Saved Bookmarks
| 1. |
Write the followingcomplex numberin polarform : (i) -3 sqrt(2) + 3 sqrt(2) i(ii) 1+i(iii) (1+7i)/(2-i)^2 |
|
Answer» Solution :(i) Let `z = - 3sqrt(2) + 3sqrt(2)i` Then, `|z| = SQRT(-3sqrt(2)^(2) + (3sqrt(2))^(2)) =6` Let `tan alpha |(Im(z))/(Re(z))| = 1 rArr alpha = (pi)/(4)` Since the pointreqresenting z lies in thesecond quadrant, the agrumentof z is givenby `theta = pi - alpha- ((pi)/(4)) = ((3pi)/(4))` So, the polarform of `z - 3sqrt(2)+3sqrt(2)`i si `z=|z|(cos theta + i SIN theta) = 6 (cos.(3pi)/(4) + i sin .(3pi)/(4))` (II)Let `z = 1+i.` Then `|z| = sqrt(1^(2)+1^(2)) = sqrt(2)`. Let `tan alpha = |(Im(z))/(Re(z))| ` Then, `tan alpha|(1)/(1)| =1 or alpha = (pi)/(4)` Sincethe point(1,1) representing z lies in the first quadrant,the argumentof z is given by `theta = aloha= pi//4`. So, thepolarform of `z =1 + i` is `z =|z| (cos theta + isin theta) = (cos.(pi)/(4) +isin.(pi)/(4))` (iii)Let`z = -1-i`. Then `|z| = sqrt((-1)^(2)+ (-1)^(2)) = sqrt(2)` Let `tan alpha = |(Im(z))/(Re(z))|` Then , `tan alpha = |(-1)/(-1)| = 1 or alpha = (pi)/(4)` Sincethe point(-1,-1) representing z lies in thethird quadrant, theargumentof z is given by `theta = -(pi -alpha) = - (pi-(pi)/(4))= (-3pi)/(4)` So, thepolarform of z = - 1- is `z = |z| (costheta + isin theta) = sqrt(2) {cos ((-3pi)/(4))+isin((-3pi)/(4)) }` (iv) Let `z = 1 - i` . Then `|z|= sqrt(1+(-1)^(2)) = sqrt(2)`. Let `tan alpha =|(Im(z))/(Re(z))|` Then, `tan alpha =|(-1)/(1)| =1 or alpha = (pi)/(4)` sincethe point (1,-1) lies in thefourthquadrant, theargument of z is givenby `theta = alpha = - pi//4`. So the polar formof z = 1 - i si `z =|z|(cos theta + isin theta) = sqrt(2) {cos((-pi)/(4)) +isin((-pi)/(4))} = sqrt(2) (cos.(pi)/(4) -isin.(pi)/(4))` (V) Let`z = (1+7i) //[(2-i)^(2)]`. Then `z = (1+7i)/(4-4i+^(2)) = (1+7i)/(3-4i) = ((1+7i)/(3-4))((3+4i)/(3+4i))=(-25+25i)/(25) = -1+i` `therefore |z| = sqrt((-1)^(2) + (1)^(2) ) = sqrt(2)` Let `alpha` be theacutue ANGLE given by `tan alpha =|(Im(z))/(Re(z))| = |(-1)/(1)| = 1` Then `alpha = pi//4`. Since the point (-1,1) represeting z lies in thesecond quadrant, we have`theta = arg(z) = pi - alpha = pi - pi//4 = 3pi//4` Hence, z in thepolar form is given by `z = sqrt(2) (cos. (3pi)/(4) + isin.(3pi)/(4))` |
|