Question 1208303
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Answer:  <font color=red>-2+2i</font>


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Quick Explanation


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, this places the complex number in the northwest quadrant (i.e. upper left corner)
Of the four answer choices, only <font color=red>-2+2i</font> is in this quadrant.


I'll let the student verify that z = -2+2i satisfies the equation abs(z^2+4) = abs(4-2iz)
abs(x) = absolute value of x


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Longer Explanation


z = x+iy
w = -3+0i
Arg(z*w) = Arg(z)+Arg(w)
Arg(z*(-3)) = Arg(z)+Arg(-3)
Arg(z*(-3)) = Arg(z)+Arg(-3+0i)
Arg(-3z) = Arg(z)+180 degrees
Arg(-3z) = arctan(y/x)+180 degrees


Let's use the given info from the instructions.
Arg(-3z) = -45 degrees
arctan(y/x)+180 = -45
arctan(y/x) = -45-180
arctan(y/x) = -225
y/x = tan(-225)
y/x = -tan(225)
y/x = -1
y = -x



The complex number
z = x+iy
updates to
z = x+i(-x)
z = x-ix


Then,
z^2+4 = (x-ix)^2+4
z^2+4 = x^2-2ix^2+i^2x^2+4
z^2+4 = x^2-2ix^2+(-1)x^2+4
z^2+4 = x^2-2ix^2-x^2+4
z^2+4 = 4+(-2x^2)i
It has real part 4 and imaginary part -2x^2
abs(z^2+4) = magnitude of z^2+4 = absolute value of z^2+4
abs(z^2+4) = sqrt( realPart^2 + imaginaryPart^2 ) 
abs(z^2+4) = sqrt( 4^2 + (-2x^2)^2 ) 
abs(z^2+4) = sqrt( 4x^4+16 ) 
Let's refer to this as equation (1) so we can use it later.


4-2iz = 4-2i(x-ix)
4-2iz = 4-2ix+2i^2x
4-2iz = 4-2ix+2(-1)x
4-2iz = 4-2ix-2x
4-2iz = (4-2x)+(-2x)i
This has real part 4-2x and imaginary part -2x
abs(4-2iz) = sqrt( realPart^2 + imaginaryPart^2 )
abs(4-2iz) = sqrt( (4-2x)^2 + (-2x)^2 )
abs(4-2iz) = sqrt( 8x^2-16x+16 )
Let's refer to this as equation (2).


The instructions mention that the left hand sides of equations (1) and (2) are equal, so that allows us equate the right hand sides.


sqrt( 4x^4+16 ) = sqrt( 8x^2-16x+16 )
4x^4+16  = 8x^2-16x+16 
4x^4-8x^2+16x+16-16 = 0
4x^4-8x^2+16x = 0
x(4x^3-8x+16) = 0
x = 0 or 4x^3-8x+16 = 0


If x = 0, then z = x-ix = 0-0i = 0, but we're specifically told in the instructions that z is nonzero.
We rule out x = 0 being allowed.


If you solve 4x^3-8x+16 = 0 through use of either<ul><li>a graphing calculator</li><li>the rational root theorem to generate the possible roots, then check each one</li></ul>then you'll find that x = -2 is a root.
4x^3-8x+16 = 0
4(-2)^3-8(-2)+16 = 0
4(-8)+16+16 = 0
-32+16+16 = 0
0 = 0



Therefore
z = x-ix
becomes
z = -2-i(-2)
z = <font color=red>-2+2i</font>
which is the final answer.



Note that Arg(-3z) = Arg(-3(-2+2i)) = Arg(6-6i) = -45 to help confirm the answer is correct.
I'll let the student verify the other requirement.


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Extra info:


4x^3-8x+16 = 4(x+2)(x^2-2x+2)
You can use synthetic division to help factor.


Solving x^2-2x+2 = 0 leads to the complex roots x=1+i and x=1-i
Use the quadratic formula or complete the square.


If x=1+i, then,
z = x-ix
z = (1+i)-i(1+i)
z = (1+i)+(-i-i^2)
z = (1+i)+(-i-(-1))
z = (1+i)+(-i+1)
z = 2
But Arg(-3z) = Arg(-3*2) = Arg(-6+0i) = 180 when we wanted -45 instead.
The angles 180 and -45 are not coterminal.
This allows us to rule out x=1+i.


If x=1-i, then,
z = x-ix
z = (1-i)-i(1-i)
z = (1-i)+(-i+i^2)
z = (1-i)+(-i-1)
z = -2i
But Arg(-3z) = Arg(-3*(-2i)) = Arg(0+6i) = 90 instead of -45 degrees. 
The angles 90 and -45 are not coterminal.
This allows us to rule out x=1-i.


Therefore, the only root we can use is x = -2.
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