Asked by David
                (2+3i)^6
Using Demoirves Theorem.
I tried solving it out, but somehow it didn't work for me.
(2+3i)^6
r=|z|=square root of 2^2 + 3^2 = square root of 13
z = square root of 13 *
(cos____ + isin_______ )
At this point I am stuck, so can someone help me?
            
        Using Demoirves Theorem.
I tried solving it out, but somehow it didn't work for me.
(2+3i)^6
r=|z|=square root of 2^2 + 3^2 = square root of 13
z = square root of 13 *
(cos____ + isin_______ )
At this point I am stuck, so can someone help me?
Answers
                    Answered by
            Reiny
            
    let z = 2 + 3i
for the angle, tanØ = 3/2
Ø = .98279 radians
so z = √13(cos .98279 + isin .98279)
then z^6
= √3^6(cos 6(.98279) + isin 6(.98279) )
= 27(cos 5.89676 + i sin 5.89676)
= appr 25 - 10.18 i
or in short form:
27 cis 5.89676
If you need the answer is terms of degrees, set your calculator to degrees and repeat my steps
    
for the angle, tanØ = 3/2
Ø = .98279 radians
so z = √13(cos .98279 + isin .98279)
then z^6
= √3^6(cos 6(.98279) + isin 6(.98279) )
= 27(cos 5.89676 + i sin 5.89676)
= appr 25 - 10.18 i
or in short form:
27 cis 5.89676
If you need the answer is terms of degrees, set your calculator to degrees and repeat my steps
                    Answered by
            David
            
    Thank you so much Reiny! I finally understand it!
    
                    Answered by
            Steve
            
    somehow the √13 became √3, but I'm sure David can fix that, eh?
    
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