|z_1-z_2|^2= (z_1-z_2)(\bar{z_1}-\bar{z_2})$

$= z_1\bar{z_1}+z_2\bar{z_2}-(z_1\bar{z_2}+z_2\bar{z_1})$

$=9+9-(tz_2\bar{z_2}+\frac{z_1\bar{z_1}}{t})$

$=18-9(t+\frac{1}{t})=18-9=\boxed{9}$ .

but ${ z }_{ 1 }\bar { { z }_{ 1 } } ={ \left| { z }_{ 1 } \right| }^{ 2 }$

Tanishq Varshney - 5 years, 5 months ago

You are absolutely correct Tanishq !

Bob Kadylo - 5 years, 4 months ago

The answer should be $\boxed 9$ because there is an exponent 2 outside the $|z_1-z_2|^2$

Bob Kadylo - 5 years, 4 months ago

Zk Lin
Feb 11, 2016

From $\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}=1$ , we have

$\frac{{z_{1}}^{2}+{z_{2}}^{2}}{z_{1}z_{2}}=1$

${z_{1}}^{2}+{z_{2}}^{2}=z_{1}z_{2}$

$(z_{1}-z_{2})^{2}=-z_{1}z_{2}$

${|z_{1}-z_{2}|}^{2}=|z_{1}||z_{2}| \ \tag{1}$

From ${z_{1}}^{2}+{z_{2}}^{2}=z_{1}z_{2}$ , we have

${z_{1}}^{2}=z_{2}z_{1}-{z_{2}}^{2}$

${z_{1}}^{2}=z_{2}(z_{1}-z_{2})$

$\frac{{|z_{1}|}^{2}}{|z_{2}|}=|z_{1}-z_{2}|$

$\frac{{|z_{1}|}^{4}}{{|z_{2}|}^{2}}={|z_{1}-z_{2}|}^{2} \ \tag{2}$

$(1)=(2)$

$\frac{{|z_{1}|}^{4}}{{|z_{2}|}^{2}}=|z_{1}||z_{2}|$

${|z_{1}|}^{3}={|z_{2}|}^{3}$

Therefore,

$|z_{1}|=|z_{2}|=3$

${|z_{1}-z_{2}|}^{2}=|z_{1}||z_{2}|=\boxed{9}$

The explanation is correct, but it seems hard to motivate such an approach.