Simplify an expression with complex numbers [duplicate]

Question

How can I get a simplify expression of following when all a,b, .., f are real numbers?

FullSimplify[Abs[(a+I b)+(c+I d)(e+I f)]^2]

$\begingroup$ Use ComplexExpand. $\endgroup$
– Henrik Schumacher
Commented Jan 18, 2018 at 22:28 — Henrik Schumacher, Commented Jan 18, 2018 at 22:28
$\begingroup$ Thanks @HenrikSchumacher $\endgroup$
– Frey
Commented Jan 18, 2018 at 22:32 — Frey, Commented Jan 18, 2018 at 22:32
$\begingroup$ You're welcome! $\endgroup$
– Henrik Schumacher
Commented Jan 18, 2018 at 22:53 — Henrik Schumacher, Commented Jan 18, 2018 at 22:53

kglr · Accepted Answer · 2018-01-19 03:18:56Z

1

PowerExpand also works in this case:

FullSimplify[PowerExpand[Abs[(a + I b) + (c + I d) (e + I f)]^2], 
  Assumptions -> Element[{a, b, c, d, e, f}, Reals]]

$(b+d e+c f)^2+(a+c e-d f)^2$

answered Jan 19, 2018 at 2:55

kglr

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David G. Stork · Accepted Answer · 2018-01-19 00:49:47Z

0

FullSimplify@Assuming[{a, b, c, d, e, f} \[Element] Reals,
  ComplexExpand[Abs[(a + I b) + (c + I d) (e + I f)]^2]]

$a^2+2 a (c e-d f)+b^2+2 b (c f+d e)+\left(c^2+d^2\right) \left(e^2+f^2\right)$

answered Jan 19, 2018 at 0:49

David G. Stork

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$\begingroup$ By default, ComplexExpand will assume that all variables are real. Consequently, the simpler FullSimplify@ComplexExpand[Abs[(a + I b) + (c + I d) (e + I f)]^2] is equivalent to your proposed solution. Further, you will get the same result with Simplify in the place of FullSimplify $\endgroup$
– Bob Hanlon
Commented Jan 19, 2018 at 5:30

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2 Answers 2