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Chapter 9: Q. 50 (page 592)

In Problems 41鈥�52, write each expression in the standard form a+bi.

50.(3-i)6

Short Answer

Expert verified

(3-i)6=-64

Step by step solution

01

Step 1. Rewrite (3-i) in the form r(cos∅+isin∅).

The magnitude of (3-i)is

(3)2+(-1)2=3+1=4=2

which gives

(3-i)=2(32-i12)=2(cos330掳-isin330掳)

02

Step 2. Using De Moivre's Theorem.

If a complex number is z=r(cos鈭�+isin鈭�)then zn=rn[cos(n鈭�)+isin(n鈭�)].

This gives

role="math" localid="1646805734516" (3-i)6=[2(cos330掳-isin330掳)]6={(2)6[cos(6脳330掳)-isin(6脳330掳)]}=64(cos1980掳-isin1980掳)=64(cos180掳-isin180掳)=64(-1+0i)=-64

So(3-i)6=-64

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