How do you simplify (2+5i)/ (1-i)2+5i1−i?

1 Answer
sente
Jan 11, 2016

(2+5i)/(1-i)= -3/2 + 7/2i2+5i1−i=−32+72i

Explanation:

The conjugate of a complex number a+bia+bi is a-bia−bi. The product of a complex number and its conjugate is a real number. We will use this fact to produce a real number in the denominator by multiplying the numerator and denominator by the conjugate of the denominator.

(2+5i)/(1-i) = (2+5i)/(1-i)*(1+i)/(1+i)2+5i1−i=2+5i1−i⋅1+i1+i

= (2 + 5i + 2i - 5)/(1 + i - i + 1)=2+5i+2i−51+i−i+1

= (-3+7i)/2=−3+7i2

= -3/2 + 7/2i=−32+72i

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