1063) think of geometrical interpretation of |z - z1| = k
1q) the value of arg(ilog(a-ib/a+ib)) where a,b are real no's is????
2q)(mod)(Z- Z1+Z2/2)^2 +(MOD)(Z1-Z2/2)^2 =????
3Q) if mod(Z-2+2i)=1 then least and greatest values of mod (z) are?
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9 Answers
71If you think first one deeply, you can get that.
Take here θ any angle, then in the numerator you will have e-iθ and in denominator eiθ. Sum BOLD! :)
See one thing:
a-iba+ib = a-ib√(a2+b2)a+ib√(a2+b2)
713.
|z-2+2i| =1
is a circle with center at 2-2i and radius 1.
Got something?
1@vivek
yep i got dat 1st one....following ur hints i got tan∂ =2ab/a^2-b^2
so i assumed that a=b then i got that angle as 0°
will it works!!!!
1@ vivek whats the use of getting the circle i lso thout in that way but nothing striking in my mind
v need its min n max values know!!!!how?
1708$\boldsymbol{Ans:(3)}$\Rightarrow |z-2+2i|=1$\\\\ Put $z=x+iy$ in This equation, We Get....\\\\ $|(x-2)+i.(y+2)|=1\Leftrightarrow \boxed{(x-2)^2+(y-(-2))^2=1}$\\\\ So This is a Equation of Circle Whose Center is at $(2,-2)$ and Radius =$1$\\\\ Now Here We have To find \underline{\underline{Max.}} and \underline{\underline{Min.}} value of $\underline{\underline{\sqrt{x^2+y^2}}}$.\\\\ Which is a Distance of any Point $\underline{\underline{P}}$(on the circle) from The Origin.\\\\ So Min. and Max. occur along the line joining $P$ and $Q$ with origin(0,0),\\\\ Where $P$ and $Q$ are \underline{\underline{Diametrically}} opposite point.\\\\ So $\underline{\underline{Min(\sqrt{x^2+y^2})}}$=$OP=OC-r=\sqrt{(2-0)^2+(2-0)^2}-1=\boxed{\boxed{2\sqrt{2}-1}}$ \\\\ Where $C$ is a Center and $O$ is Origin. Here \underline{\underline{$O,P,C$ and $Q$ are Collinear}}.\\\\ Similarly \\\\ So $\underline{\underline{Max(\sqrt{x^2+y^2})}}$=$OQ=OC+r=\sqrt{(2-0)^2+(2-0)^2}+1=\boxed{\boxed{2\sqrt{2}+1}}$
71@shree.
yep i got dat 1st one....following ur hints i got tan∂ =2ab/a^2-b^2
so i assumed that a=b then i got that angle as 0°
It isn't asked to do so. So please Don't
1708\underline{\underline{\boldsymbol{Ans(3):}}}\Rightarrow$ Here $|z-(2-2i)|=1$. Now Let $z_{1}=(2-2i)$ and $|z_{1}|=2\sqrt{2}$\\\\ Then Expression convert into $|z-z_{1}|=1......................(1)$\\\\ Now We Know That $\boxed{*|z-z_{1}|\geq ||z|-|z_{1}||}...........(2)$\\\\ (Using Triangle Inequality)\\\\ So from equation..(1) and (2),We get\\\\ $1\geq||z|-2\sqrt{2}|$\\\\ $\Leftrightarrow ||z|-2\sqrt{2}|\leq 1$\\\\ So $-1\leq|z|-2\sqrt{2}\leq 1$\\\\ $-1+2\sqrt{2}\leq |z|\leq 1+2\sqrt{2}$\\\\ $\textcolor[rgb]{0.,1.,0.}{\boxed{\boxed{\textcolor[rgb]{1.,0.,0.}{2\sqrt{2}-1}\leq \textcolor[rgb]{1.,0.,0.} {|z|}\leq \textcolor[rgb]{1.,0.,0.}{2\sqrt{2}+1}}}}$