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bleakarcher

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z^4+16i=0
=>z^4=-16i=16cis(-pi/2)
Let z=rcis(θ).
Now, [rcis(θ)]^4=r^4cis(4θ)=16cis[(-pi/2)+2kpi] for k=0,1,2,3,...
Hence, r=2 and θ=(-pi/8)+(kpi/2)
:.z=2cis(-pi/8), 2cis(3pi/8), 2cis(7pi/8), 2cis(-5pi/8)
 
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SpiralFlex

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I recall correctly this question is from Harry's. It has the incorrect answer at the back. These roots do not appear as conjugates.
 

Sanical

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You can do it via complex roots of unity except having it with 16i.

Otherwise, you can factorise it via difference of two squares. EDIT: Oops, you can't do that unless it's a real number =/.

In any case, you ask a lot of questions lol. You should get a tutor to answer all these.
 
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lolcakes52

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I recall correctly this question is from Harry's. It has the incorrect answer at the back. These roots do not appear as conjugates.
The roots are not conjugates though? The coefficient of z^0 is imaginary.
 
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All complex roots occur in conjugates. It's part of the fundamental theorem of algae.
 

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