Mathematics - Complex Numbers
Exam Duration: 45 Mins Total Questions : 30
For positive integers n1, n2 the value of the expression (1+i)n1+(1+i3)n1+(1+i5)n2+(1+i7)n2, where i=\(\sqrt { -1 } \), is a real number,if and only if,
- (a)
n1=n2+1
- (b)
n1=n2-1
- (c)
n1=n2
- (d)
n1>0,n2>0
The value of sum, \(\sum _{ n=1 }^{ 13 }{ ({ i }^{ n }+{ i }^{ n+1 }) } \), where \(i=\sqrt { -1 } \), equals
- (a)
\(i\)
- (b)
\(i-1\)
- (c)
\(-i\)
- (d)
0
If \({ x }_{ r }=cos\left( \cfrac { \pi }{ { 2 }^{ r } } \right) +isin\left( \cfrac { \pi }{ { 2 }^{ r } } \right) \), then value of \({ x }_{ 1 }.{ x }_{ 2 }.{ x }_{ 3 }.......\infty ,\) is
- (a)
\(-1\)
- (b)
\(1\)
- (c)
\(0\)
- (d)
NONE OF THESE
The equation whose roots are the \( n\)th powers of the roots of the equation \({ x }^{ 2 }-2x\ cos\ \theta +1=0\) is
- (a)
\(n{ x }^{ 2 }-2x\ cos\ \theta +n=0\)
- (b)
\({ x }^{ 2 }-2n\ cos\ \theta.x +1=0\)
- (c)
\({ x }^{ 2 }-2\ cos\ n\theta.x +1=0\)
- (d)
NONE OF THESE
For all complex numbers z1, z2 satisfying \(\left| { z }_{ 1 } \right| =12\) and \(\left| { z }_{ 2 }-3-4i \right| =5\) the minimum value of \(\left| { z }_{ 1 }-{ z }_{ 2 } \right| \), is
- (a)
0
- (b)
2
- (c)
7
- (d)
12
For a complex number z, the minimum value of \(|z|+|z-cos\alpha -isin\alpha |\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
None of these
If Z = X - iY and Z1/3 = p+iq, then \(\left( \frac { x }{ p } +\frac { y }{ q } \right) /\left( { p }^{ 2 }+{ q }^{ 2 } \right) \) is equal to
- (a)
1
- (b)
-1
- (c)
2
- (d)
-2
If a,b,c,p,q,r are six complex numbers, such that \(\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1+i\) and \(\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0\), where \(i=\sqrt{-1}\) then value of \(\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{c^2}{r^2}\) is
- (a)
0
- (b)
-1
- (c)
2i
- (d)
-2i
The least positive integer n for which \((\frac{1+i}{1-i})^n=\frac{2}{\pi}\sin^{-1}\frac{1+x^2}{2x}\),where x>0 and\(i=\sqrt{-1}\) is
- (a)
2
- (b)
4
- (c)
8
- (d)
12
If the multiplication inverse of a complex number is \((\sqrt3+4i)/19\),(where \(i=\sqrt{-1}\))then the complex number itself is
- (a)
\(\sqrt3-4i\)
- (b)
\(4+i\sqrt3\)
- (c)
\(\sqrt3+4i\)
- (d)
\(4-i\sqrt3\)
If z then n is equal to1 and\(\bar z_1\)represent adjacent vertices of a rectangular polygon of n sides whose centre is origin and if\(\frac{Im(z_1)}{Re(z_1)}=\sqrt 2-1\)
- (a)
8
- (b)
16
- (c)
24
- (d)
32
If z is any non-zero complex number, then \(arg(z)+arg(\bar z)\) is equal to
- (a)
0
- (b)
\(\pi/2\)
- (c)
\(\pi\)
- (d)
\(3\pi/2\)
If the following regions in the complex plane, the only one that does not represent a circle is
- (a)
\(z\bar z+i(z-\bar z)=0\)
- (b)
\(Re\left(\frac{1+z}{1-z}\right)=0\)
- (c)
\(arg\left(\frac{z-i}{z+i}\right)=\frac{\pi}{2}\)
- (d)
\(\left|\frac{z-i}{z+1}\right|=1\)
If \(1,\omega \ and \ \omega^2\) are the three cube roots of unity, then the roots of the equation \((x-1)^3-8=0\) are
- (a)
\(-1,-1-2\omega,-1+2\omega^2\)
- (b)
\(3,2\omega,2\omega^2\)
- (c)
\(3,1+2\omega,1+2\omega^2\)
- (d)
none of these
If \(1,\omega,\omega^2,...,\omega^{n-1}\) are n, nth roots of unity, the value of \((9-\omega)(9-\omega^2)...(9-\omega^{n-1})\) will be
- (a)
n
- (b)
0
- (c)
\(\frac{9^n-1}{8}\)
- (d)
\(\frac{9^n+1}{8}\)
If 8iz3+12z2-18z+27i=0,(where \(i=\sqrt{-1})\) then
- (a)
|z|=3/2
- (b)
|z|=2/3
- (c)
|z|=1
- (d)
|z|=3/4
The equation z-1 n-1=0 has n roots which are called the nth roots of unity. The n , nth roots of unity are \(1,\alpha ,{ \alpha }^{ 2 },....{ \alpha }^{ n-1 }\) which are in GP, where \(\alpha =cos\left( \frac { 2\pi }{ n } \right) +i\quad sin\left( \frac { 2\pi }{ n } \right) ;i=\sqrt { -1 } \) then we have following results:
(i) \(\overset { n-1 }{ \underset { r=0 }{ \Sigma } } \alpha ^{ r }=0\quad or\quad \overset { n-1 }{ \underset { r=0 }{ \Sigma } } cos\left( \frac { 2\pi r }{ n } \right) =0\quad and\quad \overset { n-1 }{ \underset { r=0 }{ \Sigma } } sin\left( \frac { 2\pi r }{ n } \right) =0\)
(ii) \({ z }^{ n }-1=\prod _{ r=0 }^{ n-1 }{ (z-\alpha ^{ r }) } \)
(iii) \(\prod _{ r=0 }^{ n-1 }{ { \alpha }^{ 2 } } =(-1)^{ n-1 }\)
(iv) \(\overset { n-1 }{ \underset { r=0 }{ \Sigma } } { \alpha }^{ kr }\)=\(\begin{cases} n,\quad if\quad k\quad is\quad multiple\quad of\quad n \\ 0,\quad if\quad k\quad is\quad not\quad multiple\quad of\quad n \end{cases}\)
If \(n\in I,n\ge 2\) then the value of \(\quad \overset { n-1 }{ \underset { r=1 }{ \Sigma } } (n-r)cos\left( \frac { 2r\pi }{ n } \right) \) is equal to
- (a)
-n
- (b)
n
- (c)
-n/2
- (d)
n/2
\(\sin^{-1}\left\{\frac{1}{i}(z-1)\right\}\), where z is non real and \(i=\sqrt{-1}\), can be the angle of a triangle if
- (a)
Re(z)=1, Im(z)=2
- (b)
Re(z)=1,\(-1\le Im(z)\le1\)
- (c)
Re(z)+Im(z)=0
- (d)
none of the above
If \(|z_1+z_2|^2=|z_1|^2+|z_2|^2\) , then
- (a)
\(\frac{z_1}{z_2} \)is purely real
- (b)
\(\frac{z_1}{z_2}\) is purely imaginary
- (c)
\(z_1\bar z_2+z_2\bar z_1=0\)
- (d)
\(amp(\frac{z_1}{z_2})=\frac{\pi}{2}\)
If \(1,\omega,\omega^2,....,\omega^{n-1}\) are the n, nth roots of unity, then \((2-\omega)(2-\omega^2)...(2-\omega^{n-1})\) equals
- (a)
2n-1
- (b)
\(^nC_1+^nC_2+....+^nC_n\)
- (c)
\([^{2n}C_0+^{2n+1}C_1+^{2n-1}C_2+....+^{2n+1}C_n]^{1/2}-1\)
- (d)
2n+1
If \(x^2+1=0\Rightarrow x^2=-1 \) or \(x=\pm\sqrt{-1}=\pm i\) (iota) is called the imaginary unit.
Also, i2=-1,i3=i2.i=(-1)i=-i and i4=(i2)2=(-1)2=1
ie, \(i^n+i^{n+1}+i^{n+2}+i^{n+3}=0\forall n\epsilon I(Interger) \) and x3-1=0\(\Rightarrow\)(x-1)(x2+x+1)=0
\(\Rightarrow (x-1)(x-\omega)(x-\omega^2)=0\)
\(\therefore x=1,\omega,\omega^2\) are the cube roots of unity. ie,\(\omega^n+\omega^{n+1}+\omega{n+2}=0\forall n\epsilon I(interger)\)
Now let z=a+ib if \(|a:b|=\sqrt{3}:1 \ or 1:\sqrt{3}\)
Then, convert z in terms of \(\omega,\ or\ \omega^2\) . Also \(|1-\omega|=|1-\omega^2|=\sqrt{3}\)
For positive interegers n1 ,n2 the value of the expression,\((1+i)^{n_1}+(1+i^3)^{n_1}+(1+i^5)^{n_2}+(1+i^7)^{n_2}\) where \(i=\sqrt {-1}\) is real if
- (a)
n1=n2+1
- (b)
n1=n2-1
- (c)
n1=n2
- (d)
n1>0,n2>0
If \(x^2+1=0\Rightarrow x^2=-1 \) or \(x=\pm\sqrt{-1}=\pm i\) (iota) is called the imaginary unit.
Also, i2=-1,i3=i2.i=(-1)i=-i and i4=(i2)2=(-1)2=1
ie, \(i^n+i^{n+1}+i^{n+2}+i^{n+3}=0\forall n\epsilon I(Interger) \) and x3-1=0\(\Rightarrow\)(x-1)(x2+x+1)=0
\(\Rightarrow (x-1)(x-\omega)(x-\omega^2)=0\)
\(\therefore x=1,\omega,\omega^2\) are the cube roots of unity. ie,\(\omega^n+\omega^{n+1}+\omega{n+2}=0\forall n\epsilon I(interger)\)
Now let z=a+ib if \(|a:b|=\sqrt{3}:1 \ or 1:\sqrt{3}\)
Then, convert z in terms of \(\omega,\ or\ \omega^2\) . Also \(|1-\omega|=|1-\omega^2|=\sqrt{3}\)
The smallest positive integer n for which \(\left(\frac{1+i}{1-i}\right)^n=1,\) where \(i=\sqrt{-1}\) is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Let \(z=a+ib=re^{i\theta}\) where a,b,\(\theta\epsilon R\) and \(i=\sqrt{1}\)
Then, \(r=\sqrt{(a^2+b^2)}=|z|\) and \(\theta=\tan^{-1}(\frac{b}{a})=arg(z)\)
Now, \(|z|^2=a^2+b^2=(a+ib)(a-ib)=z\bar z\Rightarrow\frac{1}{z}=\frac{\bar z}{|z|^2}\)
and \(|z_1z_2z_3...z_n|=|z_1||z_2||z_3|..|z_n|\)
If \(|f(z)|=1, then f(z) \) is called unimodular. In this case f(z) can always be expressed as \(f(z)=e^{i\alpha},\alpha\epsilon R\)
Also, \(e^{i\alpha}+e^{i\beta}=e^{i(\frac{\alpha+\beta}{2})}2\cos\left(\frac{\alpha-\beta}{2}\right)\) and \(e^{i\alpha}-e^{i\beta}=e^{i(\frac{\alpha+\beta}{2})}2i\sin\left(\frac{\alpha-\beta}{2}\right)\) , where \(\alpha,\beta,\epsilon R\)
If z =x+iy is a complex number with rationals z and y \(i=\sqrt{-1}\) and |z2n-1| is (\(n\epsilon N)\)
- (a)
an irrational number
- (b)
a rational number
- (c)
non terminating non recurring
- (d)
a positive real number
If(x+iy)(p+iq)=(x2+y2)i, then
- (a)
P=x2, q=y2
- (b)
x=q, y=p
- (c)
p=x, q=y
- (d)
p=-x, q=y
The real part of \(\frac { { (1+i) }^{ 2 } }{ (3-i) } \)is
- (a)
\(\frac{1}{3}\)
- (b)
\(\frac{1}{5}\)
- (c)
-\(\frac{1}{3}\)
- (d)
None of these
Evaluate:(1+i)6+(1-i)3.
- (a)
-2-10i
- (b)
2-10i
- (c)
-2+10i
- (d)
2+10i
If |z2-1|=|z|2+1, then x lies on
- (a)
imaginary axis
- (b)
real axis
- (c)
origin
- (d)
None of these
The multiplicative inverse of \(\frac{3+4i}{4-5i}\)
- (a)
\(\frac { 8 }{ 25 } -\frac { 31 }{ 25 } \)
- (b)
-\(\frac { 8 }{ 25 } -\frac { 31 }{ 25 } \)
- (c)
-\(\frac { 8 }{ 25 }+\frac { 31 }{ 25 } \)
- (d)
None of these
The square root of the number (5+12i)is
- (a)
3+2i
- (b)
±(3+2i)
- (c)
3-2i
- (d)
None of these
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