Question

Express it in Polar form:-
1+iroot 3

Answer 1

Expressing in polar form,

z = 1 + i (3)^(1/2).

Polar form be,

z = r (cos theta + i sin theta)

1 + i (3)^(1/2) = r (cos theta + i sin theta)

The real parts are 1 and r cos theta.

The imaginary parts are i (3)^(1/2) and i r sin theta.

Equating the real parts with each other,

1 = r cos theta

Now squaring om both sides,

1 = r^2 sin^2 theta

Now, equating the imaginary parts with each other,

i (3)^(1/2) = i r sin theta

Squaring on both sides,

3 = r^2 sin^2 theta

Adding the LHS and RHS of both equations,

1 + 3 = r^2 (sin^2 theta + cos^2 theta)

4 = r^2

r = 2

Therefore, Modulus (r) = 2

Now findin the values of sin theta and cos theta,

1 = 2 cos theta

cos theta = 1/2

And,

(3)^(1/2) = 2 sin theta

sin theta = (3)^(1/2) / 2

We found that both sin theta and cos theta are positive.
So, the argument is in first quadrant,

theta = 60 degrees

[Since, sin inverse of (3)^(1/2) = cos inverse of 1/2 = 60 degree]

Argument = 60 × ('pi' / 180)

= 'pi' / 3

So the polar form of 1 + i (3)^(1/2) is r(cos theta + i sin theta)

= 2 [ cos ('pi'/3) + i sin ('pi'/3)]

-WonderGirl