Mathematics: Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°). 42(cos 40° + i sin 40°) 42(cos 185° + i sin 185°) 13(cos 185° + i sin 185°) 42(cos 5800° + i sin 5800°)

Find the product of z1 and z2, where z1 = 7(cos 40° - id1217401577692, chintiffany4779

Ответ:

surfer89

15.03.2022

$\bf \qquad \textit{ product of two complex numbers}\\\\
r_1[cos(\alpha)+isin(\alpha)]\quad \cdot \quad r_2[cos(\beta)+isin(\beta)]
\\\\\\
~~~~~~~~~r_1\cdot r_2[cos(\alpha + \beta)+isin(\alpha + \beta)]\\\\
-------------------------------\\\\
\stackrel{z1}{7[cos(40^o)+i~sin(40^o)]}~~\cdot ~~\stackrel{z2}{6[cos(145^o)+i~sin(145^o)]}
\\\\\\
~~~~~~~~~7\cdot 6[cos(40^o+145^o)+i~sin(40^o+145^o)]
\\\\\\
~~~~~~~~~42[cos(185^o)+i~sin(185^o)]$

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Ответ:

yuhhhhh2801

03.03.2022

Mathematics

it is 11

Step-by-step explanation:

There is geometric theorem that states that when two points in triangles are the midpoints of the side lengths there is a line drawn between them

the line is half of the base

any acute triangle

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Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°). 42(cos 40° + i sin 40°) 42(cos 185° + i sin 185°) 13(cos 185° + i sin 185°) 42(cos 5800° + i sin 5800°)

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