P(n): (cos x + i sin x)n = cos(nx) + isin(nx) ⇢ (1)
For n = 1, we have
P(1) = (cos x + i sin x)1
P(1) = cos(1x) + i sin(1x)
P(1) = cos(x) + i sin(x)
That is true and thus, P(1) is true.
Assuming P(k) is true, i.e.
P(k) = (cos x + i sin x)k = cos(kx) + i sin(kx) ⇢ (2)
Now, we just have to prove that the P(k+1) is also true.
P(k+1) = (cos x + i sin x)k+1
⇒ P(k+1) = (cos x + i sin x)k (cos x + i sin x)
⇒ P(k+1) = (cos (kx) + i sin (kx)) (cos x + i sin x) [Using (i)]
⇒ P(k+1) = cos (kx) cos x − sin(kx) sinx + i (sin(kx) cosx + cos(kx) sinx)
⇒ P(k+1) = cos {(k + 1)x} + i sin {(k + 1)x}
⇒ P(k+1) = (cos x + i sin x)k+1 = cos {(k + 1)x} + i sin {(k + 1)x}
Thus, P(k+1) is also true, thus by the principal of mathematical induction, P(n) is true.
Hence the result is proved.
Given,
- r = √(12 + 12) = √2
- θ = π/4
Polar form of (1 + i) =(2\sqrt{2})^{4}-(\sqrt{2})^{4}i
According to De Moivre's Theorem
(cosθ + sinθ)n = cos(nθ) + i sin(nθ)
Thus,
(1 + i)5 = [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^5
⇒ (1 + i)5 = (\sqrt{2})^{5}[cos(\frac{5\pi}{4})+i\ sin(\frac{5\pi}{4})]\\ =(\sqrt{2})^{5}[cos(\pi+\frac{\pi}{4})+i\ sin(\pi+\frac{\pi}{4})]\\ =(\sqrt{2})^{5}[-cos(\frac{\pi}{4})-i\ sin(\frac{\pi}{4})]\\ =(\sqrt{2})^{5}[-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i]\\ =(\sqrt{2})^{4}-(\sqrt{2})^{4}i
⇒ (1 + i)5 = −4 − 4i
Here, r = \sqrt{(2^2+2^2)} = 2\sqrt{2}
, θ = π/4
The polar form of (2 + 2i) = [2\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]
According to De Moivre's Theorem: (cosθ + sinθ)n = cos(nθ) + i sin(nθ).
Thus, (2 + 2i)6 = [2\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^6
⇒ (2 + 2i)6 = (2\sqrt{2})^{6}[cos(\frac{6\pi}{4})+i\ sin(\frac{6\pi}{4})]\\ =(2\sqrt{2})^{6}[cos(\frac{3\pi}{2})+i\ sin(\frac{3\pi}{2})]\\ =(2\sqrt{2})^{6}[cos(\pi+\frac{\pi}{2})+i\ sin(\pi+\frac{\pi}{2})]\\ =(2\sqrt{2})^{6}[cos(\frac{\pi}{2})-i\ sin(\frac{\pi}{2})]\\ =(2\sqrt{2})^{6}[0-i]\\ =-(2\sqrt{2})^{6}i
⇒ (2 + 2i)6 = 512 (-i) = −512i
Here, r = \sqrt{(1^2+1^2)} = \sqrt{2}
, θ = π/4
The polar form of (1+i) = [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]
According to De Moivre's Theorem: (cosθ + sinθ)n = cos(nθ) + isin(nθ).
Thus, (1 + i)18 = [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{18}
⇒ (1 + i)18= (\sqrt{2})^{18}[cos(\frac{18\pi}{4})+i\ sin(\frac{18\pi}{4})]^{18}\\ =(\sqrt{2})^{18}[cos(\frac{9\pi}{2})+i\ sin(\frac{9\pi}{2})]\\ =(\sqrt{2})^{18}[cos(4\pi+\frac{\pi}{2})+i\ sin(4\pi+\frac{\pi}{2})]\\ =(\sqrt{2})^{18}[cos(\frac{\pi}{2})+i\ sin(\frac{\pi}{2})]\\ =(\sqrt{2})^{18}[0+i]\\ =(\sqrt{2})^{18}i
⇒ (1 + i)18 = 512i
Here, r = \sqrt{((-\sqrt{3})^2+3^2)} = 2\sqrt{3}
, θ = 2π/3
The polar form of (-√3 + 3i) = [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]
According to De Moivre's Theorem: (cosθ + sinθ)n = cos(nθ) + i sin(nθ).
Thus, (-√3 + 3i)31= [2\sqrt{3}(cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4}))]^{31} = (2\sqrt{3})^{31}[cos(\frac{31\pi}{4})+i\ sin(\frac{31\pi}{4})]\\ =(2\sqrt{3})^{31}[cos(8\pi-\frac{\pi}{4})+i\ sin(8\pi-\frac{\pi}{4})]\\ =(2\sqrt{3})^{31}[cos(\frac{\pi}{4})-i\ sin(\frac{\pi}{4})]\\ =(2\sqrt{3})^{31}[\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i]
r = \sqrt{(1^2+(-1)^2)} = \sqrt{2}
, θ = π/4
The polar form of (1 - i) = \sqrt{2}[cos(\frac{\pi}{4})+i \ sin(\frac{\pi}{4})]
According to De Moivre's Theorem: (cosθ + sinθ)n = cos(nθ) + i sin(nθ).
Thus, (1 - i)10 = [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{10}
= [\sqrt{2}(cos(\frac{π}{4})+ i sin(\frac{π}{4}))]^{10}\\ = (\sqrt2)^{10}[cos(\frac{10π}{4})+i\ sin(\frac{10π}{4})]\\ =(\sqrt2)^{10}[cos(\frac{5π}{2})+i\ sin(\frac{5π}{2})]\\ =(\sqrt2)^{10}[cos(2\pi+\frac{π}{2})+i\ sin(2\pi+\frac{π}{2})]\\ =(\sqrt2)^{10}[cos(\frac{π}{2})-i\ sin(\frac{π}{2})]\\
= 32 [0 + i(-1)]
= 32 (-i)
= -32i
Modulus of (1 + √3i)6 = \sqrt{1^2+(\sqrt{3})^2} = 2
Argument = tan-1(√3/1) = tan-1(√3) = π/3
⇒ Polar form = 2[cos(\frac{\pi}{3})+i\ sin(\frac{\pi}{3})]
Now, (1 + √3i)6 = [2(cos(\frac{\pi}{3})+i\ sin(\frac{\pi}{3}))]^6
As per DeMoivre's theorem, (cos x + isinx)n = cos(nx) + isin(nx).
⇒ [2(cos(\frac{\pi}{3})+i\ sin(\frac{\pi}{3}))]^6
= 2^6(cos(\frac{6\pi}{3})+i\ sin(\frac{6\pi}{3}))
= 64 (cos 2π + i sin 2π)
= 64(1 + 0)
= 64
Modulus = r = \sqrt{0^2+1^2}
= 1
Argument = tan-1[1/0] = π/2
Polar Form = r[cosθ + isinθ] = 1[cos(\frac{\pi}{2}) +i\ sin(\frac{\pi}{2})]
Now, i^{√3} = [cos(\frac{\pi}{2}) + i\ sin(\frac{\pi}{2})]^{\sqrt3}
As per DeMoivre's theorem: (cosθ + isinθ)n = cos(nθ) + isin(nθ).
⇒ [cos(\frac{\pi}{2}) + i\ sin(\frac{\pi}{2})]^{\sqrt3}
= [cos(\frac{\sqrt3\pi}{2}) + i\ sin(\frac{\sqrt3\pi}{2})].