Re: (1+i)^16/(1-i)^15
---------------------------------------------------------------------------------------------------------------------------
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 ei 16(π/4) = (√2)16 ei (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 ei 15(-π/4) = (√2)15 ei (-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?
Re: (1+i)^16/(1-i)^15
---------------------------------------------------------------------------------------------------------------------------
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 ei 16(π/4) = (√2)16 ei (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 ei 15(-π/4) = (√2)15 ei (-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.