# Complex numbers

#### Glitch

The question:

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

My attempt:

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

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

= 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!

#### tonio

The question:

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

My attempt:

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

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

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

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

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

Tonio

Glitch

#### HallsofIvy

MHF Helper
The question:

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

My attempt:

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

$$\displaystyle 3 + 4i + 18 + 32 + 3 - 4i$$
The middle term is wrong. You should still have a square root- it should be $$\displaystyle 2\sqrt{9- 16i^2}= 2(5)$$, 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!

Glitch

#### Glitch

Thanks! I can't believe I missed that!

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