# 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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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:

`r=sqrt(80^2+18^2)=82`

`Theta=atan(18/80)=12.68`

`z=82(cos12.68-isin12.68)`

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.

`z_(1)=((4-16i)+sqrt((-256-128i+16)-(1040+160i)))/8=((4-16i)+4sqrt(-80-18i))/8`

`z_(2)=((4-16i)-sqrt((-256-128i+16)-(1040+160i)))/8=((4-16i)-4sqrt(-80-18i))/8`

`z_(1)=(1-4i+9+i)/2=5-1.5i`

`z_(2)=(1-4i-9-i)/2=-4-2.5i`

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Let

`sqrt(-80-18i)=a+ib`

squaring both side

`-80-18i=a^2-b^2+2abi`

coparing real and magnary parts both side

`a^2-b^2=-80`

`2ab=-18`

But

`(a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2`

`(a^2+b^2)=(-80)^2+(-18)^2`

`=6400+324=6724`

`a^2+b^2=+-82`

Now solve

`a^2-b^2=-80`

`a^2+b^2=+-82`

`If`

`a^2-b^2=-80`

`a^2+b^2=82`

`Then`

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

`if`

`a^2-b^2=-80`

`a^2+b^2=-82`

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

`Thus`

`sqrt(-80-18i)=+-1+-9i`

`or`

`sqrt(-80-18i)=+-9+-i`

`4z^2+(16i-4)z+(65+10i)=0`

`z=(-(16i-4)+-sqrt((16i-4)^2-16(65+10i)))/8`

`=(-(16i-4)+-sqrt(-256+16-128i-1040-160i))/8(`

`=(-(16i-4)+-sqrt(-1280-288i))/8`

`=(-(4i-1)+-sqrt(-80-18i))/2`

`Thus`

`z=(-4i+1+-1+-9i)/2`

`and`

`z=(-4i+1+-9+-i)/2`