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(-81)^\frac{1}{4}

 Feb 10, 2017

Best Answer 

 #9
avatar+26397 
+15

Hello Melody,

 

i am not certain,

if      z=i

then  z2=(i)(i)=(i)2=i2=1

 

if      z2=1

then z=1=i, but z can also be -i

                                 this  is not equivalent (<=>)

 

laugh

 Feb 10, 2017
 #1
avatar+37168 
0

81^(1/4)  =  481 = 3

 Feb 10, 2017
edited by ElectricPavlov  Feb 10, 2017
 #2
avatar+118710 
+10

(-81)^\frac{1}{4}

 

Firstly there is no real solution for this because no real number to the power of 4 is going to be negative.

 

But I will give you the complex solution :)

 

 

(81)14=(1)14(81)14=(1)143=((1)12)123=i123

 

          Now

          i=1+2i12=1+2i+i22=(1+i)22=1+i2

 

=3i=3(1+i)2=32(1+i)2

 Feb 10, 2017
 #3
avatar+37168 
0

Ooops.....I missed the negative sign !    Thanx, Melody !  

ElectricPavlov  Feb 10, 2017
 #4
avatar+118710 
+5

I forgot to finish it............this is one answer but there are 3 more...

I see Heureka is answering so I expect he will give the other ones.  :)

Thanks in advance Heureka  cool

Melody  Feb 10, 2017
 #6
avatar+26397 
+10

Hello Melody,

 

you set -1 to i2 but i think this is not allowed, if (-1) is a complex number.

only (-1) real is i^2 and (-1) complex is eiπ.

 

This way is okay i21
This way is not allowed 1i2

 

 

laugh

heureka  Feb 10, 2017
 #7
avatar+118710 
0

Hi Heureka,

 

I certainly accept that your knowledge of mathematics is far superior to mine but I do not get why I cannot do this.

Our answers are the same. 

 

I suppose it is some mathematics nuance that has gone over my head :(

 

But thank you   frown

Melody  Feb 10, 2017
 #9
avatar+26397 
+15
Best Answer

Hello Melody,

 

i am not certain,

if      z=i

then  z2=(i)(i)=(i)2=i2=1

 

if      z2=1

then z=1=i, but z can also be -i

                                 this  is not equivalent (<=>)

 

laugh

heureka  Feb 10, 2017
 #10
avatar+118710 
+11

Oh ok Heureka, I can see what you mean :)

Melody  Feb 10, 2017
 #5
avatar+26397 
+15

(-81)^\frac{1}{4}

 

x=481=(81)14=(81)14(1)14=3(1)14|eiπ=1 !(Euler's Formula for Complex Numbers)x1=3(eiπ)14=3eiπ4=3[cos(π4)+isin(π4)]|cos(π4)=sin(π4)=22=3(22+i22)=322(1+i)x2=3ei(π4+π2)x2=3e34iπx3=3ei(π4+2π2)x3=3e54iπx4=3ei(π4+3π2)x4=3e74iπ

 

laugh

 Feb 10, 2017
 #8
avatar+26397 
+15

Hello Melody,

 

i am not certain,

if z =-i then z2 = -1

 

if z2 = -1 then z = 1 = i, but z can also be -i

 

laugh

 Feb 10, 2017

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