Complex Analysis Properties Question

[RJLiberator]

Use properties to show that:
(question is in the attached picture)

Now, it is my understanding that due to properties you can express (sqrt(5)-i) as the sqrt((sqrt(5))^2+(-1)^2) which equals sqrt(6).
And (2zbar+5) can be represented as (2z+5).

But this would be sqrt(6)*(2z+5) which is NOT sqrt(3)*(2z+5)

Was there a typo in this problem? Or am I not thinking of something?

Thank you.

 

[Dick]

Use properties to show that:
(question is in the attached picture)

Now, it is my understanding that due to properties you can express (sqrt(5)-i) as the sqrt((sqrt(5))^2+(-1)^2) which equals sqrt(6).
And (2zbar+5) can be represented as (2z+5).

But this would be sqrt(6)*(2z+5) which is NOT sqrt(3)*(2z+5)

Was there a typo in this problem? Or am I not thinking of something?

Thank you.

There are some absolute value signs missing, but yes, I think there is a typo.

 

[RJLiberator]

Excellent. After performing the operations, I felt like I had a good understanding of it. This seems to confirm my suspicions :). Kind regards for your help tonight.

 

[Fredrik]

Now, it is my understanding that due to properties you can express (sqrt(5)-i) as the sqrt((sqrt(5))^2+(-1)^2) which equals sqrt(6).

What you wrote here is very far from true. The real number $\sqrt{6}$ obviously can't be equal to $\sqrt{5}-i$, which isn't even real. But Dick said something about missing absolute value signs, so I suppose you could be talking about something like this: For all $z,w\in\mathbb C$, if $\operatorname{Re}\bar z w=0$, then $|z+w|^2=|z|^2+|w|^2$. This implies that $|\sqrt{5}-i|^2=|\sqrt{5}|^2+|-i|^2=5+1=6$, and this implies that $\sqrt{6}=|\sqrt{5}-i|$.