SOLUTION: What are the roots of (z^2) - z(1 + i) + 5i = 0?

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Question 1181502: What are the roots of (z^2) - z(1 + i) + 5i = 0?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

roots of
z%5E2+-+z%281+%2B+i%29+%2B+5i+=+0......... substitute z=x%2Byi

%28x%2Byi%29%5E2+-+%28x%2Byi%29%281+%2B+i%29+%2B+5i+=+0...expand

x%5E2%2B2xy%2Ai%2B%28yi%29%5E2+-+%28x%2Bx%2Ai%2Byi%2By%2Ai%5E2%29%2B+5i+=+0

x%5E2%2B2xy%2Ai%2B%28yi%29%5E2+-+%28x%2Bx%2Ai%2Byi-y%29%2B+5i+=+0

x%5E2%2B2xy%2Ai%2B%28y%5E2i%5E2%29+-+x-x%2Ai-yi%2By%2B+5i+=+0

x%5E2%2B2xy%2Ai-y%5E2+-+x-x%2Ai-yi%2By%2B+5i+=+0...........combine real parts and imaginary parts

%28x%5E2-y%5E2+-x%2By%29%2B2xy%2Ai-x%2Ai-yi%2B+5i=0

%28x%5E2-y%5E2+-x%2By%29-i%282xy-x-y%2B5%29=0

Complex numbers can be equal only if their real and imaginary parts are equal.
x%5E2-y%5E2+-x%2By=0.....eq.1
2xy-x-y%2B5=0.....eq.2
-------------------solve this system
2xy-x=y-5....eq.2, solve for x
%282y-1%29x=y-5
x=%28y-5%29%2F%282y-1%29
substitute in eq.1
%28%28y-5%29%2F%282y-1%29%29%5E2-y%5E2+-%28y-5%29%2F%282y-1%29%2By=0
%28%28y+-+2%29+%28y+%2B+1%29%29%2F%282+y+-+1%29+=+0} .........zeros are in numerator
%28y+-+2%29+%28y+%2B+1%29=0
y=2+or y=-1
then
x=%282-5%29%2F%282%2A2-1%29=-3%2F3=-1
x=%28-1-5%29%2F%282%28-1%29-1%29=-6%2F-3=2
solutions:

x=2, y=-1
or
x=-1, y=2
Substitute back z=x%2Byi, and roots are
z%5B1%5D=2-i
z%5B2%5D=-1%2B2i


Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
What are the roots of (z^2) - z(1 + i) + 5i = 0?
~~~~~~~~~~~~~~~~~ 

What you see in front of you, is a quadratic equation over complex numbers with complex coefficients.

Exactly as usual quadratic equation over real numbers, you can solve it over complex numbers using the Quadratic Formula.


    z%5B1%2C2%5D = %28-b+%2B-+sqrt%28b%5E2-4ac%29%29%2F%282a%29 = %28%281%2Bi%29+%2B-+sqrt%28%281%2Bi%29%5E2+-+4%2A%285i%29%29%29%2F2 = %28%281%2Bi%29+%2B-+sqrt%281%2B2i%2Bi%5E2+-20i%29%29%2F2 = 

  = %28%281%2Bi%29+%2B-+sqrt%28-18i%29%29%2F2 = %28%281%2Bi%29+%2B-+3%2Asqrt%28-2i%29%29%2F2.    (1)


Now notice that  -2i = 2*cis(270°);  therefore,  sqrt%28-2i%29 = sqrt%282%29%2Acis%28135%5Eo%29 = (-1+i).

Having it, we can continue the formula (1) this way


    z%5B1%2C2%5D = %28%281%2Bi%29+%2B-+3%2A%28-1%2Bi%29%29%2F2.


Therefore,


    z%5B1%5D = %28%281%2Bi%29+%2B+3%2A%28-1%2Bi%29%29%2F2 = %281%2Bi+-+3+%2B+3i%29%2F2 = %28-2+%2B+4i%29%2F2 = -1 + 2i;

    z%5B2%5D = %28%281%2Bi%29+-+3%2A%28-1%2Bi%29%29%2F2 = %281%2Bi+%2B+3+-+3i%29%2F2 = %284+-+2i%29%2F2 = 2 - i.


ANSWER.  The roots are  -1 + 2i  and  2 - i.

Solved.

Compare the volume of my calculations with that by @MathLover1.

-------------------

There is a bunch of my lessons on complex numbers
    - Complex numbers and arithmetical 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
    - How to take a square root of a complex number
    - Solution of the quadratic equation with complex 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
    - Solved problems on de'Moivre formula
in this site.

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".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.