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Chapter 9: Q. 53 (page 592)

In Problems 53鈥�60, find all the complex roots. Leave your answers in polar form with the argument in degrees.

53. The complex cube root of1+i.

Short Answer

Expert verified

The complex cube roots of1+iarerole="math" localid="1646812515277" 26(cos15+isin15),26(cos135+isin135),26(cos255+isin255).

Step by step solution

01

Step 1. Rewrite 1+i in the polar form.

The magnitude of 1+iis

12+12=2

which gives

1+i=2(12+i12)=2(cos45掳+isin45掳)

02

Step 2. Using the theorem of complex roots.

We get

zk=23[cos(453+360k3)+isin(453+360k3)];k=0,1,2=26[cos(15+120k)+isin(15+120k)]

03

Step 3. Substitute k=0,1,2 in zk=26(cos(15°+120°k)+isin(15°+120°k))

We get

z0=26(cos15+isin15)z1=26(cos(15+120)+isin(15+120))=26(cos135+isin135)z2=26(cos(15+240)+isin(15+240))=26(cos255+isin255)

So we have found the complex cube roots of1+i.

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