#3**+5 **

Providing I know what you are talking about this is how I will solve it. If I misinterpret the question, please forgive me.

Here are the answers:

$$2 sqrt4(-1) \approx 1.4142+1.4142i$$

$$2(-1)^{\frac{3}{4}}\approx -1.4142+1.4142i$$

$$-2 sqrt4(-1)\approx 1.4142-1.4142i$$

$$-2(-1)^{\frac{3}{4}}\approx 1.4142-1.4142i$$

This is only correct if I understand the question >_<

TakahiroMaeda
Aug 12, 2014

#5**0 **

Takahiro I think you may have forgotten one minus sign. Two of your answers are the same. (The last 2)

Also your answers are approximations. You would be better off to stick to exact answers.

I gave you thumbs up though. Thanks

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NOW I am looking at these answers and I can certainly see that they work BUT to find these for this or a similar question what do I need to know or keep in mind?

$$[\pm(1\pm i)]^2=\pm2i\;\rightarrow \;[\pm2i]^2=-4$$

Do I need to commit this to memory?

Melody
Aug 12, 2014

#6**0 **

Melody asked: *Do I need to commit this to memory?*

Probably best to remember it in terms of polar representation of complex numbers:

etc.

Alan
Aug 12, 2014

#7**0 **

Hi Sir Alan,

Please do not be offended but I am amused by what you think is good idea.

BUT I will still try and work out what you are telling me.

the first line is good I am glad i can understand that much.

But, where did the 2n+1 that you have everywhere come from?

Melody
Aug 12, 2014

#8**0 **

Hi Melody. The n in (2n+1) is just an integer.

For a square root there are two solutions, so n takes on two values, say 0 and 1, and 2n+1 becomes 1 and 3.

These give two values of -1 when plugged in to e^{i(2n+1)pi}. The two square roots of -1 are then found by taking the square root of these (e^{i(2n+1)pi})^{1/2} = e^{i(2n+1)pi/2}. When n = 0, e^{i(2n+1)pi/2}=^{ }e^{ipi/2} = i, and when n = 1, e^{i(2n+1)pi/2}=^{ }e^{i3pi/2} = -i. Hence we have two values (i and -i) for the square root of -1, one for each value of n.

Similarly, we need three values of n (0, 1 and 2) for finding the three cube roots of -1, and four values for the four fourth roots etc.

Alan
Aug 12, 2014