complex numbers q (1 Viewer)

sssona09 · Nov 6, 2017

so theres this q,
am^2+am+c=0 where m is complex number and a,b,c are real
so we need to use conjugate theorems to show that a(m conjugate)^2 +b(m conjugate) +c=0

i thought we cn just say that conjugate of m is a root, but how do we use conjugate theorems :/

1729 · Nov 6, 2017

sssona09 said:
so theres this q,
am^2+am+c=0 where m is complex number and a,b,c are real
so we need to use conjugate theorems to show that a(m conjugate)^2 +b(m conjugate) +c=0

i thought we cn just say that conjugate of m is a root, but how do we use conjugate theorems :/

$$\noindent Take the conjugate of both sides and use $\overline{z_1+z_2+\ldots +z_n}=\overline{z_1} + \overline{z_2} + \ldots + \overline{z_n}$ and $\overline{z_1 \times z_2 \times \ldots \times z_n} = \overline{z_1} \times \overline{z_2} \times \ldots \times \overline{z_n}$ and the fact that the complex conjugate of a real number is itself.$

sssona09 · Nov 6, 2017

1729 said:
$$\noindent Take the conjugate of both sides and use $\overline{z_1+z_2+\ldots +z_n}=\overline{z_1} + \overline{z_2} + \ldots + \overline{z_n}$ and $\overline{z_1 \times z_2 \times \ldots \times z_n} = \overline{z_1} \times \overline{z_2} \times \ldots \times \overline{z_n}$ and the fact that the complex conjugate of a real number is itself.$

thank you <3

complex numbers q (1 Viewer)

sssona09

Member

1729

Active Member

sssona09

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)