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Question: The greatest positive argument of complex number satisfying \(\left| {z - 4} \right| = \operatorname...

The greatest positive argument of complex number satisfying z4=Re(z)\left| {z - 4} \right| = \operatorname{Re} \left( z \right) is-
A. π3\dfrac{\pi }{3}
B. 2π3\dfrac{{2\pi }}{3}
C. π2\dfrac{\pi }{2}
D. π4\dfrac{\pi }{4}

Explanation

Solution

We know that Re means the real part of the equation then we know that z=x+iyz = x + iy where x is the real part and y is the imaginary part of the equation. And we also know that z=x2+y2\left| z \right| = \sqrt {{x^2} + {y^2}} so using this puts the value of the given equation with real part in place of x and imaginary part as y. Solve the equation. You’ll see that it forms the equation of parabola. Now we have to find θ\theta which is called the argument. So the value of tanθ\tan \theta which has the highest positive value will give you the argument.

Complete step by step solution:
Given, z4=Re(z)\left| {z - 4} \right| = \operatorname{Re} \left( z \right)
Here the Re means real part of z hence we can write given equation as-
(x4)2+y2=x\Rightarrow \sqrt {{{\left( {x - 4} \right)}^2} + {y^2}} = x
Where x is the real part of the complex number and y is the imaginary part.
On squaring both sides we get,
(x4)2+y2=x2\Rightarrow {\left( {x - 4} \right)^2} + {y^2} = {x^2}
Now we know that a22ab+b2=(ab)2{a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}
So on applying this formula we get,
(x2+422×4×x)+y2=x2\Rightarrow \left( {{x^2} + {4^2} - 2 \times 4 \times x} \right) + {y^2} = {x^2}
On solving we get,
x2+168x+y2=x2\Rightarrow {x^2} + 16 - 8x + {y^2} = {x^2}
On cancelling the same terms on both side we get,
168x+y2=0\Rightarrow 16 - 8x + {y^2} = 0
Now on adjusting the above equation we get,
y2+168x=0\Rightarrow {y^2} + 16 - 8x = 0
On rearranging we can write,
y2=8x16\Rightarrow {y^2} = 8x - 16
On taking 88 common from right side of the equation we get,
y2=8(x2)\Rightarrow {y^2} = 8\left( {x - 2} \right) --- (i)
This equation is in the form of parabola.
It represents the parabola with focus (4,0)\left( {4,0} \right) lying above the x-axis as both y and x coordinates are positive in it. The imaginary axis is the directrix.

Then the two tangents from the directrix will be at right angles.
And then the greatest positive argument of z will be π4\dfrac{\pi }{4} because at other angles the value of tanθ\tan \theta is less than 11.

Hence the correct answer is D.

Note:
If the question had asked for the imaginary part instead of the real part then we would have put y in the place of x in the right side and solved the equation. Here parabola is upwards as it has both the coordinates on positive x-axis and positive y- axis.