Question

z= -16/1+i√3 find in polar form​

Answer 1

Answer:

z = 8 ( cos(2$\pi$/3)  + i sin (2$\pi$/3) ) or

z = 8 ( cos (120°) + i sin (120°) )

Step-by-step explanation:

Given, z = -16 / ( 1+i$\sqrt{3}$ )

First convert it into standard form,

multiply and divide with ( 1 - i$\sqrt{3}$ )

z = -16( 1 - i$\sqrt{3}$ ) / ( 1 + i$\sqrt{3}$ ) * ( 1 - i$\sqrt{3}$ )

z = -16 + 16$\sqrt{3}$ i / 1 - (i$\sqrt{3}$)^2

z = -16 + 16$\sqrt{3}$ i / 1 - ( 3$i^{2}$)

z = -16 + 16$\sqrt{3}$ i / 1 - (-3)

z = -16 + 16$\sqrt{3}$ i / 4

z = -4 + 4$\sqrt{3}$ i

compare it with z = a + ib

a = -4 ; b = 4$\sqrt{3}$

converting into polar form,

r = $\sqrt{a^{2} + b^{2} }$

r = $\sqrt{(-4)^{2}+ (4\sqrt{3}) ^{2} }$

r = $\sqrt{16 + ( 16*3 ) }$

r = $\sqrt{16 + 48}$

r = $\sqrt{64}$

r = 8

x = tan-1(b/a) + $\pi$

x = tan-1(4$\sqrt{3}$ / -4) + $\pi$

x = tan-1(-$\sqrt{3}$) + $\pi$

x = - $\pi$ / 3 + $\pi$

x = 2$\pi$ / 3

Now the polar form is ;

z = r(cos x + i sin x)

z = 8 ( cos(2$\pi$/3)  + i sin (2$\pi$/3) ) or

z = 8 ( cos (120°) + i sin (120°) )

I hope this answer is helpful :)