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1.Algebra Booster

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Complex Numbers 4.9<br />

i.e. z represents the point P and wz represents the point<br />

2p<br />

Q, where – POQ = .<br />

3<br />

7. Relation between z and w 2 z<br />

Let z = re iq .<br />

4<br />

2 Ê4pˆ Ê4pˆ<br />

i<br />

w = cos isin<br />

e p<br />

Á + =<br />

Ë<br />

˜ Á ˜<br />

3<br />

3 ¯ Ë 3 ¯<br />

Then<br />

( )<br />

2 i( ) ( )<br />

Now, 4 p<br />

i 4 p<br />

3 iq<br />

q +<br />

3<br />

w z = e ◊ re = re<br />

i.e. z represents the point P and w 2 z represents the point<br />

4p<br />

Q, where – POQ = .<br />

3<br />

X¢<br />

4p<br />

3<br />

O<br />

Y<br />

q<br />

P( z)<br />

X<br />

fi<br />

Ê2z3-2z1- z2+<br />

z1ˆ 3<br />

Á<br />

=-<br />

Ë 2( z2-<br />

z1) ˜<br />

¯ 4<br />

fi (2z 3<br />

– z 1<br />

– z 2<br />

) 2 = –3(z 2<br />

– z 1<br />

) 2<br />

fi<br />

2 2 2<br />

(4z3 + z1 + z2 - 4zz 1 3- 4z2z3+<br />

2 zz 1 2)<br />

2 2<br />

=- z2 + z1 - z1z2<br />

fi<br />

2 2 2<br />

4( z1 + z2 + z3 - z1z2- z2z3- z1z3) = 0<br />

fi 2 2 2<br />

( z1 + z2 + z3 - z1z2- z2z3- z1z3) = 0<br />

2<br />

3( 2 )<br />

which is the required condition.<br />

2 2 2<br />

Also, ( z1 + z2 + z3 - z1z2- z2z3- z1z3) = 0<br />

fi (z 1<br />

– z 2<br />

)(z 2<br />

– z 3<br />

) + (z 2<br />

– z 3<br />

)(z 3<br />

– z 1<br />

)<br />

+ (z 3<br />

– z 1<br />

)(z 1<br />

– z 2<br />

) = 0<br />

Dividing both the sides by<br />

(z 1<br />

– z 2<br />

)(z 2<br />

– z 3<br />

)(z 3<br />

– z 1<br />

), we get,<br />

fi<br />

fi<br />

1 1 1<br />

+ + = 0<br />

z3- z1 z1- z2 z2-<br />

z3<br />

1 1 1<br />

+ + = 0<br />

z - z z - z z - z<br />

1 2 2 3 3 1<br />

which is also a condition for an equilateral triangle.<br />

9. Condition of an Isosceles Right-angled Triangle<br />

Cz ( 3)<br />

w 2<br />

Q( z)<br />

8. Condition of an Equilateral Triangle<br />

Y¢<br />

Cz ( 3)<br />

60°<br />

Az ( 1)<br />

Bz ( 2)<br />

Here, DABC is an equilateral triangle.<br />

z3 z1 z3 z1<br />

i p<br />

Ê - ˆ -<br />

3<br />

Á = ¥ e<br />

Ë z2- z ˜<br />

1¯<br />

z2-<br />

z1<br />

fi<br />

z3 z1 | z3 z1|<br />

i p<br />

Ê - ˆ -<br />

e<br />

3<br />

Á = ¥<br />

Ë z2- z ˜<br />

1¯<br />

| z2-<br />

z1|<br />

fi<br />

z3 z1<br />

i p<br />

Ê - ˆ<br />

3<br />

Êpˆ Êpˆ<br />

Á = e = cosÁ ˜ + isin<br />

Á ˜<br />

Ë z2-<br />

z ˜ Ë<br />

1¯<br />

3¯ Ë 3¯<br />

fi<br />

Ê z3-<br />

z1ˆ 1 3<br />

Á = + i<br />

Ë z2-<br />

z ˜<br />

1¯<br />

2 2<br />

fi<br />

2<br />

2<br />

Ê z3-<br />

z1<br />

1ˆ<br />

Ê 3ˆ<br />

Á - =Ái<br />

˜<br />

Ë z - z 2˜ ¯ Ë 2 ¯<br />

2 1<br />

Bz ( 1)<br />

Bz ( 2)<br />

z3 z2 z3 z2<br />

i p<br />

Ê - ˆ - -<br />

We have e<br />

2<br />

Á = ¥<br />

Ë z1- z ˜<br />

2¯<br />

z1-<br />

z2<br />

z3 z2 | z3 z2|<br />

i p<br />

Ê - ˆ - -<br />

fi e<br />

2<br />

Á = ¥<br />

Ë z - z ˜<br />

¯ | z - z |<br />

1 2 3 2<br />

Ê z - z ˆ = = - i<br />

3 2 -i<br />

p<br />

fi e<br />

2<br />

Á<br />

Ë z1-<br />

z ˜<br />

2¯<br />

fi (z 3<br />

– z 2<br />

) = –i(z 1<br />

– z 2<br />

)<br />

fi (z 3<br />

– z 2<br />

) 2 = –(z 1<br />

– 2) 2<br />

fi<br />

fi<br />

2 2 2 2<br />

3 + 2 - 2 2 3= -( 1 + 2 -2 1 2)<br />

2 2 2<br />

1+ 2- 2 1 2= 2 1 3+ 2 2 3-2 1 2-2<br />

3<br />

z z z z z z z z<br />

z z zz zz z z zz z<br />

fi (z 1<br />

– z 2<br />

) 2 = 2(z 1<br />

– z 3<br />

)(z 3<br />

– z 2<br />

)<br />

which is the required condition.<br />

10. Condition of Circumcentre w.r.t. an Equilateral<br />

Triangle<br />

Az ( 1)<br />

Bz ( ) 3<br />

Oz ( 0)<br />

Cz ( 2)

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