Question

The greatest positive argument of complex number satisfying |z-4|=re(z)

Answer 1

Solution:

A complex number, Z = x  + i y

has, Modulus (Z)= $\sqrt{x^2 +y^2}$

And Principal Argument =

$Tan A=\frac{y}{x}$

Greatest positive argument , when A= 90°, that is when , y= $\pm A$ and x=0.

Now,it is given that

$\left | z-4 \right |= Real(Z)\\\\ \left | x + i y-4 \right |= x\\\\ \left | x-4 + yi \right |= x\\\\ \sqrt{(x-4)^2+y^2}=x$

Squaring both sides

$x^2- 8 x +16+y^2= x^2\\\\ y^2= 8 x - 16$

The argument will be greatest ,that is of 90° when x=0 and , y= $\pm4i$

So, Z= 0  $\pm4i$

Answer 2

Answer:

The greatest positive argument is $Z= 0 \pm4i$

Step-by-step explanation:

We have to find : The greatest positive argument of complex number satisfying |z-4|=re(z)

Solution :

A complex number is in the form,

$Z = x+ iy$

Modulus (Z) is

$|Z|=\sqrt{x^2 +y^2}$

And Principal Argument is  

$\tan A=\frac{y}{x}$

Greatest positive argument , when A= 90°, that is when, $y=\pm A$ and x=0.

Now,

$\left | z-4 \right |= Real(Z)\\\\ \left | x + i y-4 \right |= x\\\\ \left | x-4 + yi \right |= x\\\\ \sqrt{(x-4)^2+y^2}=x$

Squaring both sides,

$x^2- 8 x +16+y^2= x^2\\\\ y^2= 8 x - 16$

The argument will be greatest,

that is of 90° when x=0 and ,  $y= \pm4i$

So, $Z= 0 \pm4i$

Therefore, The greatest positive argument is $Z= 0 \pm4i$