+0

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

0
421
9

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

math
Guest Aug 12, 2014

#4
+26412
+5

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

Alan  Aug 12, 2014
Sort:

#1
+676
0

TakahiroMaeda  Aug 12, 2014
#2
0

Guest Aug 12, 2014
#3
+676
+5

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 >_<

TakahiroMaeda  Aug 12, 2014
#4
+26412
+5

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

Alan  Aug 12, 2014
#5
+91517
0

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

----------------------------------------------

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
+26412
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
+91517
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
+26412
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 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.

Alan  Aug 12, 2014
#9
+91517
0

okay thanks Alan.

Melody  Aug 12, 2014

### 9 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details