Question

Find the modules and argument of the complex no. (1+i)(1+√3i)/1-i​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

(1+i)(1+√3i)/(1-i)= -√3+i

Step-by-step explanation:

(1+i)(1+√3i)/(1-i)

=1(1+√3i)+i(1+√3i)/(1-i)

=1+√3i+i+√3i^2/(1-i)

=1+√3i+i-√3/(1-i)

=(1-√3)+(1+√3)i/(1-i)

On Rationalization,

=(1-√3)+(1+√3)i × (1+i)/(1-i)×(1+i)

=1{(1-√3)+(1+√3)i}+i{(1-√3)+(1+√3)i}/1^2-i^2

=1-√3+i+√3i+i-√3i+i^2+√3i^2/1+1

=1-√3+i+√3i+i-√3i-1-√3/2

= -2√3+2i/2

=2(-√3+i)/2

= -√3+i