How do you write the following expression in standard form #(1+i)/i-3/(4-i)#?

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Answer 1 · 2016-10-22T12:14:21

The standard form is #5/17-(20i)/17#

le #z=(1+i)/i-3/(4-i)#

Reducing to the same denominator

#z=((1+i)(4-i)-3i)/(i(4-i))#

Recall that #i^2=-1#

Then #z=(4+3i-i^2-3i)/(4i-i^2)=5/(1+4i)#

We simplify further by multiplying by the conjugate of the denominator #1-4i#

#z=(5*(1-4i))/((1+4i)(1-4i))=(5-20i)/(1-16i^2)=(5-20i)/17#

#z=5/17-(20i)/17#