# Math Help - Complex numbers

1. ## Complex numbers

The question:

Simplify $(\sqrt{3 + 4i}+\sqrt{3-4i})^2$ where we assume $\sqrt{z}$ has a non negative real part.

My attempt:

$(3+4i) + 2(9-16i^2) + (3-4i)$ (perfect square)

$3 + 4i + 18 + 32 + 3 - 4i$

= 56

That's not the answer though, it's supposed to be 16. I'm not sure what I'm going wrong. Any assistance would be great!

2. Originally Posted by Glitch
The question:

Simplify $(\sqrt{3 + 4i}+\sqrt{3-4i})^2$ where we assume $\sqrt{z}$ has a non negative real part.

My attempt:

$(3+4i) + 2(9-16i^2) + (3-4i)$ (perfect square)

$3 + 4i + 18 + 32 + 3 - 4i$

= 56

That's not the answer though, it's supposed to be 16. I'm not sure what I'm going wrong. Any assistance would be great!

Who knows what you did here: $(\sqrt{a}+\sqrt{b})^2=a+2\sqrt{ab}+b^2$ , and you did something different (I think though

that you forgot the square root in the middle term), so

$(\sqrt{3 + 4i}+\sqrt{3-4i})^2=3+4i+2\sqrt{9+16}+3-4i=3+10+3=16$ , since $9-16i^2=9+16$

Tonio

3. Originally Posted by Glitch
The question:

Simplify $(\sqrt{3 + 4i}+\sqrt{3-4i})^2$ where we assume $\sqrt{z}$ has a non negative real part.

My attempt:

$(3+4i) + 2(9-16i^2) + (3-4i)$ (perfect square)

$3 + 4i + 18 + 32 + 3 - 4i$
The middle term is wrong. You should still have a square root- it should be $2\sqrt{9- 16i^2}= 2(5)$, not 2(5)(5).

= 56

That's not the answer though, it's supposed to be 16. I'm not sure what I'm going wrong. Any assistance would be great!

4. Thanks! I can't believe I missed that!