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avatar+91 

Find \(\left|\left(1 + \sqrt{3}i\right)^4\right|\)

 Thanks :)

 Jun 19, 2019
 #1
avatar
+2

Simplify the following:
abs((sqrt(3) i + 1)^4)

(sqrt(3) i + 1)^4 = ((sqrt(3) i + 1)^2)^2:
abs(((sqrt(3) i + 1)^2)^2)

(sqrt(3) i + 1)^2 = 1 + i sqrt(3) + i sqrt(3) - 3 = 2 i sqrt(3) - 2:
abs((2 i sqrt(3) - 2)^2)

 

Factor 2 out of 2 i sqrt(3) - 2 giving 2 (i sqrt(3) - 1):
abs((2 (i sqrt(3) - 1))^2)

(2 (i sqrt(3) - 1))^2 = 2^2 (i sqrt(3) - 1)^2:
abs(2^2 (i sqrt(3) - 1)^2)
abs(4 (i sqrt(3) - 1)^2)

(i sqrt(3) - 1)^2 = 1 - i sqrt(3) - i sqrt(3) - 3 = -2 i sqrt(3) - 2:
abs(4 -2 i sqrt(3) - 2)

 

Factor 2 out of -2 i sqrt(3) - 2 giving 2 (-i sqrt(3) - 1):
abs(4×2 (-(i sqrt(3)) - 1))

Factor -1 from -(i sqrt(3)) - 1:
abs(4×2×-(i sqrt(3) + 1))
abs(-8 (i sqrt(3) + 1))

 

abs(-8 (1 + i sqrt(3))) = abs(-8) abs(1 + i sqrt(3)):
abs(-8) abs(1 + i sqrt(3))
Since -8<=0, then abs(-8) = 8:
8 abs(1 + i sqrt(3))

abs(1 + i sqrt(3)) = 2:


= 16

 Jun 19, 2019
 #2
avatar+91 
+1

Wow that's complicated... but thanks for helping me and taking the time to do it :)

 Jun 19, 2019
 #3
avatar+104688 
+1

Here's another way

 

( 1 + √3 i)^4  =

 

(1 + √3 i)^2  *  ( 1 + √3 i)^2  =

 

(1 + 2√3 i  + 3i^2)     * ( 1 + √3 i)^2   =

 

(2√3 i  + 1 - 3)   *   ( 1 + √3 i)^2   =

 

(2√3 i  - 2)    ....so we have

 

(2√3 i - 2) * ( 2√3 i - 2)  =

 

4*3 i^2 - 8√3 i + 4  =

 

-12 + 4 - 8√3 i  =

 

-8 - 8√3 i  =

 

-8 (1 + √3 i )   

 

And the absolute value of - 8  = 8

 

And the absolute value of     (1 + √3 i)   =  √ [ 1^2 + (√3)^2 ]   = √[1 + 3 ]  = √4  = 2

 

So

 

8 * 2   =

 

16

 

 

 

cool cool cool

 Jun 20, 2019
 #4
avatar+23273 
+2

Find

\( \left|\left(1 + \sqrt{3}i\right)^4\right|\)

 

\(\begin{array}{|rcll|} \hline \tan{60^\circ} &=& \dfrac{\sqrt{3}}{1} \\ |1 + \sqrt{3}i| &=& \sqrt{1^2 +\sqrt{3}^2 } \\ |1 + \sqrt{3}i| &=& 2 \\\\ 1 + \sqrt{3}i &=& 2\cdot e^{i\frac{60^\circ \pi}{180^\circ}} \\ (1 + \sqrt{3}i)^4 &=& 2^4\cdot e^{4i\frac{60^\circ \pi}{180^\circ}} \\ |(1 + \sqrt{3}i)^4| &=& 2^4 \\ \mathbf{|(1 + \sqrt{3}i)^4|} &=& \mathbf{16} \\ \hline \end{array}\)

 

laugh

 Jun 20, 2019

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