Mathematics: If z=-1-sq root 3i and z^6=a+bi then a = and b=

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If z=-1-sq root 3i and z^6=a+bi then a = and b=

jdodger5165

10.03.2022

a=64 and b=0

Step-by-step explanation:

$z=-1-\sqrt{3}i$

use $z=e^{ix}=cosx+isinx$

Now we find out x using given z

$z=-1-\sqrt{3}i$

Multiply and divide the right side by -2

$z=-2(\frac{1}{2} +\frac{\sqrt{3}}{2})$

We know cos(pi/3) = 1/2

sin(pi/3)= sqrt(3)/2

$z=-2(\frac{1}{2} +\frac{\sqrt{3}}{2})$

$z=-2(cos(\frac{\pi}{3})+sin(\frac{\pi}{3})$

compare the above equation with use $z=e^{ix}=cosx+isinx$

x= pi/3

so $z=-2e^{\frac{\pi}{3}i}$

$z^6=64(e^{\frac{\pi}{3}i})^6$

$z^6=64e^{\frac{6\pi}{3}i}$

$z^6=64e^{2\pi*i}$

Plug in 2pi in the z equation

$z=64e^{2\pi*i}=64(cos(2\pi)+isin(2\pi))$

So z= 64(1 + i(0))

z=64

So a= 64 and b=0

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