If z_1=1-2 i, z_2=1+i and z_3=3+4 i,then left | vedclass

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(C) Given $\left|\left(\frac{1}{z_1}+\frac{2}{z_2}\right) \frac{z_3}{z_2}\right| = \left|\frac{1}{z_1}+\frac{2}{z_2}\right| \cdot \frac{|z_3|}{|z_2|}$

Vedclass · Accepted Answer

$\sqrt{\frac{45}{2}}$

Vedclass · Answer

(C) Given $\left|\left(\frac{1}{z_1}+\frac{2}{z_2}ight) \frac{z_3}{z_2}ight| = \left|\frac{1}{z_1}+\frac{2}{z_2}ight| \cdot \frac{|z_3|}{|z_2|}$ Substituting the values: $\left|\frac{1}{1-2 i}+\frac{2}{1+i}ight| \cdot \frac{|3+4 i|}{|1+i|}$ $= \left|\frac{1+2 i}{1^2+2^2} + \frac{2(1-i)}{1^2+1^2}ight| \cdot \frac{\sqrt{3^2+4^2}}{\sqrt{1^2+1^2}}$ $= \left|\frac{1+2 i}{5} + \frac{2-2 i}{2}ight| \cdot \frac{5}{\sqrt{2}}$ $= \left|\frac{2+4 i + 10-10 i}{10}ight| \cdot \frac{5}{\sqrt{2}}$ $= \left|\frac{12-6 i}{10}ight| \cdot \frac{5}{\sqrt{2}} = \frac{6}{10} |2-i| \cdot \frac{5}{\sqrt{2}}$ $= \frac{3}{5} \sqrt{2^2+(-1)^2} \cdot \frac{5}{\sqrt{2}} = \frac{3 \sqrt{5}}{\sqrt{2}} = \sqrt{\frac{9 	imes 5}{2}} = \sqrt{\frac{45}{2}}$

If $z_1=1-2 i, z_2=1+i$ and $z_3=3+4 i$,then $\left|\left(\frac{1}{z_1}+\frac{2}{z_2}\right) \frac{z_3}{z_2}\right|=$

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