Find the modulus and argument of 1 + 2 i / 1 - 3 i ???

Hi!
Here is the answer to your question.
 
 
 
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1+2i / 1-3i * 1+3i / 1+3i (rationalise it)

=1-5i+6i2 / 12-(3i)2

=1-6+5i / 1+9

=-1+i / 2  =-1/2 + i/2

now let polar form be 1/2 + i/2 = r(cos x + i sin x )

so, r = root ((-1/2)2+(1/2)2)

=root( 1/4+1/4)  =1/root 2

so, -1/2 + i/2 = 1/root 2 (cos x + i sin x )

=1/root 2 cos x +1/root 2 i sin x

on comparing,

1/ root 2 cos x = 1/2  and  1/root 2 sin x = 1/2

so sin x = cos x = 1/root 2

therefore, theta=pie/4 or 45 degree

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hope it helps........  : )

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