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### solve z^4=2(1+isqrt(3)) giving roots in form r(cos(theta)+isin(theta))

Sorry but it's a little hard to write the question out! I have the working on paper but I can't upload it.

Ok so first you need to multiply out the brackets to make it a little easeier to look at and obtain the form z^4 = 2+2sqrt(3)i.

From here you can get a modulus of 4 and an angle of 1/3pi rad using a^2+b^2=C^2 and tan^-1 of b/a respectivley.

Remember that a+bi = mode^argi where mod is the modulus (sqrt(a^2+b^2)) and arg is the argument (tan^-1(b/a)).

From here use the identity to get exponential form and calculate Z, Z^4 = 4e^(1/3pi*i) so Z=4^(1/4)exp((1/3pi+2kpi)*i/4) = sqrt(2)*(e^(1/12pi*i), e^(7/12pi*i),e^(13/12pi*i),e^(19/12pi*i) This all looks very difficult especially without the use of an argand diagram and with only text to illustrate the equations but remember it all breaks down quite easily and all we've done is convert it to polar.

Lastly use cos(theta)+isin(theta) identity with calculated thetas to give requested form.

1 year ago

Answered by Robert, a GCSE Maths tutor with MyTutor

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