Given 𝑧1 =3(cos28°+𝑖sin28°) and 𝑧2 =9(cos14°+𝑖sin14°), find each of the following. Leave your answers in polar form.

Question

Given 𝑧1 =3(cos28°+𝑖sin28°) and 𝑧2 =9(cos14°+𝑖sin14°), find each of the following. Leave your answers in polar form.

a) 𝑧1𝑧2

b) 𝑧1/𝑧2

Hello! I am so confused about this problem please help me.

Answer 1

Hi,

𝑧1 =3(cos28°+𝑖sin28°) and 𝑧2 =9(cos14°+𝑖sin14°)

𝑧1 * 𝑧2 = 3(cos28°+𝑖sin28°) 9(cos14°+𝑖sin14°) = 27 [cos28cos14 + isin14cos28 +isin28cos14-sin28sin14]

as you know: cosa cosb - sina sinb = cos(a+b) AND sina cosb + cosa sinb = sin(a+b)

𝑧1 * 𝑧2 = 27 [ cos(28+14) + i sin( 28+14] = 27 [cos 42 + isin42]

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𝑧1/𝑧2 = 3(cos28°+𝑖sin28°) / 9(cos14°+𝑖sin14°)

As you know in division must be time of conjugant of denominator IT MEANS TIMES (cos14°- 𝑖sin14°), ]

1/3 [ (cos28°+𝑖sin28°) / (cos14°+𝑖sin14°)* [(cos14°-𝑖sin14°) /(cos14°- 𝑖sin14°))]

after multiplication nominators and denominators you find the answer be carful after you did time denominators you will have only 1 and final answer is:

𝑧1/𝑧2 = [cos 14 - i sin 14] /3 = 1/3 [cos14 - sin 14]

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I hope it is useful,

Minoo

Answer 2

z1=3*e^i(28π/180)

z2=9*e^i(14π/180)

z1*z2=27*e^i(42π/180),,,same as 27*(cos42+isin42) in degrees

z1/z2=(1/3)*e^i(14π/180),,,same as (1/3)*(cos14+isin14) in degrees