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a) Find the square roots of $3+4i$.

 

b) Find all cube roots of 8i.
 

where i=sqrt-1

 Dec 12, 2017
 #2
avatar+26396 
+2

Algrebra Help

 

a) Find the square roots of 3+4i.

z=a+bi=3+4i|z|=a2+b2=32+42=5sin(φ)=b|z|=45=0.8φ=arcsin(0.8)+2kπz=|z|eiφz=5ei(arcsin(0.8)+2kπ)

 

3+4i=(3+4i)12=512(ei(arcsin(0.8)+2kπ))12=5ei(arcsin(0.8)+2kπ2)=5ei(arcsin(0.8)2+kπ)|k=(0,1)3+4i=5ei(arcsin(0.8)2)|k=0=5(cos(arcsin(0.8)2)+isin(arcsin(0.8)2))=2.23606797750(0.89442719100+i0.44721359550)=2+i13+4i=5ei(arcsin(0.8)2+π)|k=1=5(cos(arcsin(0.8)2+π)+isin(arcsin(0.8)2+π))=5(cos(arcsin(0.8)2)isin(arcsin(0.8)2))=2.23606797750(0.89442719100i0.44721359550)=2i1

 

All 2nd roots of 3+4i :

2+i2i

 

laugh

 Dec 12, 2017
 #3
avatar+26396 
+1

Algrebra Help

 

b) Find all cube roots of 8i.

38i=383i=3233i=23i

 

z=a+bi=i|z|=02+12=1=1sin(φ)=b|z|=11=1φ=arcsin(1)+2kπφ=π2+2kπz=|z|eiφ=1ei(π2+2kπ)z=ei(π2+2kπ)

 

3i=(i)13=(ei(π2+2kπ))13=ei(π2+2kπ3)=ei(π6+23kπ)|k=(0,1,2)3i=ei(π6)|k=0=cos(π6)+isin(π6)|π6=30=cos(30)+isin(30)=32+i1223i=2(32+i12)23i=3+i3i=ei(π6+23π)|k=1=ei(5π6)=cos(5π6)+isin(5π6)|5π6=150=cos(150)+isin(150)=cos(30)+isin(30)=32+i1223i=2(32+i12)23i=3+i3i=ei(π6+223π)|k=2=ei(3π2)=cos(3π2)+isin(3π2)|3π2=270=cos(270)+isin(270)=0+isin(90)=i23i=2(i)23i=2i


All 3rd roots of 8i :
3+i3+i2i

 

laugh

 Dec 12, 2017
 #4
avatar
0

Simplify the following:
sqrt(4 i + 3)

 

Express 4 i + 3 as a square using 4 i + 3 = 4 + 4 i + i^2, then look to factor.
3 + 4 i = 4 + 4 i - 1 = 4 + 4 i + i^2 = (i + 2)^2:
 sqrt((i + 2)^2 ) 

                                                                                                                                                     

For all complex z with Re(z)>0, sqrt(z^2) = z.
Cancel exponents. sqrt((2 + i)^2) = i + 2:
2 + i    and    -2 - i

 Dec 12, 2017
 #5
avatar
0

Simplify the following:
(8 i)^(1/3)
 
Express 8 i as a cube using 8 i = 3 sqrt(3) + 9 i + 3 sqrt(3) i^2 + i^3, then look to factor.

8 i = 3 sqrt(3) + 9 i - 3 sqrt(3) - i = 3 sqrt(3) + 9 i + 3 sqrt(3) i^2 + i^3 = (sqrt(3))^3 + 3 (sqrt(3))^2 i + 3 sqrt(3) i^2 + i^3 = (sqrt(3) + i)^3:
 ((sqrt(3) + i)^3 )^(1/3)
 

For all complex z with -π/3 Cancel exponents. ((sqrt(3) + i)^3)^(1/3) = sqrt(3) + i:

sqrt(3) + i      -sqrt(3) + i        -2i

 Dec 12, 2017

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