(1-i√3)(√3+1)?

Question

how do you find (1-i√3)(√3+1)

in standard form?
and in trigonometric form?

Answer 1

That is SIMPLE!!! It's C. It's always C.

Answer 2

Standard form

(1-i√3)(√3+1) = √3+1 -3i-√3i

= (√3+1) -(3+√3)i

REAL part + Imaginary part

Divide it & multiply it by SQRT( REAL^2+ IMAG^2)

Then it can be shown that 8(2+√3){ Cos z- iSinz}

where Z is ARC SIN [(√3-1)/8]

Answer 3

(1-i√3)(√3+1)
= 1*√3 +1*1 -i√3*√3-i√3
= √3 +1 -9i -i√3
=√3 +1 -i(9+√3)

r^2 = (√3+1)^2 +(9+√3)^2
r^2 =3+2√3+1 +81+18√3 +3
r^2 = 88 +20√3
r =√(88 +20√3)
Tan theta = (9+√3)/(1+√3)
So theta = arctan[(9+√3)/(1+√3)] = 75.72 degrees
So in trigonometric form the answer is:
√(88 +20√3)(cos 75.72 +isin75.72)

Answer 4

= √3 + 1 - 3i - √3.i
= (√3 + 1) - (3 + √3).i

Answer 5

(1-i√3)(√3+1)
= (√3+1) - (3+√3)i
---------------
Standard form is: a+bi