SOLUTION: z-|z|=8+4i Find all complex numbers z that satisfy the equation:

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Question 1073485: z-|z|=8+4i
Find all complex numbers z that satisfy the equation:

Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let z = a + bi be our complex number under the question.
We need to find its real and imaginary parts "a" and "b".

As you, probably, know  |z| = sqrt%28a%5E2+%2B+b%5E2%29.

So, the given equation becomes

a + bi - sqrt%28a%5E2+%2B+b%5E2%29 = 8 + 4i.


It actually deploys in two independent equations for the real part and the imaginary part separately:

a - sqrt%28a%5E2+%2B+b%5E2%29 = 8     (1)     and
b            = 4.    (2)

Equation (1) is equivalent to

a - 8 = sqrt%28a%5E2+%2B+b%5E2%29,

%28a-8%29%5E2 = a%5E2+%2B+b%5E2,

a%5E2+-+16a+%2B+64 = a%5E2+%2B+4%5E2,    (I substituted b= 4)

-16a + 64 = 16  --->  -16a = 16-64  --->  -16a = -48  ---> a = 3.


So, the logic leads us to the solution  a = 3, b = 4, i.e. z = 3 + 4i.


Now CHECK this solution: 3 + 4i - sqrt%283%5E2+%2B+4%5E2%29 = 3 + 4i - 5 = -2 + 4i.


It doesn't check !!!


The conclusion is: The given equation HAS NO solutions.

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When I got this result,  I was very surprised:  it is for the first time in my life I see the equation for complex numbers,
which has no solution.

But reviewing the problem and the solution,  I got the understanding WHY it happened in this case.

The equation of the form  z - |z| = u + iw  CAN NOT have a solution if the real part "u" of the complex number
in the right hand side is positive.

Actually,  it is OBVIOUS !

Now I am calm.  Because I know the reason.
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See my lessons on complex numbers in this site
    - Complex numbers and arithmetic operations on them
    - Complex plane
    - Addition and subtraction of complex numbers in complex plane
    - Multiplication and division of complex numbers in complex plane
    - Raising a complex number to an integer power
    - How to take a root of a complex number
    - Solution of the quadratic equation with real coefficients on complex domain

    - Solved problems on taking roots of complex numbers
    - Solved problems on arithmetic operations on complex numbers
    - Solved problem on taking square root of complex number
    - Miscellaneous problems on complex numbers
    - Advanced problem on complex numbers


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Complex numbers".



Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

let z = x + yi, where x and y are real

Substitute in

z+-+abs%28z%29+%22%22=%22%22+8+%2B+4i

x+%2B+yi+-+abs%28x+%2B+iy%29+%22%22=%22%22+8+%2B+4i%0D%0A%0D%0A%7B%7B%7Bx+%2B+yi+-+sqrt%28x%5E2+%2B+y%5E2%29+%22%22=%22%22+8+%2B+4i

Set real parts equal, and imaginary parts equal:

x-sqrt%28x%5E2%2By%5E2%29%22%22=%22%228,  yi+%22%22=%22%22+4i
                   y+%22%22=%22%22+4

Substitute y=4

x-sqrt%28x%5E2%2B4%5E2%29%22%22=%22%228

x-sqrt%28x%5E2%2B16%29%22%22=%22%228

x-8%22%22=%22%22sqrt%28x%5E2%2B16%29

Square both sides:

x%5E2-16x%2B64%22%22=%22%22x%5E2%2B16

Cancel the x2's

-16x%2B64%22%22=%22%2216

-16x%22%22=%22%22-48

x%22%22=%22%223

But we must check to see if 3 is 
a solution or only an extraneous answer.
Substitute in:

x-sqrt%28x%5E2%2B16%29%22%22=%22%228
3-sqrt%283%5E2%2B16%29%22%22=%22%228
3-sqrt%289%2B16%29%22%22=%22%228
3-sqrt%2825%29%22%22=%22%228
3-5%22%22=%22%228
-2%22%22=%22%228

That's false so 3 is extraneous and there
is no solution.

Edwin