If 5z27z1 is purely imaginary, then 2z1 + 3z22z1 - 3z2 =

5z27z1= purely imaginary 5z27z1=k i ⇒z2z1=7ki5 ∴2z1+3z22z1-3z2=2+3z2z12-3z2z1 =2+21ki52-21ki5=1 since z=z ∴2z1+3z22z1-3z2=1

If 5z27z1 is purely imaginary, then 2z1 + 3z22z1 - 3z2 =

5z27z1= purely imaginary 5z27z1=k i ⇒z2z1=7ki5 ∴2z1+3z22z1-3z2=2+3z2z12-3z2z1 =2+21ki52-21ki5=1 since z=z ∴2z1+3z22z1-3z2=1

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If $\frac{5 z_{2}}{7 z_{1}}$ is purely imaginary, then $|\frac{2 z_{1} + 3 z_{2}}{2 z_{1} - 3 z_{2}}| =$

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Detailed Solution

$\frac{5 z_{2}}{7 z_{1}} =$ purely imaginary

$\frac{5 z_{2}}{7 z_{1}} = k i \Rightarrow \frac{z_{2}}{z_{1}} = \frac{7 ki}{5} ∴ |\frac{2 z_{1} + 3 z_{2}}{2 z_{1} - 3 z_{2}}| = |\frac{2 + 3 \frac{z_{2}}{z_{1}}}{2 - 3 \frac{z_{2}}{z_{1}}}| = |\frac{2 + \frac{2_{1} ki}{5}}{2 - \frac{2_{1} ki}{5}}| = 1 since |z| = |z| ∴ |\frac{2 z_{1} + 3 z_{2}}{2 z_{1} - 3 z_{2}}| = 1$