How Do Complex Number Operations Affect Results?

Ry122
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How do I express the following in the form a+ib?

(\sqrt{-4} - 1) (\sqrt{-9})

My attempt:
(2i - 1)(3i)
6i-3i = 3i
 
Last edited:
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Close, but (2i)(3i)=6i2=-6, not 6i.
 
Can you tell me why the answer -6-+3i has plus or minus 3i?

Also can you tell me the answer to (3+i)^2 - (3+i)^2
The answer I got was 0.
9+6i+i^2-9-6i-i^2=0
 
Last edited:
You have to define what you mean by sqrt(-n) for n>0. You could choose it to be either i*sqrt(n) or -i*sqrt(n). That's your sign ambiguity.
 
Ry122 said:
Can you tell me why the answer -6-+3i has plus or minus 3i?
What was that the answer to?? Surely not the (2i-1)(3i) problem you were just talking about!

Also can you tell me the answer to (3+i)^2 - (3+i)^2
The answer I got was 0.
9+6i+i^2-9-6i-i^2=0
The answer to a2- a2 is 0 no matter what a is!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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