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Mathematics 20 Online
snowflake0531:

healp

snowflake0531:

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snowflake0531:

it could only be 4 5 or 6 but 1. idk how to get those solutions 2. idk which one it would be

snowflake0531:

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surjithayer:

let \[z=r(\cos \theta +\iota \sin \theta)\]\[z^7=r^7(\cos \theta+\iota \sin \theta)^7\]\[z^7=r^7(\cos 7\theta+\iota \sin 7\theta)=128 \iota =2^7(0+\iota)\] \[r^7=2^7,r=2\] \[\cos 7\theta=0,\sin 7\theta=1\] \[as~\cos 7\theta=0,\sin 7\theta=1\] \[7\theta ~lies~on~y-axis .\] also 7 theta lies in first or second quadrant. \[\sin 7\theta=1=\sin 90=\sin (360n+90)\] \[7 \theta=90+360n\] where n is an integer. \[\theta=\frac{ 90 }{ 7 }+\frac{ 360 }{ 7 }n\] n=1 \[\theta=\frac{ 90 }{ 7 }+\frac{ 360 }{7 }\times 1=\frac{ 450 }{ 7 }=64\frac{ 2 }{ 7 }\] n=2 \[\theta=\frac{ 90 }{ 7}+\frac{ 360 }{ 7 }\times2=\frac{ 810 }{ 7 }=115\frac{ 5 }{ 7 }\] n=3 \[\theta=\frac{ 90 }{ 7 }+\frac{ 360 }{ 7 }\times3=\frac{ 90+1080 }{ 7 }=\frac{ 1170 }{ 7 }=167\frac{ 1 }{ 7 }\] n=4 \[\theta=\frac{ 90 }{ 7}+\frac{ 360\times4 }{ 7 }=\frac{ 90+1440 }{ 7}=\frac{ 1530 }{ 7 }=218\frac{ 4 }{ 7}\] \[n=5\] \[\theta=\frac{ 90 }{ 7 }+\frac{ 360\times5 }{ 7}=\frac{ 90+1800 }{ 7 }=\frac{ 1890 }{ 7 }=270\] n=6 \[\theta=\frac{ 90 }{ 7 }+\frac{ 360\times6 }{ 7 }=\frac{ 90+2160 }{ 7 }=\frac{ 2250 }{ 7 }=321\frac{ 3 }{ 7 }\] only \[\theta=270^\circ ~satisfies ~the~given~condition.\] so \[z=2(\cos 270+ \iota \sin 270)=2(0-1 \iota)=0+(-2)\iota \] \[\cos 270=\cos (180+90)=-\cos 90=0\] \[\sin 270=\sin (180+90)=-\sin 90=-1\]

snowflake0531:

tytytytyttytyttytyty o-0

Florisalreadytaken:

its kinda weird working it like that -- here is another way of doing it though so your answer should be \( \frac{\pi}{14}+\frac{2n\pi}{7} \) \( for \ \ n=0 \Rightarrow \frac{\pi }{14} \) \(for \ \ n=1 \Rightarrow \frac{5\pi }{14} \) \(for \ \ n=2 \Rightarrow \frac{9\pi }{14} \) \(for \ \ n=3 \Rightarrow \frac{13\pi }{14} \) \(for \ \ n=4 \Rightarrow \frac{17\pi }{14} \) \(for \ \ n=5 \Rightarrow \frac{3\pi }{2} \) \(for \ \ n=6 \Rightarrow \frac{25\pi }{14} \) now you know that the answer should be between whatever in degrees which you multiply by \( \frac{pi}{180} \) and you get \( \frac{5\pi }{4} \) and \(\frac{7\pi }{4} \) so n=6 is right, yes plug that in the main formula and you will get the same though, i like the way they did it in the above reply -- i really havent seen anyone do it like that

Florisalreadytaken:

i mean n=5* : )

surjithayer:

because the angle was in degrees so i solved in degrees. If it is in radians then i would have solved like this.

Florisalreadytaken:

easy guessable — though, it looks unnatural.

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