Complex numbers

Apr 2010
487
9
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!
 
Oct 2009
4,261
1,836
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
 
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HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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!
 
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Apr 2010
487
9
Thanks! I can't believe I missed that!
 
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