Roots of an equation.

[Sonal7]

I rearranged the equation and made it to be z=-1 +- 2i. I don't understand the options given. I drew and argand diagram but still clueless.

 

[Romsek]

\(\displaystyle
(z+1)^4 = -16 = 16 e^{i\pi}\\

(z+1) = 2 e^{i k \pi/4},~k=0,1,2,3\\

z = 2e^{ik\pi/4} - 1,~k=0,1,2,3 \\

\text{Can you finish from here?}
\)

 

[Dr.Peterson]

View attachment 15338
I rearranged the equation and made it to be z=-1 +- 2i. I don't understand the options given. I drew and argand diagram but still clueless.

There are four fourth roots of -16; but -2i is not one of them. How did you get your result?

What have you learned about complex roots? It can be done without Romsek's suggestion (though that is easiest); but it depends on what you know.

 

[Sonal7]

I have I need to use the r(cos theta+i sin theta) then apply De Moivres theorem. Thanks for the reply.

 

[Sonal7]

As the mod is 16 r is 2. Romek is using eulers formula

 

[Dr.Peterson]

I have I need to use the r(cos theta+i sin theta) then apply De Moivres theorem. Thanks for the reply.

What Romek wrote is related to De Moivre's theorem.

Express -16 as r(cos theta+i sin theta) and then take the root.

Please show us what you have done with this method.

 

[pka]

View attachment 15338
I rearranged the equation and made it to be z=-1 +- 2i. I don't understand the options given. I drew and argand diagram but still clueless.

\(\displaystyle -16=16\exp(\pi i)\) write in polar form.
\(\displaystyle (z+1)=2\exp\left(\dfrac{\pi i}{4}\right)\) find the fourth root.
\(\displaystyle z+1=2\left(\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2}\right)\) convert from polar.
\(\displaystyle z+1=\sqrt 2+i\sqrt 2\) multply through by \(\displaystyle 2\)
\(\displaystyle z=\sqrt 2-1+i\sqrt 2\)

 

[Steven G]

View attachment 15338
I rearranged the equation and made it to be z=-1 +- 2i. I don't understand the options given. I drew and argand diagram but still clueless.

(-2i)^4 = [(-2i)(-2i)][(-2i)(-2i)]= [-4][-4] = 16 NOT -16

 

[Sonal7]

Thank you very much. I got it now. Its just applying De M's theorem