How do you divide #(4+i)/(2-5i)#?

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Answer 1 · 2016-08-28T16:32:53

#(4+5i)/(2-5i)=-17/29+30/29i#

We should always remember two things while dividing a complex number by another.

One - multiply numerator and denominator each by complex conjugate of the Divisor.

Two - #i^2=-1#.

Hence, #(4+5i)/(2-5i)#

= #((4+5i)(2+5i))/((2-5i)(2+5i)#

= #(4×2+4×5i+5i×2+25i^2)/(2×2+2×5i-5i×2-25i^2)#

= #(8+20i+10i-25)/(4+10i-10i-25×(-1))#

= #(-17+30i)/(4+25)#

= #(-17+30i)/29#

= #-17/29+30/29i#