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# How do i solve the sqrt4(-16)? (complex numbers; 4 different roots)

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How do i solve the sqrt4(-16)? (complex numbers; 4 different roots)

math
Aug 12, 2014

#4
+33053
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You should check the above by raising each of them to the power 4.

Aug 12, 2014

#1
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Aug 12, 2014
#2
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Aug 12, 2014
#3
+676
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Providing I know what you are talking about this is how I will solve it. If I misinterpret the question, please forgive me.

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

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

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

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

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

Aug 12, 2014
#4
+33053
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You should check the above by raising each of them to the power 4.

Alan Aug 12, 2014
#5
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Takahiro I think you may have forgotten one minus sign.  Two of your answers are the same. (The last 2)

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?

Aug 12, 2014
#6
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Melody asked: Do I need to commit this to memory?

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

etc.

Aug 12, 2014
#7
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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?

Aug 12, 2014
#8
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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 ei(2n+1)pi. The two square roots of -1 are then found by taking the square root of these (ei(2n+1)pi)1/2 = ei(2n+1)pi/2. When n = 0, ei(2n+1)pi/2= eipi/2 = i, and when n = 1, ei(2n+1)pi/2= ei3pi/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.

Aug 12, 2014
#9
+117762
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okay thanks Alan.

Aug 12, 2014