Mathematics 12 TRICKS FOR JEE MAINS L FOR COMPLEX VARIABLE PROBLEMS

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12 TRICKS FOR JEE MAINS L FOR COMPLEX VARIABLE PROBLEMS

First Set of 5 Problems

`(|z-z_1|)/(|z- z_2|) = 1`

`| z -z_1| = | z - z_2|`

• z moves in such a way that it is always equidistant with `z_1 ` & `z_2`

• z lies on perpendicular bisector of line joining AB

• z will lie on CD

if `|z - i| = |z + i | `

`=>` z will lie on CD i.e. on x -axis

if `| z - 2 | = | z -4| `

`| z -2 | > | z -4 |`

`=>` Right side of x =3

`| z -2 | < | z -4 | => ` left side of x =3

` (| z - z _1 |)/( | z - z_2 |) = 0 => | z -z_1 | = 0`

` (| z - z _1 |)/( | z - z_2 |) = k ` say `3/5`

put `z = x +iy` and solve

`25 [ (x -x_1 )^2 + (y -y_1)^2 ] = 9 [ ( x -x_2 )^2 + (y -y_2 )^2 ]`

`=> 16x^2 + 16^2 + ax + by + c = 0`

as coefficient of `x^2 = ` coefficient of `y^2` and there is no xy terms

`:.` ` (| z - z _1 |)/( | z - z_2 |) = k` will represent a circle for all value of `k != 0, 1`

Q 2670091816

Match the column if `|(z_1z-z_2)/(z_1z+z_2)|=k, (z,z_2 ne 0)` then locus of `z` is

Column I		Column II
(A)	`k = 1`	(P)	line segment
(B)	`k = 0`	(Q)	Parabola
(C)	`k notin (0,1)`	(R)	point
(D)	`k=2`	(S)	straight line
		(T)	circle

(A)

`A-> q, quad B-> r, quad C->s, quad D->t`

(B)

`A-> t, quad B-> r, quad C->t, quad D->s`

(C)

`A-> s, quad B-> r, quad C->t, quad D->t`

(D)

`A-> t, quad B-> r, quad C->s, quad D->t`

Solution:

`|(z_1z-z_2)/(z_1z+z_2) |=k => | (z-z_2/z_1)/(z+z_2/z_1) |=k`

Clearly if `k ne 0, 1`, then `z` would lie on a circle

If `k = 1, z` would lie on a perpendicular bisector of the line segment joining `z_2/z_1` and `(-z_2)/z_1`

If `k = 0, z` represents a point.

Correct Answer is `=>` (C) `A-> s, quad B-> r, quad C->t, quad D->t`

Q 1187323287

P is a set consisting of those points which lie on the portion of a curve given by `arg((3z−6−3i)/(2z−8−6i))=π/4`, Q is a set consisting of those points which lie on the curve given by ` | z - 3 + i | = 3`, then

(This question may have multiple correct answers)

(A) `4+4/sqrt5+i(1−2/sqrt5)` is the complex number associated to both the sets

(B) Complex number z in the set Q for which `| | z - ( 2 + i ) | - | z - ( 4 + 3i ) | |` takes maximum value is `– i`

(D) Complex number z in the set Q or which `| | z - ( 2 + i ) | - | z - ( 4 + 3i ) | |` takes minimum values are `(2+5/2i)` and `(3+5/2i)`

Solution:

Equation of `C_2`

`( x - 3 )^2 + ( y + 1)^2 = 9`

`x^2 + y^2 - 6x + 2y + 1 = 0` ..........(i)

For `C_1 , sin 45^0=sqrt(((3−2)^2+(2−1)^2)/r)`

`r=sqrt2sqrt2=2`

Centre (4,1)

Equation of curve `C_1 ( x - 4 )^2 + ( y - 1 )^2 = 4`

`⇒ x^2 + y^2 - 8x - 2y + 13 = 0`

Solving `C_1` and `C_2`

`2x + 4y - 12 = 0`

`x = 6 - 2y.`

Put value of x in (i) and after solving

`y=1−2/sqrt5x=4+4/sqrt5 `

Now, `| | z - ( 2 + i ) | - | z - ( 4 + 3i ) | |` will take maximum value

if `z, 2 + i ,4 + 3i ` are collinear.

Equation of line joining `( 2 + i )` and `( 4 + 3i )` is `y - 1 = 1 ( x - 2 )`

`⇒ x - y - 1 = 0`

Intersection of `x - y - 1 = 0` with `C_2 ( x - 3 )^2 + ( x - 1 + 1 )^2 = 9`

`⇒ x = 3, 0`

`x = 0 ⇒ y = - 1`

Also, the point will lie on `C_2` and on the line which is perpendicular bisector of line joining `(2,1)` and `(4,3)` then the value will be zero.

One of the possible value is `x < 3, y = 2`

Correct Answer is `=>` (A)

Q 1679734616

For any integer `k`, let `\alpha _{ k }=\cos \frac { k\pi }{ 7 } +i\sin\frac { k\pi }{ 7 } `, where `i=\sqrt { -1 }`. The value of the expression `\frac { \sum _{ k=1 }^{ 12 } | \alpha _{ k+1 } - \alpha _{ k }| }{ \sum _{ k=1 }^{ 3 } | \alpha _{ 4k-1 } - \alpha _{ 4k-2 }| }` is
JEE ADVANCED 2015

Solution:

`| \alpha_k - \alpha_{k+1} |` represents the side length of a 14 sided polygon (tetradecagon) inscribed in a unit circle.

Similarly, `|\alpha_{4k-1} - \alpha_{4k-2} | ` represents the side length of a 14 sided polygon inscribed in a unit circle.

Let l be the length of the side.

Hence, the required ratio `= frac{ 12 l}{3 l} =4`

Correct Answer is `=>` 4

Q 1107112988

If `z` is a complex number with modulus `1`, then in the equation `((1+ia)/(1−ia))^4=z`, unknown ‘a’ has

(A)

All roots real and distinct

(B)

Three roots real and one roots imaginary

(C)

One roots real and three roots imaginary

(D)

Two roots real and two roots imaginary

Solution:

we have `((1+ia)/(1−ia))^4=z` and `| z | = 1.`

Let `((1+ia)/(1−ia))^4=cos⁡θ+isin⁡θ`

`⇒(1+ia)/(1−ia)=(cos⁡θ+isin⁡θ)^(1/4) `

`=(cos(2kπ+θ)+isin(2kπ+θ))^(1/4), kϵZ`

`=cos((2kπ+θ)/4) + isin((2kπ+θ)/4),k=0,1,2,3`

`⇒(1+ia)/(1−ia)=cos⁡α+isin⁡α` where `α=(2kπ+θ)/4 `

`⇒2/(2ia)=(cos⁡α+isin⁡α+1)/(cos⁡α+isin⁡α−1)=(2cos^2(α/2)+2isin(α/2)cos⁡ α/2)/(−2sin^2 (α/2)+2isin (α/2)cos⁡ α/2) `

`=(2cos(α/2)(cos(⁡α/2)+isin(α/2)))/(2isin(α/2)(cos⁡(α/2)+isin(α/2)))=1/icot⁡(α/2)`

`∴ia=itan⁡(α/2)⇒a=tan⁡(α/2) `

`∴α=tan((⁡2kπ+θ)/4),k=0,1,2,3 `

`≡tan⁡(θ/8),tan(π/4+θ/8),tan(π/2+θ/8),tan((3π)/4+θ/8) `

Correct Answer is `=>` (A) All roots real and distinct

Q 1177012886

If the imaginary part of the expression `(z−1)/e^(θt)+(e^(θt))/(z−1)` be zero, then the locus of z can be

(A)

A parabola

(B)

A straight line parallel to x - axis

(C)

None of these

(D)

A circle of radius 1

Solution:

Let `u=(z−1)/e^(θi)⇒(e^(θi))/(z−1)=1/u`, Now

`(u+1/u)−(baru+1/(baru))=0 `

`⇒(u−baru)(1−1/(ubaru))=0`

If u is not purely real, then

`ubaru=1⇒|(z−1)/e^(θi)|=1⇒|z−1|=1`

Correct Answer is `=>` (D) A circle of radius 1

Second Set of 5 Problems

`|z| = 1 => | z|^2 = 1 => z bar z =1 `

`=> z = 1/(bar z)`

`|z| = 1 => z = e ^(i theta )= cos theta + i sin theta`

`| z | = 1 => sqrt (a^2 + b^2) = 1 => a^2 + b^2 =1`

where `z = a + i b`

Additionally note that

` cos theta + i sin theta = 1/ ( cos theta - i sin theta )`

Q 1416156979

If `| z | = max {| z - 1|, | z + 1|}`, then

(A)

`|z + bar z | = 1`

(B)

`|z + bar z | = - 1`

(C)

`|1 + bar z | = 1`

(D)

None of theses

Solution:

If `|z | =| z - 1| => | z |^2 =| z - 1|^2`

`=> z bar z = (z -1)(bar z -1)`

`=> z. + bar z = 1`

and if `| z|= |z + 1|=> | z|^2 = | z + 1|^2`

`=> z bar z = (z + 1) (bar z + 1)`

`=> z+bar z= - 1`

`:. z+ bar z=±1`

`:. |z + bar z| = 1`.

Correct Answer is `=>` (A) `|z + bar z | = 1`

Q 1426167071

The equation ` | z + i | -| z - i | = k` (where `i = R`)
represents a hyperbola if

(A)

`- 2 < k < 2`

(B)

`- 1 < k < 1`

(C)

` k < 2`

(D)

`- 2 < k `

Solution:

For hyperbola `| k | <| i- (- i)|`

`=> |k| < 2`

`=> -2 < k < 2`

Correct Answer is `=>` (A) `- 2 < k < 2`

Q 1486201177

Let ` z_1 = 6 + i` and `z^2 = 4- 3i` (where `i = R`). Let `z ` be a complex number such that
`arg ((z-z_1)/(z_2-z))=pi/2`, the n `z` sati.sfies

(A)

`|z-(5-i)|=pi/2`

(B)

`|z-(5-i)|=sqrt5`

(C)

`|z-(5+i)|=5`

(D)

`|z-(5+i)|=sqrt 5`

Solution:

`arg ((z-z_1)/(z_2-z))=pi/2`

`:. z_1, z_2, z_3` lie on a circle.

`=> z_1` and `z_2` are the end points of diameter

`:.` centre `(z_0)= (z_1 + z_2)/2`

`:. z_0 = 5-i`

and radius `=| z_1 - z_0 |`

`=|1 + 2i|`

`r=sqrt 5`.

`:.` Equation circle is `| z - z_0 | = r`

or `| z - (5 - i) | = sqrt 5`

Correct Answer is `=>` (D) `|z-(5+i)|=sqrt 5`

Q 1435791662

If `|z- 3 + 2i | le 4`, (where `i =sqrt (-1`) then the difference of greatest and least values of `| z |` is

(A)

`2 sqrt 11`

(B)

`3 sqrt 11`

(C)

`2 sqrt 13`

(D)

`3 sqrt 13`

Solution:

`| z - 3 + 2i | le 4`

`=> | z - (3 - 2i) le 4`

Least value of `| z | =| z - 0 |`

`=OP`

`=CP -CO`

`=4- sqrt(13)`

and greatest value of `| z |`

`=|z-0|`

`=OQ`

`=OC +CQ`

`=sqrt 13+4`

`:.` Difference `= (4 + sqrt 3)- (4-sqrt 13) = 2 sqrt 13`

Correct Answer is `=>` (C) `2 sqrt 13`

Q 2615534460

Separate into real and imaginary parts of log (- 5)

Solution:

`log (- 5) = log (- 5) + 2n pi i`

Now `log (- 5) = log (- 5 + i.0)`

`= log (rcos theta + i r sin theta)`

where `rcos theta = -5, r sin theta = 0`.

`:. r = sqrt ([ (-5) ^2 + 0^2]) = 5 , r sin theta = 0`

`:. log (- 5) = log 5 (cos pi + i sin pi)`

`= log (5e^(i pi ))`

`= log 5 + i pi`

Hence `log (- 5) = log 5 + i pi + 2n pi i`

`= log 5 + i (2n + 1) pi`

Third Set of 5 Problems

Some key Method to solve complex variable problems

(i) Graphically

(ii) `z = r e^(itheta)`

(ii) ` z = r ( cos theta + i sin theta )`

(vi) ` z = x +i y`

(v) Triangle in equality

`=> | | z_1 | - | z_2 | | <= | z_1 pm z_2 | <= | z_1 | + z_2 | `

Q 2371880726

If `a, b, c` are integers and not all are equal, then the least value of `| a + b omega + c omega^2|`

is (where `omega` and `omega^2` are non-real cube roots of unity)
BITSAT Mock

(A)

`0`

(B)

`sqrt (3)/2`

(C)

`1`

(D)

`1/sqrt (3)`

Solution:

`| a + b omega + c omega^2 |^2`

`= (a + b omega + c omega^2 )(a + b omega^2 + c omega)`

`= a^2 + b^2 + c^2 - ab - ac - bc`

`= 1/2 [ (a - b)^2 + (b - c)^2 + (c - a)^2]`

` ge 1`

as `a, b, c` are not all equal

Correct Answer is `=>` (C) `1`

Q 2651401324

If `|z| =1` and `Arg (z + z omega) = pi/2` , then one of the arguments of `z` is `pi/6` (where `omega` is a non-real cube root of unity) what is is the other value of argument of `z`

(A)

True

(B)

False

Solution:

`z+ z omega =e^(i theta)( 1+ omega)= e^(i theta)(1+ (-1 pm sqrt 3i)/2)=e^(i theta) ((1 pm sqrt 3 i)/2)`

`=e^(i theta) e ^(pm i pi/3)= e^((theta pm pi/3))`

`:. theta pm pi/3 =pi/2 `

`:. theta=pi/2 pm pi/3 =pi/6 , (5 pi)/6`

Correct Answer is `=>` (A)

Q 1137480382

If `4ac > b^2` and `a + c > b` for real numbers a, b and c, then which of the following is true ?

(This question may have multiple correct answers)

(A) `a + b + c >0`

(B) `a > 0`

(D) `c > 0`

Solution:

Let `f (x) = ax^2 + bx + c`

its discriminant `D = b^2 - 4ac < 0`

and, `f (- 1) = a - b + c > 0`

So, `f (x) > 0` for all x ∈ R

`f (0) > 0`

`⇒ a + b+ c > 0`

`f (- 2) > 0`

`⇒ 4a - 2b + c > 0`

The correct answer is: `a > 0, c > 0, a + b + c >0, 4a +c > 2b`

Correct Answer is `=>` (A)

Q 1117323280

If `z_1,z_2` and `z_3` be the vertices of ∆ABC and `z_0` be the circum centre, then `((z_0−z_1)/(z_0−z_3))(sin⁡2A)/(sin⁡2C)+((z_0−z_2)/(z_0−z_3))(sin⁡2B)/(sin⁡2C)` is equal to

(A)

`0`

(B)

`1`

(C)

`2i`

(D)

`-1`

Solution:

Taking rotation at O

`(z_1−z_0)/(z_3−z_0)=(|z_1−z_0|)/(|z_3−z_0|)e^(i2B)`

`⇒(z_0−z_1)/(z_0−z_3)=cos⁡2B+isin⁡2B` ............(i)

Also `(z_2−z_0)/(z_3−z_0)=e^(−i2A)`

`⇒(z_0−z_2)/(z_0−z_3)=cos⁡2A−isin⁡2A` ...........(ii)

Now `((z_0−z_1)/(z_0−z_3))(sin⁡2A)/(sin⁡2C)+((z_0−z_2)/(z_0−z_3))(sin⁡2B)/(sin⁡2C)`

`=((sin⁡2Acos2B+isin⁡2Asin⁡2B+sin⁡2Bcos⁡2A−isin⁡2Asin⁡2B))/(sin⁡2C)`

`=(sin(2A+2B))/(sin⁡2C)=−1`

Correct Answer is `=>` (D) `-1`

Q 1167112985

Let `z_1, z_2` be two complex numbers represented by point on the circles `| z | = 1` and `| z | = 2` respectively then

(This question may have multiple correct answers)

(A) None of these

(B) `|z_2+1/z_1| ≤ 3`

(D) Min `| z_1 - z_2 | = 1`

Solution:

`| 2z_1 + z_2 | ≤ | 2z_1 | + | z_2 | = 2 | z_1 | + | z_2 | = 2 × 1 + 2 = 4`

`| z_1 - z_2 |` is least when `0, z_1, z_2` are collinear.

`∴ | z_1 - z_2 | = 1`

Again `|z_2+1/z_1| ≤ |z_2|+|1/z_1|=2+|1/z_1|=2+1/1=3`

There fore : Max `| 2z_1 =+ z_2 | = 4`, Min `| z_1 - z_2 | = 1`, `|z_2+1/z_1| ≤ 3`

Correct Answer is `=>` (B)

Fourth Set of 5 Problems

Catching car other way round

`| i z + 3 | = | z -3i|`

We known that `i^4 =1`

but keep in mind `- i^2 =1`

`cos theta + i sin theta = e ^(i theta)`

`sin theta + i cos theta = cos (pi/2 - theta ) + i sin (pi/2 - theta)`

`= e^(i ( pi/2 - theta ))`

`sin theta - i cos theta = sin (pi -theta ) + cos ( pi - theta )`

`= e^(i ( pi - theta ))`

`cos theta - i sin theta = 1/ ( cos theta + i sin theta ) = 1/ (e^(i theta)) = e^(-i theta)`

Q 2371880726

If `a, b, c` are integers and not all are equal, then the least value of `| a + b omega + c omega^2|`

is (where `omega` and `omega^2` are non-real cube roots of unity)
BITSAT Mock

(A)

`0`

(B)

`sqrt (3)/2`

(C)

`1`

(D)

`1/sqrt (3)`

Correct Answer is `=>` (C) `1`

Q 2651401324

If `|z| =1` and `Arg (z + z omega) = pi/2` , then one of the arguments of `z` is `pi/6` (where `omega` is a non-real cube root of unity) what is is the other value of argument of `z`

(A)

True

(B)

False

Correct Answer is `=>` (A)

Q 1137480382

If `4ac > b^2` and `a + c > b` for real numbers a, b and c, then which of the following is true ?

(This question may have multiple correct answers)

(A) `a + b + c >0`

(B) `a > 0`

(D) `c > 0`

Correct Answer is `=>` (A)

Q 1117323280

If `z_1,z_2` and `z_3` be the vertices of ∆ABC and `z_0` be the circum centre, then `((z_0−z_1)/(z_0−z_3))(sin⁡2A)/(sin⁡2C)+((z_0−z_2)/(z_0−z_3))(sin⁡2B)/(sin⁡2C)` is equal to

(A)

`0`

(B)

`1`

(C)

`2i`

(D)

`-1`

Correct Answer is `=>` (D) `-1`

Q 1167112985

Let `z_1, z_2` be two complex numbers represented by point on the circles `| z | = 1` and `| z | = 2` respectively then

(This question may have multiple correct answers)

(A) None of these

(B) `|z_2+1/z_1| ≤ 3`

(D) Min `| z_1 - z_2 | = 1`

Correct Answer is `=>` (B)