Let O be the origin and A be the point z_1=1+2i. If B is the point z_2,Re(z_2)<0, such that OAB is a right angled isosceles triangle with OB as…

Given $O$O be the origin and $A$A be the point ${z}_{1}=1+2i$z1​=1+2i,

$B$B is the point ${z}_{2},Re({z}_{2})<0$z2​,Re(z2​)<0,

Also $OAB$OAB is a right angled isosceles triangle with $OB$OB as hypotenuse,

Plotting the diagram we get,

Now by using concept of rotation we get,

$$\frac{{z}_{2}-0}{(1+2i)-0}=\frac{|OB|}{|OA|}{e}^{\frac{i\pi }{4}}$$(1+2i)−0z2​−0​=∣OA∣∣OB∣​e4iπ​

$$\Rightarrow \frac{{z}_{2}}{1+2i}=\sqrt{2}{e}^{\frac{i\pi }{4}}$$⇒1+2iz2​​=2​e4iπ​

$$Or{z}_{2}=(1+2i)(1+i)$$Orz2​=(1+2i)(1+i)

$$=-1+3i$$=−1+3i

Now finding $\mathrm{arg}{z}_{2}=\pi -{\mathrm{tan}}^{-1}\frac{|3|}{|1|}=\pi -{\mathrm{tan}}^{-1}3$argz2​=π−tan−1∣1∣∣3∣​=π−tan−13

Now $|{z}_{2}|=\sqrt{{1}^{2}+{3}^{2}}=\sqrt{10}$∣z2​∣=12+32​=10​

Now solving ${z}_{1}-2{z}_{2}=(1+2i)+2-6i=3-4i$z1​−2z2​=(1+2i)+2−6i=3−4i

So, $\mathrm{arg}({z}_{1}-2{z}_{2})=-{\mathrm{tan}}^{-1}\frac{4}{3}$arg(z1​−2z2​)=−tan−134​

Now solving $|2{z}_{1}-{z}_{2}|=|2+4i+1-3i|=|3+i|$∣2z1​−z2​∣=∣2+4i+1−3i∣=∣3+i∣

$$=\sqrt{10}$$=10​

So, $|2{z}_{1}-{z}_{2}|=5$∣2z1​−z2​∣=5 is not true.