Prove that the statements|z+ 1|>|z−1|and Re(z)>0 are equivalent.

Question

Please show step by step.

Answer 1

Let z = a+bi  Re(z) = a.

 

l z+1l = l (a+1) + bi l = √[a²+2a+1+b²]

 

l z-1l = l (a-1) + bi l = √[a²-2a+1+b²]

 

So, if l z+1 l > l z-1 l, then a²+2a+1+b² > a²-2a+1+b²

 

   Simplify to obtain 4a > 0.  Therefore, a > 0.

 

Answer 2

Before working out an algebraic proof of the statement maybe try viewing it another way.

 

Remember that |z₁ - z₂| = distance between z₁ and z₂.

 

|z + 1| = |z - (-1)| = the distance between z and -1.

 

|z - 1| = the distance between z and 1.

 

So the statement that |z + 1| > |z - 1| is equivalent to the statement that z is closer to 1 than -1. Think on why this is equivalent to Re(z) > 0.

 

Picture the points in the complex plane that are closer to 1 than -1. Picture those points that are equidistant.

 

I hope this helps you work out the proof yourself, but if you'd be interested in some more, please let me know. I'll be happy to explain further. Thank you!