Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
+3
2525
3
avatar+26 

Find (1+i2)46. Any help is appreciated! Thanks!

 Aug 29, 2018
 #1
avatar+818 
+1

In (1+i2)46 , we can apply exponent rule, (ab)c=acbc . If we do it here, we will get: (1+i)46223 .

 Aug 30, 2018
 #2
avatar+9488 
+3

Find   (1+i2)46 .

 

Let's convert  1 + i  to polar form so we can use de Moivre's Theorem.

 

1+i=2(cosπ4+isinπ4)

 

Now lets raise both sides of that equation to the power of  46 .

 

(1+i)46=[2(cosπ4+isinπ4)]46 (1+i)46=(2)46(cosπ4+isinπ4)46

 

And divide both sides by  (2)46

 

(1+i)46(2)46=(cosπ4+isinπ4)46 (1+i2)46=(cosπ4+isinπ4)46

 

De Moivre's Theorem says....

 

(cos(x)+isin(x))n=cos(nx)+isin(nx)    So....

 

(cosπ4+isinπ4)46=cos(46π4)+isin(46π4) (1+i2)46=cos(46π4)+isin(46π4) (1+i2)46=cos(23π2)+isin(23π2)

 

Now let's find a reference angle that is coterminal with  23π2  by subtracting  2π  five times.

 

23π22π2π2π2π2π=23π210π=23π220π2=3π2

 

So....

 

(1+i2)46=cos(3π2)+isin(3π2) (1+i2)46=0+i(1) (1+i2)46=i

 Aug 30, 2018
 #3
avatar+26397 
+7

Find

(1+i2)46

\left(\frac{1+i}{\sqrt{2}}\right)^{46}.

 

(1+i2)46=(1+i)46(2)46=((1+i)2)232462|(1+i)2=2i=(2i)23223=223i23223=i23=(i2)11i|i2=1=(1)11i=i

 

laugh

 Aug 30, 2018

3 Online Users

avatar