Mathematics

Complex Variables and Numbers

206 Questions

Complex numbers and variables form a crucial part of advanced mathematics syllabi. This topic covers roots of unity, Argand plane geometry, and Z-transforms. These concepts are frequently tested in engineering entrance exams and UPSC mathematics optional papers.

Roots of unityArgand plane geometryZ-transform sequencesExponential complex formsComplex number quadrantsRiemann hypothesis

Complex Variables and Numbers Questions

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $\left| {z - 1} \right| + \left| {z + 3} \right| \le 8$ then the range of values of $\left| {z - 4} \right|$

  1. $[1,\,7]$
  2. $[1,\,8]$
  3. $[1,\,9]$
  4. $[2,\,5]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$|z-1|+|z+3|\le 8$

Using the triangle inequality 
$|z _1\pm z _2|\le |z _1|+|z _2|$
We have $|z-1|+|z+3|\le 8$
$\implies |z-1+z+3|\le 8$
$\implies |z+1|\le 4$
Using triangle inequality again 
$|z|+1\le 4\implies |z|\le 3$
So, the maximum value of $|z _1+ _2|$ is $|z _1|+|z _2|$
And  the minimum value of $|z _1+ _2|$ is $|z _1|-|z _2|$
Hence the maximum value of $|z-4|$ is $|z|+|4|=3+4=7$
 the minimum value of $|z-4|$ is $|z|-|4|=3-4=-1$
Hence the range of $|z-4|$
$1\le|z-4|\le 7$

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $\left|z\right| <\sqrt{2} -1$, then $\left|z^2 + 2 z  cos  \alpha \right|$ is

  1. less than 1
  2. $\sqrt{2} + 1$
  3. $\sqrt{2} -1$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\left| z \right| <\sqrt { 2 } -1\ \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| \le \left| { z }^{ 2 } \right|+ \left| 2z\cos { \alpha  }  \right| \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| { z } _{ 1 }+{ z } _{ 2 } \right| \le \left| { z } _{ 1 } \right| +\left| { z } _{ 2 } \right|  \right} \ \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| \le |z|(|z|+2) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| \cos { \alpha  }  \right| \le 1\quad  \right} \ \Rightarrow \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| <(\sqrt{2}-1){ \left( \sqrt { 2 } +1 \right)  }<1\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| z \right| <\sqrt { 2 } -1\quad  \right} \ \therefore \quad \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| <1\ $
Hence, option 'A' is correct.

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If z be a complex number for which $|2z  cos  \theta + z^2| = 1$, then the minimum value of |z|
 is ......................

  1. $\sqrt{3} -1$
  2. $\sqrt{3} +1$
  3. $\sqrt{2} -1$
  4. $\sqrt{2} +1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$|z^{2}+2zcos\theta|$
$=|z(z+2cos\theta)|$
$=|z|.|z+2cos\theta|$
$=1$
Now 
$|z|=1$ and 
$|z+2cos\theta|=1$
Now 
$|z+2cos\theta|\leq |z|+|2cos\theta|$
Considering 
$|z+2\cos\theta|=|z|+|2cos\theta|=1$
Hence
$|z|=|2cos\theta|\pm1$
Considering $z=|2cos\theta|-1$ we get the minimum value at multiples of $\theta=45^{0}$
Hence
$z=\sqrt{2}-1$.

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

$sin^{-1}\left { \frac{1}{i} (z-1)\right }$ ,Where Z is non - real, can be the angle  of a triangle, if 

  1. $Re(z)=1, Im(z)=2$
  2. $Re(z)=1,-1\leq Im(z)\leq 1$
  3. $Re(z)=1,Im(z)=0$
  4. $Re(z)=1,Im(z)=-2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let theta be a real angle of a triangle, so 0 < theta < pi. We are given sin(theta) = (1/i)(z - 1), which means z - 1 = i * sin(theta), so z = 1 + i * sin(theta). Since sin(theta) is real and lies between -1 and 1, the real part of z is 1 and the imaginary part satisfies -1 <= Im(z) <= 1.

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $|z-4+3i|\le 1$ and $m$ and $n$ are the least and greatest values of $|z|$ and $k$ is the least value of $\displaystyle \frac { { x }^{ 4 }+{ x }^{ 2 }+4 }{ x } $ on the interval $(0,\infty)$, then $k$ is equal to

  1. $m$
  2. $n$
  3. $m+n$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have, 

$1\ge \left| z-\left( 4-3i \right)  \right| $
$\Rightarrow 1\ge \left| z \right| -\left| 4-3i \right| \quad ,\quad \left| 4-3i \right| -\left| z \right| $
$\Rightarrow 1\ge \left| z \right| -5\quad ,\quad 5-\left| z \right| $
$\left| z \right| \le 6,\left| z \right| \ge 4\Rightarrow 4\le \left| z \right| \le 6\Rightarrow m=4,n=6$
Let $y=\displaystyle\frac { 4+{ x }^{ 2 }+{ x }^{ 4 } }{ x } ={ x }^{ 3 }+x+\displaystyle\frac { 4 }{ x } ={ x }^{ 3 }+x+\frac { 1 }{ x } +\frac { 1 }{ x } +\frac { 1 }{ x } +\frac { 1 }{ x } $
$\because x\in \left( 0,\infty  \right) $, then ${ x }^{ 3 },x,\frac { 1 }{ x } ,\frac { 1 }{ x } ,\frac { 1 }{ x } ,\frac { 1 }{ x } $ are all positive numbers whose product is 1.
Thus their sum y will be least when 
${ x }^{ 3 }=x=\displaystyle\frac { 1 }{ x } \Rightarrow x=1$
So least value of $y=6,k=6$
So $k=n$

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles


$|\mathrm{z} _{1}-\mathrm{z} _{2}|=$

  1. $\geq||z _{1}|-|z _{2}||$
  2. $\leq|z _{1}|-|z _{2}|$
  3. $=|\mathrm{z} _{1}|-|\mathrm{z} _{2}|$
  4. $\geq|\mathrm{z} _{1}|-|\mathrm{z} _{2}|$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $ argz _1=\theta _1  \quad argz _2=\theta _2$
we know that
$|z _{1}-z _{2}|^{2}=|z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})$


now, $+1\geq cos(\theta _{1}-\theta _{2})\geq -1$


$-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})\geq -2|z _{1}||z _{2}|$


$\therefore |z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})\geq |z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|$


$\therefore |z _{1}-z _{2}|^{2}\geq (|z _{1}|-|z _{2}|)^{2}\Rightarrow |z _{1}-z _{2}|\geq ||z _{1}|-|z _{2}||$


Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles


 lf $|\mathrm{z} _{1}|=2,\ |\mathrm{z} _{2}|=3$, then $|\mathrm{z} _{1}+\mathrm{z} _{2}+5+12\mathrm{i}|$ is less than or equal to

  1. $8$
  2. $18$
  3. $10$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that $|z _{1}+z _{2}+ z _3 |\leq |z _{1}|+|z _{2}| + | z _3| $
$|z _{1}|+|z _{2}|=5$
$z _3 = 5+12 i $

$|z _3 | = 13 $
$\therefore 18\geq |z _{1}+z _{2}+5+12i|$
Hence, option B is correct

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles


 Let $z _{1}=24+7i$ and $z _{2}$ be complex number whose magnitude is unity, then

  1. Maximum value of $|z _{1}+z _{2}|$ is 26
  2. Maximum value of $|z _{1}+z _{2}|$ is 31
  3. Minimum value of $|z _{1}+z _{2}|$ is 24
  4. Minimum value of $|z _{1}+z _{2}|$ is 19
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$ |z _1 | =  25 $

$| z _2 | = 1$

We have, 
$\left| |z _1| - |z _2| \right| \leq |z _1+z _2 | \leq \left | |z _1| + |z _2| \right |$
$\Rightarrow 24 \leq |z _1+z _2| \leq 26$
Hence, options A and C are correct 

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $|{z _1}| = |{z _2}| = |{z _3}| = 1$ and ${z _1} + {z _2} + {z _3} = 0$ then the area of the triangle whose vertices are $z _1, z _2, z _3$ is

  1. $\frac{3\sqrt{3}}{4}$
  2. $\frac{\sqrt{3}}{4}$

  3. 1

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

given  $|Z _1|=|Z _2|=|Z _3|=1$     and     $Z _1+Z _2+Z _3=0$
$\Rightarrow |Z _1 -Z _2|=2 (Cos 30)=\sqrt{3}$
$\Rightarrow  area =\frac{\sqrt{3}}{4}a^2$    &     $ a=\sqrt{3}$
So, area $=\frac{\sqrt{3}}{4}\cdot 3 =\frac{3\sqrt{3}}{4}$

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

$z _0$ is a root of the equation $z^n cos \theta _o+z^{n-1} cos\theta _1+....+z cos\theta _{n-1}+cos\theta _n=2$, where $\theta, \epsilon R$, then

  1. $|z _0| > 1$
  2. $|z _0| > \dfrac {1}{2}$
  3. $|z _0| > \dfrac {1}{4}$
  4. $|z _0| > \dfrac {3}{2}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

$z^n cos\theta _0+z^{n-1} cos\theta _1+.....+z cos\theta _{n-1}+cos\theta _n=2$

or $2=|z _0^n cos\theta _0+z _0^{n-1} cos\theta _1+....+z _0 cos\theta _{n-1}+cos\theta _n|$

or $2\leq |z _0|^n |cos\theta _0|+|z|^{n-1}|cos\theta _1|+....+|z _0||cos\theta _{n-1}|+|cos\theta _n|$

or $2\leq |z _0|^n+|z _0|^{n-1}+|z _0|^{n-2}+.....+|z _0|+1$

which is clearly satisfied for $|z _0| \geq 1$. If $|z _0| < 1$, then

$2 < 1+|z _0|+|z _0|^2+.....+|z|^n+....\infty$

$\Rightarrow 2 < \dfrac {1}{1-|z _0|}$

$\Rightarrow |z _0| > \dfrac {1}{2}$

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $z _{1},\ z _{2}--,\ z _{n}$ are complex numbers such that $|z _{i}|<\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{d}\lambda _{i}>0$ for $i=1,2,---n$ and $\lambda _{1}+\lambda _{2}+--+\lambda _{n}=1$ then $|\lambda _{1}z _{1}+\lambda _{2}z _{2}+--+\lambda _{n}\mathrm{z} _{1}|?$

  1. $=1$
  2. $<1$
  3. $>1$
  4. $=n$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given:

$\lambda _i>0$  and  $\lambda _1+\lambda _2+...+\lambda _n=1$
$\therefore 0<\lambda _i<1$
Also,  $|z _i|<1$
$\therefore \lambda _1|z _1|+\lambda _2|z _2|+.....+\lambda _n|z _n|<1$                ......( 1 )

$\therefore |\lambda _1z _1+\lambda _2z _2+.....+\lambda _nz _n|$
$\leq |\lambda _1z _1|+|\lambda _2z _2|+.....+|\lambda _nz _n|$
$\leq \lambda _1|z _1|+\lambda _2|z _2|+.....+\lambda _n|z _n|$
$<1$

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

Simlify: $\sqrt{\dfrac{-17}{144}-i}$

  1. $ \pm \left( {\dfrac{3}{2} - \dfrac{i}{3}} \right)$
  2. $ \pm \left( {\dfrac{3}{4} - \dfrac{{2i}}{3}} \right)$
  3. $ \pm \left( {\dfrac{3}{5} - \dfrac{{5i}}{6}} \right)$
  4. $ \pm \left( {\dfrac{2}{3} - \dfrac{{3i}}{4}} \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that,

${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$

 

Now, let,

$ -2ab=-i $

$ ab=i $

 

Now, consider $\dfrac{-17}{144}$. We can write it as,

$\dfrac{-17}{144}=\dfrac{64-81}{9\times 16}=\dfrac{4}{9}-\dfrac{9}{16}$

 

Thus,

$ \sqrt{\dfrac{-17}{144}-i}=\sqrt{\dfrac{4}{9}-\dfrac{9}{16}-i} $

$ =\sqrt{{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( i \right)}^{2}}{{\left( \dfrac{3}{4} \right)}^{2}}-2\times \left( \dfrac{2}{3} \right)\times \left( \dfrac{3i}{4} \right)} $

$ =\sqrt{{{\left( \dfrac{2}{3}-\dfrac{3i}{4} \right)}^{2}}} $

$ =\pm \left( \dfrac{2}{3}-\dfrac{3i}{4} \right) $

 

Hence, this is the required result.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If $1+\surd {3}i/2$ is a root of equation $x^{4}-x^{3}+x1=0$ then its real roots are 

  1. $1,1$
  2. $-1,-1$
  3. $1,-1$
  4. $1,2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that 1 + i√3/2 is a root, its complex conjugate 1 - i√3/2 is also a root (coefficients are real). The sum of all roots is 1 (from x³ coefficient with opposite sign). If the remaining two roots are real and equal to r, then: 2(1) + 2r = 1, giving r = ±1. Testing shows x = 1 and x = -1 satisfy the equation. Therefore, the real roots are 1 and -1.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If a,b,c and d are the real roots of the equation : $x^{4}+p _{1}x^{1}+p _{2}x^{2}+p _{3}x+p _{4}=0$ and $(1+a^{2})(1+b^{2})(1+c^{2})(1+d^{2})=k(1-p _{2}+p _{4})^{2}+(p _{3}-p _{1})^{2}$ then the value f k is:

  1. -1

  2. 1

  3. 2

  4. -2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the relationship between roots and coefficients, and the identity that (1+a²)(1+b²)(1+c²)(1+d²) can be expressed in terms of the polynomial's coefficients, we can derive that k = 1. This involves substituting the roots into the given expression and using Vieta's formulas to relate it to the coefficients p₁, p₂, p₃, p₄.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $z _{1}$ is a root of the equation $a^{n} _{0}z^{n}+a _{1}z^{n-1}+....+a _{n-1^{z}}+a _{n}=3$, where $|a _{i}|<2$ for $i=0,1,....,n.$ Then,

  1. $|z _{1}|>\dfrac {1}{3}$
  2. $|z _{1}|<\dfrac {1}{4}$
  3. $|z _{1}|>\dfrac {1}{4}$
  4. $|z|<\dfrac {1}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to question,

${l} { a _{ 0 } }{ z^{ n } }+{ a _{ 1 } }{ z^{ n-1 } }+.............+{ a _{ n-1 } }z+{ a _{ n } }=3 \ \Rightarrow \left| { { a _{ 0 } }{ z^{ n } }+{ a _{ 1 } }{ z^{ n-1 } }+.............+{ a _{ n-1 } }z+{ a _{ n } } } \right| =\left| 3 \right|  \ \Rightarrow \left| { { a _{ 0 } } } \right| \, { \left| z \right| ^{ n } }+\left| { { a _{ 1 } } } \right| \, { \left| z \right| ^{ n-1 } }\, +..........+\left| { { a _{ n-1 } } } \right| \, \left| z \right| \, +\left| { { a _{ n } } } \right| \ge 3 \ \Rightarrow 2\, ({ \left| z \right| ^{ n } }+{ \left| z \right| ^{ n-1 } }+...........\left| z \right| +1)\, \, >\, 3 \ \Rightarrow (1+\left| z \right| +{ \left| z \right| ^{ 2 } }+...........+{ \left| z \right| ^{ n } })\, \, >\, \frac { 3 }{ 2 }  \ \Rightarrow \frac { { \, \, \, \, 1-{ { \left| z \right|  }^{ n+1 } } } }{ { 1-\left| z \right|  } } \, \, >\, \frac { 3 }{ 2 }  \ \Rightarrow 2-2{ \left| z \right| ^{ n+1 } }\, >\, 3-3\left| z \right|  \ \Rightarrow 2{ \left| z \right| ^{ n+1 } }<3\left| z \right| \, -1 \ \Rightarrow 3\, \left| z \right| -1\, >0 \ \, \, \, \, \, \therefore \, \, \, \left| z \right| \, >\, \frac { 1 }{ 3 }  \ so\, the\, correct\, option\, is\, \, A$