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Question: Find the complex number z, the least in absolute value which satisfy the given Condition \( \left...

Find the complex number z, the least in absolute value which satisfy the given
Condition ∣z−2+2i∣\left| {z - 2 + 2i} \right| =1

Explanation

Solution

First we will substitute z= x + iy and then we’ll find the magnitude of the given complex number and with the help of that we’ll have the value of x and y. Later on doing the differentiation and equating it to zero, we’ll have the value of θ\theta and finally on putting the value of θ\theta we’ll have our answer.

Complete step-by-step answer:
In this question we need to find the complex number z,
Which is satisfying the above condition so let say z = x + iy
And hence on putting the value, we have
∣x+iy−2+2i∣=1\left| {x + iy - 2 + 2i} \right| = 1
So now in order to remove modulus we have to find the magnitude of the given equation and hence,
(x−2)2+(y+2)2=1\sqrt {{{(x - 2)}^2}} + \sqrt {{{\left( {y + 2} \right)}^2}} = 1
And hence on squaring both sides, we have
(x - 2)2+(y+2)2=1{{\text{(x - 2)}}^2} + {(y + 2)^2} = 1
We know that equation of a circle with centre (h, K) and radius ‘r’ is given as:
(x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}
And hence we can conclude that the above equation is an equation of a circle.
So, we can write:
x - 2 = cosθ{\text{x - 2 = cos}}\theta
Now on rearranging we have
x = 2 + cosθ{\text{x = 2 + cos}}\theta
Similarly we can say that
y = - 2 + sinθ{\text{y = - 2 + sin}}\theta
And hence on putting the value of x and y we have
z = 2 - 2i + (cos⁡θ+isin⁡θ){\text{z = 2 - 2i + }}\left( {\cos \theta + i\sin \theta } \right)
And hence ∣z∣2=K=(cos⁡θ+2)2+(sin⁡θ−2)2{\left| z \right|^2} = K = {(\cos \theta + 2)^2} + {(\sin \theta - 2)^2}
And hence on doing the simplification we have,
⇒1 + 8 + 4(cos⁡θ−sin⁡θ)\Rightarrow {\text{1 + 8 + 4}}\left( {\cos \theta - \sin \theta } \right)
⇒9+4(cos⁡θ−sin⁡θ)=f(θ)\Rightarrow 9 + 4\left( {\cos \theta - \sin \theta } \right) = f(\theta )
Now on doing the differentiation and equating it to zero, we have,
df(θ)dθ = 0\dfrac{{df(\theta )}}{{d\theta }}{\text{ = 0}}
d(9+4(cos⁡θ−sin⁡θ))dθ = 0 \dfrac{{d(9 + 4\left( {\cos \theta - \sin \theta } \right))}}{{d\theta }}{\text{ = 0 }}
⇒4(−sin⁡θ−cos⁡θ)=0\Rightarrow 4\left( { - \sin \theta - \cos \theta } \right) = 0
∴cos⁡θ=−sin⁡θ\therefore \cos \theta = - \sin \theta
And on simplification, we have
⇒tanθ = - 1\Rightarrow {\text{tan}}\theta {\text{ = - 1}}

∴θ = 3π4\therefore \theta {\text{ = }}\dfrac{{3\pi }}{4}
And hence on putting the value of θ\theta , we have
z = 2(1−i)+(−1+i2){\text{z = 2}}\left( {1 - i} \right) + \left( {\dfrac{{ - 1 + i}}{{\sqrt 2 }}} \right)
Now on separating the real and imaginary part we have,
z = 0 + (1 - i)[2−12]{\text{z = 0 + (1 - i)}}\left[ {2 - \dfrac{1}{{\sqrt 2 }}} \right]

Note: In this question, you should have in mind that you have to convert the given expression in terms of a single variable function. This is only possible if we use the parametric form of a circle. Once you find the equation in one variable, you should know how to find the maximum or minimum value of a function. The maxima or minima of a function occur at critical points or at the end points.