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

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May 3, 2014

#1
+8

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 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?   May 4, 2014

#1
+8

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?   CPhill May 4, 2014
#2
+5

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∠(θ12), for division z1/z2=r1/r2∠(θ12) 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.

May 4, 2014