complex numbers

Question: Consider the set $H = \{z \in C: z=x-1+xi ; x \in R\}$ .

Prove that there is a unique number $z \in H$ such that $|z| \leq |w|$ for all $w \in H$ .

Solution:

Let $w=y-1+yi$ with $y \in R$ . It suffices to prove that there is unique number $x \in R$ such that

$(x-1)^{2}+x^{2} \leq (y-1)^{2}+y^{2}$ for all $y \in R$ .

In other words, x is the minimum point of the function

$f: R \longrightarrow R$ , $f(y)=(y-1)^{2}+y^{2}=2y^{2}-2y+1=2(y-1/2)^{2}+1/2$ .

Hence, $x=1/2$ and $z=-1/2+i/2$ .

Crisp and clear …right?

More later…

Nalin Pithwa

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Google account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Connecting to %s