How do you multiply (9-4i)(2-i) (94i)(2i) in trigonometric form?

1 Answer
Apr 25, 2018

color(red)(z=22.0*[cos(-50.5^@)+i*sin(-50.5^@)]z=22.0[cos(50.5)+isin(50.5)]

Explanation:

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Multiply the two complex numbers (9-4i)*(2-i)(94i)(2i) using FOILFOIL Method.

(9-4i)*(2-i)(94i)(2i)

rArr18-9i-8i+4i^2189i8i+4i2

rArr 18-17i+4*(-1)1817i+4(1)

Note: i^2=(-1)i2=(1)

rArr 18-17i-41817i4

rArr14-17i1417i

Hence, color(red)((9-4i)*(2-i)=14-17i(94i)(2i)=1417i Intermediate Result 1

Convert this intermediate result to Trigonometric Form.

For a complex number in standard form: color(blue)(z=a+biz=a+bi,

r=sqrt(a^2+b^2r=a2+b2

theta =tan^(-1)(b/a)θ=tan1(ba)

cos (theta) = a/rcos(θ)=ar

rArr a=r*cos(theta)a=rcos(θ)

sin(theta)=b/rsin(θ)=br

rArr b=r*sin(theta)b=rsin(θ)

:. z=r*[cos(theta)+i*sin(theta)]

Using Intermediate Result 1,

r=sqrt(a^2+b^2

rArr sqrt(14^2+(-17)^2

rArr sqrt(196+289

rArr sqrt(485

:. r ~~ 22.02272

theta = tan^(-1)(b/a)

rArr tan^(-1)((-17)/14)

theta=tan^(-1)(-1.21429)

theta ~~-50.52763939^@

theta ~~ -50.5^@

Hence,

color(red)(z=22.0*[cos(-50.5^@)+i*sin(-50.5^@)]

Hope it helps.