(1+i)^16/(1-i)^15

Re: (1+i)^16/(1-i)^15

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Let's simplify these separately.  We need to remember how to write complex numbers in exponential form. This is given by:

And

Where      z= ( a + bi)       r= l z l = √(a² + b²)       tan^-1(b/a) = θ

Let's do the first one

z= (1 + i)   r = √(1² + 1²)  = √2     tan^-1(1/1) = π/4

So we have  (√2)¹⁶ e^{i 16(π/4)}   = (√2)¹⁶ e^{i (4π)}  =  (2)⁸ [ cos (4π) + i sin (4π)] = 128 [1 + 0i] = 128

The second one (1 - i) is similar except that tan^-1(-1/1) = -π/4

(√2)¹⁵ e^{i 15(-π/4)}   = (√2)¹⁵ e^{i (-15π/4)}  =  (√2)¹⁵ [ cos (-15π/4) + i sin (-15π/4)] =

64√2[ (1/√2) + (1/√2) i ] = 64 (1 + i)

Putting all this together, we have  (128) / [64 ( 1 + i)] = 2/(1 +i) and multiplying by the conjugate (1 -i) on top and bottom, we have  [2 (1- i)] /2 = (1 - i)

Whew!!.....that was a lot of work just for that, huh?

If you're multiplying, dividing or raising complex numbers to a power, it's best to work with polar form.

If z₁=r₁∠θ₁ and z₂=r₂∠θ₂, then the rules are, for multiplication, z₁z₂=r₁r₂∠(θ₁+θ₂), for division z₁/z₂=r₁/r₂∠(θ₁-θ₂) and raising to a power, zⁿ=rⁿ∠nθ.

So, (1+i)¹⁶=(√2∠(π/4))¹⁶=(√2)¹⁶∠(4π),

(1-i)¹⁵=(√2∠(-π/4))¹⁵=(√2)¹⁵∠(-15π/4), 

and ∴ (1+i)¹⁶/(1-i)¹⁵=((√2)¹⁶∠(4π))/((√2)¹⁵∠(-15π/4))=√2∠(4π+15π/4)=√2∠(-π/4),

and switching back to algebraic form, the result is 1-i.