How do you evaluate #(2-5i)(p+q)i# when #p=2# and #q=5i#?

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Answer 1 · 2016-08-31T13:34:18

#(2-5i)(p+q)i=29i#

While multiplying complex numbers we should remember that #i^2=-1#.

As #p=2# and #q=5i#,

#(2-5i)(p+q)i#

= #(2-5i)(2+5i)i#

= #(2×2+2×5i-5i×2-5i×5i)×i#

= #(4+10i-10i-25×i^2)×i#

= #(4+cancel(10i)-cancel(10i)-25×(-1))×i#

= #(4+25)×i#

=#29i#