Too Complex!

Let $z=a+bi$ , so replace all the values: $\dfrac{\pi}{4}=\arg\left(\dfrac{a+bi-(10+6i)}{a+bi-(4+6i)}\right)$ Now, order the real part and the imaginary part: $\dfrac{\pi}{4}=\arg\left(\dfrac{(a-10)+i(b-6)}{(a-4)+i(b-6)}\right)$ And rationalize: $\dfrac{\pi}{4}=\arg\left(\dfrac{(a-10)(a-4)+(b-6)^{2}+i((a-4)(b-6)-(a-10)(b-6))}{(a-4)^{2}+(b-6)^{2}}\right)$ The argument of a complex number $z=a+bi$ is: $\arg(z)=\arctan\frac{b}{a}$ , so: $\dfrac{\pi}{4}=\arctan\dfrac{(a-4)(b-6)-(a-10)(b-6)}{(a-10)(a-4)+(b-6)^{2}}$ $1=\dfrac{(a-4)(b-6)-(a-10)(b-6)}{(a-10)(a-4)+(b-6)^{2}}$ $(a-10)(a-4)+(b-6)^{2}=(a-4)(b-6)-(a-10)(b-6)$ Now, try to obtain $a^{2}+b^{2}$ from that expression. The result is: $a^{2}+b^{2}=14a+18b-112$ As we can see, that value is the square of the absolute value of $z$ , but that still depends of some variables. However, the value we're looking for is $|z-7-9i|$ , so let $m$ be that value and have the form $m=c+di$ . Comparing the real parts and the imaginary parts we obtain: $c=a-7 \leftrightarrow a=c+7$ $d=b-9 \leftrightarrow b=d+9$ Substitute: $(c+7)^{2}+(d+9)^{2}=14(c+7)+18(d+9)-112$ And again try to obtain $c^{2}+d^{2}$ , which is the square of the absolute value of $m$ . The result is: $c^{2}+d^{2}=18$ Here, we do have the square of our desired value in numerical value, take square root to both sides to obtain the absolute value of $m$ : $\sqrt{c^{2}+d^{2}}=\boxed{3\sqrt{2}}\approx\boxed{4.2426}$

arg [(z-z1)/(z-z2)]= pi/4 means if z,z1,z2 forms a triangle , angle between z-z1 and z-z2 is pi/4 and it is asked to find value of |z-7-9i| let it be constant k i.e; |z-7-9i|=k is a circle with centre (7,9) and k radius and we can observe (7,9) is equidistant to z1 , z2 we know angle formed by a chord in a circle is constant it means z1 z2 are on circle and radius is answer i.e; 3*sqrt(2)

The locus of $z$ is major arc of circle containing points $(4,6i),(10,6i)$ on its periphery.

The angle that the chord joining these points subtend an angle of $\pi/2$ at its centre $C$ .

Now drop a perpendicular from $C$ to the chord joining $z_1$ and $z_2$ . Let the coordinates of $C$ is ( $h,ki$ . Since chord is parallel to real axis, we find

$h=7$

Now by using $|C-z_1|= |C-z_2|$

We get $k=9$ .

Hence radius of circle $R= 3\sqrt{2}$

Since $z$ lies on circle , answer is $|z-C|= R= \boxed{3\sqrt{2}}$

We can do it all these problem by creating a geometrical analogy( for example here a circle) ...and hey I am unable to post any image of my solution please help me!!