Question

13. Express -1+I in polar form. Hence show that(-1+i)¹⁶ is real and that 1/(-+i)⁶ is purely imaginary, giving the value for each

Answer 1

a = r cos θ

b = r sin θ

a+bi= r (cos θ+i sin θ)

here a=-1 and b=1

here r=$\sqrt{a^2+b^2}=\sqrt2$

θ=$\tan^-(b/a)=\tan^-(-1)=3\pi/4$

so in polar form it becomes $\sqrt2(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$

$((-1+i)^2)^8=(1-1-2i)^8=(-2)^8.(i)^8=2^8.1=2^8$

Hence it is purely real

Let's calculate the value of $\frac{1}{(-1+i)^6}$

$\frac{1}{(1+i^2-2i)^3}=\frac{1}{(-2i)^3}=\frac{1}{8i}=-\frac{i}{8}$

which is purely imaginary