Question

how does one go about solving something such as:

conjugate(z+2-3i) = z+5+10i

Conjugates of actual complex numbers are very simple, but the inclusion of z trips me up

Answer 1

Let z = x+iy
z+2-3i = x+2 + (y-3)i
So, its conjugate is x+2 - (y-3)i
and your equation becomes

x+2 - (y-3)i = x+5 + (y+10)i

since the real and imaginary parts must be equal, there is no solution, since for no x,y is x+2=x+5 and y-3=y+10

I suspect some kind of typo...

Answer 2

I assume we are solving for z.


when we multiply conjugates we are to get a real number
(z+2-3i)(z+5+10i)
= z^2 + 5z + 10iz + 2z + 10 + 20i - 3iz - 15i - 30i^2
= z^2 + 7z + 7iz + 40
let z = a+bi

z^2 + 7z + 7iz + 40
= (a^2 + 2bi + b^2 i^2) + 7(a + bi) + 7i(a+bi) + 40
= a^2 + 2bi - b^2 + 7a + 7bi + 7ai - 7b + 40
= (a^2 - b^2 + 7a - 7b + 40) + (6b+7a)i
for this to be real, 6b+7a = 0, 6b = -7a
b = -7a/6
then the real part
= a^2 - 49a^2/36 + 7a + 49a/6 + 40
= (36a^2 - 49a^2 + 252a + 294a + 1440)/36
= (-13a^2 + 546a + 1440)/36


but the right side was
a+bi + 5 + 10i
= (a+5) + (b+10)i

then: (-13a^2 + 546a + 1440)/36 = a+5
-13a^2 + 546a + 1440 = 36a + 180
-13a^2 + 510a + 1260 = 0
13a^2 - 510a - 1260 = 0

I get a = appr 41.563 or a = appr -2.332
then the corresponding
b= -48.49 and b = 2.72

z = 41.563 - 48.49i OR z = -2.332 + 2.72i

let's check the 2nd one:
first part = z +2-3i
= -2.332 + 2.72i + 2+3i
= -.332 + 5.72i

right side = z+5+10i
= -2.332 + 2.72i + 5 + 10i
= 2.668 - 12.72i

OH NO, I was expecting them to be conjugates.
Where did i go wrong ????

(I will copy/paste my reply and study it carefully. Hopefully I will get back to you if I find my error)

Answer 3

I started off the same way as Steve and then ...
....you never know with complex numbers,

as it stands, I noticed a real silly error on my part.
I dropped the 20i - 15i in the 3rd line.

No fun correcting and seeing where it leads me