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1.Algebra Booster

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Complex Numbers 4.87<br />

58. Let z = cos q + i sin q<br />

z cos q + isin<br />

q<br />

fi =<br />

2<br />

1 - z 1 - ( cos q + isin<br />

q) 2<br />

cos q + isin<br />

q<br />

=<br />

1 - (cos2q<br />

+ isin 2 q)<br />

cos q + isin<br />

q<br />

=<br />

(1 -cos 2 q) -isin 2q<br />

cos q + isin<br />

q<br />

=<br />

2<br />

2sin q - i 2sinqcosq<br />

(cos q + isin q)<br />

=<br />

–2 isin q(cos q + isin q)<br />

1<br />

=<br />

–2 isin<br />

q<br />

i<br />

=<br />

2sinq<br />

z<br />

Thus, the complex number lies on the imaginary<br />

axis.<br />

1 - z<br />

2<br />

59.<br />

Y<br />

X¢<br />

Qz ( 2)<br />

90°<br />

O<br />

Y¢<br />

1<br />

3<br />

Pz ( 1)<br />

1 1<br />

È Êpˆ Êpˆ˘<br />

Here, z1 = Í6+ 2 cos Á ,5+<br />

2sin<br />

Ë<br />

˜ Á ˜<br />

4¯ Ë 4¯˙<br />

Î<br />

˚<br />

= (7, 6)<br />

By rotation theorem, about the origin<br />

z2-0 z2-0<br />

ip/2 ip/2<br />

= e = e = i<br />

z1-0 z1-0<br />

z2= iz1= i(7 + 6 i) = - 6 + 7 i = (-6, 7)<br />

3 3<br />

60. Given zz + zz = 350<br />

fi<br />

fi<br />

2 2<br />

zz◊ z + zz ◊ z = 350<br />

2 2 2 2<br />

|| z ◊ z + || z ◊ z = 350<br />

2 2 2<br />

fi | z| ( z + z ) = 350<br />

fi (x 2 + y 2 )(2x 2 – 2y 2 ) = 350<br />

fi (x 2 + y 2 )(x 2 – y 2 ) = 175<br />

fi (x 2 + y 2 )(x 2 – y 2 ) = 25 ¥ 7<br />

fi (x 2 + y 2 ) = 25, (x 2 – y 2 ) = 7<br />

It is possible only when x = 4, y = 3.<br />

Thus, the area of the rectangle = (2 ¥ 4) ¥ (2 ¥ 3)<br />

= 48 sq unit.<br />

61. Given,<br />

15<br />

Â<br />

m=<br />

1<br />

2m-1<br />

Im( z )<br />

X<br />

3 5 29<br />

= Im( z) + Im( z ) + Im( z ) + … + Im( z )<br />

= sin q + sin 3q + sin 5 q + … + sin 29q<br />

= sin q + sin( q + 2 q) + sin( q + 2.2 q)<br />

+º+ sin( q + (15 – 1)2 q)<br />

Ê15.2q<br />

ˆ<br />

sin Á<br />

Ë<br />

˜<br />

2 ¯ Ê Ê2q<br />

ˆˆ<br />

= ¥ sin q + (15 -1)<br />

2q<br />

Á ˜<br />

Ê ˆ Á<br />

Ë Ë 2 ¯ ˜<br />

¯<br />

sin Á<br />

Ë<br />

˜<br />

2 ¯<br />

sin (15 q)<br />

= ¥ sin (15 q)<br />

sin ( q)<br />

2<br />

sin (15 q)<br />

=<br />

sin ( q)<br />

2<br />

sin (15 ¥ 2°)<br />

=<br />

sin (2°)<br />

2<br />

sin (30°)<br />

=<br />

sin (2°)<br />

1<br />

=<br />

4sin(2°)<br />

62. (A) z = 0 clealrly satisfies |z – i|z|| = |z + i|z||<br />

For z π 0, we can write<br />

z z<br />

- i = + i<br />

|| z || z<br />

fi<br />

z<br />

is equidistant from –i and i<br />

|| z<br />

fi<br />

z<br />

lies on the real axis<br />

|| z<br />

fi z lies on the real axis<br />

fi Im (z) = 0 and Im(z) £ 1.<br />

(B) Given |z + 4| + |z – 4| = 10 represents an ellipse<br />

whose foci are –4 and 4 and the length of the major<br />

axis = 10.<br />

Thus, 2ae = 8 and 2a = 10<br />

fi ae = 4 and a = 5<br />

fi e = 4/5.<br />

(C) Given |w| = 2<br />

fi w = 2e iq = 2(cos q + i sin q)<br />

Now,<br />

-iq<br />

1 iq<br />

e<br />

z = w- = 2e<br />

-<br />

w 2<br />

fi<br />

3 5<br />

x = cos q,<br />

y = sin q<br />

2 2<br />

fi<br />

2 2<br />

x y<br />

+ = 1<br />

9/4 25/4<br />

which represents an ellipse and its eccentricity<br />

=<br />

2<br />

b 9 4<br />

1- = 1- =<br />

2<br />

a 25 5<br />

(D) Given |w| = 1<br />

fi w = e iq

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