# Multiplying & Dividing Complex Numbers

Discussion in 'Geometry and Trigonometry' started by nycmathguy, Feb 19, 2022.

1. ### nycmathguy

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Section 6.6

Can you do 48 and 50 in step by step fashion? Thank you very much.

nycmathguy, Feb 19, 2022
2. ### MathLover1

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48 .
3i(1-sqrt(2)*i)
=3i*1-3i*sqrt(2)*i
=3i-3sqrt(2)*i^2
=3i-3sqrt(2)*(-1)
=3i+3sqrt(2)
=3sqrt(2) + 3i

50 .

(1+sqrt(3)*i)/(6-3i).........rationalize

((6+3i)(1+sqrt(3)*i))/((6-3i)(6+3i))

((6+3i)(1+sqrt(3)*i))/(6^2-(3i)^2)

(6+6sqrt(3)*i+3i+3i*sqrt(3)*i)/(36-(9(-1))

((3 + 6sqrt(3))*i - 3sqrt(3) + 6)/(36+9)

(3(1 + 2sqrt(3)) i - sqrt(3) + 2)/45

((1 + 2sqrt(3)) i - sqrt(3) + 2)/15

((1 + 2sqrt(3)) i/15 - (sqrt(3) - 2)/15

-(1/15) (sqrt(3) - 2) +(1/15)(1 + 2sqrt(3))*i

MathLover1, Feb 19, 2022
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3. ### nycmathguy

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Thank you so much. We continue with Section 6.6 today.

nycmathguy, Feb 20, 2022
4. ### nycmathguy

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Each problem has three parts. You only did part (a). Can you do parts (b) & (c)?

nycmathguy, Feb 25, 2022
5. ### MathLover1

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48.

b. product in trigonometric form

3i(1-sqrt(2)*i)
use 3i and (1-sqrt(2)*i),

in trig. form they are

z=3(cos(pi/2)+i*sin(pi/2))..........1)

z=sqrt(3)(cos(-0.955316618)+i*sin(-0.955316618))..........2)

multiply 1) and 2)

3(cos(pi/2)+i*sin(pi/2))*sqrt(3)(cos(-0.955316618)+i*sin(-0.955316618))=4.24264+3* i

c.

product in standard form (I actually did it first time)

3i(1-i*sqrt(2)*i)=3i-3i*sqrt(2)*i=3sqrt(2) + 3 i=4.24264+ 3i

so, result is same

you can do 50 same way

MathLover1, Feb 26, 2022
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