Find the square roots of the complex number `-80-18i ` and solve the following quadratic equation. `4z^2+(16i-4)z+(65+10i) = 0`



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aruv's profile pic

Posted on (Answer #1)



squaring both side


coparing real and magnary parts both side








Now solve







`a=+-1 and b=+-9`




`a=+-9 and b=+-1`














flbyrne's profile pic

Posted on (Answer #2)

Two square roots of -80-18i

The nth-root theorem states that every complex number except zero has exactly n distinct nth roots. In this case, we are looking for 2 square roots.

Converting -80-18i to polar form:





Applying the n-root theorem:

`z^(1/2)=82^(1/2)(cos(12.68/2+(360k)/2)+isin(12.68/2+(360k)/2))`` <span class="AM"></span>`

= `9.06(cos(6.34+180k)+isin(6.34+180k))`

for k=0

First root = `9.06(cos(6.34)+isin(6.34)) = 9+i`


for k=1

Second root = `9.06(cos(6.34+180)+isin(6.34+180))`

= `9.06(cos(186.34)+isin(186.34))=-9-i`

The first root is (9+i) and the second root is (-9-i)


The roots of the equation `4z^2+(16i-4)+65+10i=0 ` can be found using the quadratic formula and then again, using the nth-root theorem.

Applying the quadratic formula:





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