Question 1208303
.
let z = x + iy , |z^2 + 4| = |4 - 2 iz|, z ≠ 0 and Arg(-3z) = -45 , then z = ....,
(A) 2 - 2i , (B) 3i - 3, (C) -2 + 2i, (D) 3 - 3i.
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        In his short solution, tutor @math_tutor2020 is not precisely correct.

        He said that only one of four optional numbers is in the second quadrant.

        It is not so.  Together with -2+2i, the number 3i-3 is also in the second quadrant.


        So, I came to bring another, really short, simple and complete solution.


        First 4 lines of my solution repeat first 4 lines of the solution by @math_tutor2020. 
        But what follows after that, is different.



<pre>
Arg(kz)   =  Arg(z)  + 180  when  k < 0
Arg(-3z) =  Arg(z)  + 180
-45 = Arg(z) + 180
Arg(z) = -225 degrees which is coterminal to -225+360 = 135 degrees.


Since Arg(z) = 135 degrees, complex number z is in the second quadrant and has the form  

               z = -a + ia,  where "a" is a positive real number.


In the optional list, we have two and only two such numbers, (B) 3i-3,  and  (C) -2+2i.


So, we will consider two cases.


Case (C):  z = -2+2i.

           Then it is clear that z^2 will be pure imaginary.  Indeed,  z^2 = (-2+2i)^2 = 4 - 8i -4 = -8i.

           Hence,  z^2+4 = 4-8i  and  |4-8i| = {{{sqrt(4^2+8^2)}}} = {{{sqrt(16+64)}}} = {{{sqrt(80)}}}.


           From the other side,  4-2iz = 4 - 2i*(-2+2i) = 4 + 4i + 4 = 8+4i.

           Hence,  |4-2iz| = {{{sqrt(8^2+4^2)}}} = {{{sqrt(64+16)}}} = {{{sqrt(80)}}}.


           Thus we see that  |z^2+4| = |4-2iz|,  so case (C) is the solution.

            


Case (B):  z = 3i-3.

           Again, it is clear that z^2 will be pure imaginary.  Indeed,  z^2 = (3i-3)^2 = -9 -18i + 9 = -18i.

           Hence,  z^2+4 = 4-18i  and  |4-18i| = {{{sqrt(4^2+18^2)}}} = {{{sqrt(16+324)}}} = {{{sqrt(340)}}}.


           From the other side,  4-2iz = 4 - 2i*(3i-3) = 4 + 6 + 6i = 10+6i.

           Hence,  |4-2iz| = {{{sqrt(10^2+6^2)}}} = {{{sqrt(100+36)}}} = {{{sqrt(136)}}}.


           Thus we see that  |z^2+4| =/= |4-2iz|,  so case (B) is not the solution.


Thus, from the four given options, only one,  z = -2 + 2i,  is the solution to the problem.


<U>ANSWER</U>.  From the four given options, only one,  z = -2 + 2i,  is the solution to the problem.
</pre>

Solved completely, in a short and simple way.


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I do not think that long solution by the other tutor is a right approach: it makes the entire job 
unjustifiably long and complicated.


I also do not believe that this complicated way is the right way to teach.


And, finally, I do not think that that long complicated solution is what is expected.


The work can be done in much shorter and simpler way.