An Equation With Nothing | Problem 451

[@SIB1963]

You're Turkish? Very cool.

[@chaosredefined3834]

Using c(z) to represent the conjugate... i z / (z - 1) = c(z) i z = c(z) (z - 1) i z = z c(z) - c(z) c(z) + i z = z c(z) Take the conjugate of both sides z - i c(z) = c(z) z Note that the right hand side of both is the same c(z) + i z = z - i c(z) c(z) + i c(z) = z - i z c(z) (1 + i) = z (1 - i) c(z) (1 + i)(1 - i) = z (1 - i)(1 - i) Note that (1 + i)(1 - i) = 2, and (1 - i)(1 - i) = -2i 2 c(z) = -2i z c(z) = -iz Substitute this back into the original equation i z / (z - 1) = -iz iz = -iz(z - 1) iz = -i z^2 + iz 0 = -i z^2 0 = z^2 0 = z So, the solution is... z = 0.

[@NadiehFan]

We have (1) iz = zz̅ − z̅ and conjugating both sides gives (2) −iz̅ = zz̅ − z Subtracting (2) from (1) gives i(z + z̅ ) = (z − z̅ ) 2i·Re(z) = 2i·Im(z) so (3) Re(z) = Im(z) Adding (1) and (2) gives i(z − z̅ ) = 2·zz̅ − (z + z̅ ) i·2i·Im(z) = 2·((Re(z))² + (Im(z))²) − 2·Re(z) −2·Im(z) = 2·((Re(z))² + (Im(z))²) − 2·Re(z) Since, in accordance with (3), Re(z) = Im(z) this gives −2·Im(z) = 4·Im(z))² − 2·Im(z) 4·Im(z))² = 0 and therefore (4) Im(z) = 0 In accordance with (3) this implies that we also have (5) Re(z) = 0 From (5) and (4) it follows that z = 0 is indeed the sole solution of (1).

[@moeberry8226]

Finally I know your Turkish. You have been hiding this for a while bro lol.

[@scottleung9587]

I got z=1+i as another solution.

[@NadiehFan]

That is not correct. If you claim some solution, then please show your work.

[@Etienne-pq3dx]

1+z is not equal to 1-z

[@wes9627]

Let z=a+ib, z*=a-ib, and iz=ia-b. iz=(z-1)z*. ia-b=(a+ib)(a-ib)-(a-ib)=a^2+b^2-a+ib. Equating real and imaginary parts: a^2+b^2-a+b=0 and a=b. Thus, a=b=0, so z=0.

[@tod9141]

sybermath???