Question

Find a complex Z such that : |Z|+Z=4(2+i).

Answer 1

since |Z| = the distance from the origin

id propose something like:

\[\sqrt{a^2+b^2}+a+bi=8+4i\]

Answer 2

\[\sqrt{a^2+b^2}+a+bi=8+4i\]
\[\sqrt{a^2+b^2}=8+4i-a-bi\]
\[a^2+b^2=(8+4i-a-bi)^2\]
\[a^2+b^2=a^2+2abi-(16+8 i) a-b^2+(8-16i) b+(48+64i)\]
\[b^2=2abi-(16+8 i) a-b^2+(8-16i) b+(48+64i)\]
\[0=2abi-(16+8 i) a-2b^2+(8-16i) b+(48+64i)\]

that might be going someplace, but it doesnt seem like it

Answer 3

maybe:
\[|Z|=8+4i=\sqrt{8^2+4^2}\]
\[|Z|=\sqrt{80}\]
\[|Z|=4\sqrt{5}\]

\[4\sqrt{5}+Z=8+4i\]
\[Z=(8-4\sqrt{5})+4i\]
maybe?

Answer 4

z=3+4i according to the wolf

Answer 5

\[Z = a + ib\]
\[ \sqrt{a^2 + b^2} + a + ib = 8 + i4\]
\[b=4\]
\[\sqrt{a^2 + b^2} + a = 8 \]
\[\sqrt{a^2 + 16} = 8 -a\]
\[a^2 + 16 = 64 + a^2 - 16a\]
Maybe like this I am not sure

Answer 6

"according to the the wolf"

LOL

Answer 7

seems like (|Z|+a) = 8

Answer 8

\[16a = 64 - 16\]
\[4a = 16 - 4\]
\[a = 4-1\]
\[a = 3\]
\[a + ib = z= 3 + i4\]

Answer 9

Yeah exactly!

Answer 10

comparing coeffs, good idea

Answer 11

In case it is not obvious |Z| is by definition a real number.
This means you can break the problem into 2 parts:
let Z= a+bi
Im(Z)= Im(8+4i)= 4
so b= 4
and Re(|Z|+Z)= Re(8+4i)= 8
|Z| + a = 8
or |Z|= 8-a
substitute for |Z|= sqrt(a^2+b^2) = sqrt(a^2+16), and solve for a
As noted in previous posts, a= 3

Answer 12

Solution : Z=3+4i.

Answer 13

And note it is easily checked. from pythagoras |Z|= 5 and 5+3+4i= 8+4i