please help
determine M(z) as : z is a complex nember and
/z/=z+conjugated(z)
NOTE:
(/z/=module of z)
+118677 Thanks Shaomada, you are ofcourse correct. The question just confused me.
when I said that $$|z|\ne z+\bar z$$ I really meant that they are not the same thing.
You are right - they are equal sometimes.
$$\\\sqrt{a^2+b^2}=2a\qquad a \ge 0\\
a^2+b^2=4a^2\\
b^2=3a^2$$
They will be equal whenever this condition is met.
+118677 I do not understand this question so I am just going to prattle about it.
$$\\if\;\;z=a+bi\\\\
\mbox{Then conjugate of z }=\bar z = a-bi\\\\
and\\\\
|z|=|\bar z|=\sqrt{a^2+b^2}\\\\
|z|$ is the distance of z from the origin in the complex number plane
$\\\\
|z|$ is the modulus of z$\\\\
|z| \ne z+\bar z\\\\
z+\bar z=a+bi+a-bi = 2a\\\\
$Hence I am confused $$$
+66 Shouldnt exactly the complex numbers of the form $$z=a+i\cdot b$$ for which $$\lvert b \rvert = \sqrt{3} \cdot a$$ fullfill this? You want $$\sqrt{a^2+b^2} = \lvert z \rvert = z + \overline{z} = (a + i \cdot b) + (a - i \cdot b) = 2 \cdot a = \sqrt{4a^2}$$. If we assume $$a$$ to be not negative this is equivalent to $$a^2 + b^2 = 4a^2$$ or to $$b^2 = 3 \cdot a^2$$.
Also Melody, i don't get why $$\lvert z \rvert \neq z + \overline{z}$$ should hold for all complex numbers $$z$$.
+118677 Thanks Shaomada, you are ofcourse correct. The question just confused me.
when I said that $$|z|\ne z+\bar z$$ I really meant that they are not the same thing.
You are right - they are equal sometimes.
$$\\\sqrt{a^2+b^2}=2a\qquad a \ge 0\\
a^2+b^2=4a^2\\
b^2=3a^2$$
They will be equal whenever this condition is met.
