-
question_answer1)
\[\sqrt{-2}\,\sqrt{-3}=\] [Roorkee 1978]
A)
\[\sqrt{6}\] done
clear
B)
\[-\sqrt{6}\] done
clear
C)
\[i\sqrt{6}\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer2)
If \[n\] is a positive integer, then which of the following relations is false
A)
\[{{i}^{4n}}=1\] done
clear
B)
\[{{i}^{4n-1}}=i\] done
clear
C)
\[{{i}^{4n+1}}=i\] done
clear
D)
\[{{i}^{-4n}}=1\] done
clear
View Solution play_arrow
-
question_answer3)
If \[n\] is a positive integer, then \[{{\left( \frac{1+i}{1-i} \right)}^{4n+1}}\]=
A)
1 done
clear
B)
- 1 done
clear
C)
\[i\] done
clear
D)
\[-i\] done
clear
View Solution play_arrow
-
question_answer4)
If \[{{\left( \frac{1+i}{1-i} \right)}^{m}}=1,\]then the least integral value of \[m\] is [IIT 1982; MNR 1984; UPSEAT 2001; MP PET 2002]
A)
2 done
clear
B)
4 done
clear
C)
8 done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer5)
If \[{{(1-i)}^{n}}={{2}^{n}},\]then \[n=\] [RPET 1990]
A)
1 done
clear
B)
0 done
clear
C)
\[-1\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer6)
The value of \[{{(1+i)}^{5}}\times {{(1-i)}^{5}}\] is [Karnataka CET 1992]
A)
- 8 done
clear
B)
\[8i\] done
clear
C)
8 done
clear
D)
32 done
clear
View Solution play_arrow
-
question_answer7)
\[{{\left( \frac{1+i}{1-i} \right)}^{2}}+{{\left( \frac{1-i}{1+i} \right)}^{2}}\]is equal to
A)
\[2i\] done
clear
B)
\[-2i\] done
clear
C)
\[-2\] done
clear
D)
\[2\] done
clear
View Solution play_arrow
-
question_answer8)
The value of \[\frac{{{i}^{592}}+{{i}^{590}}+{{i}^{588}}+{{i}^{586}}+{{i}^{584}}}{{{i}^{582}}+{{i}^{580}}+{{i}^{578}}+{{i}^{576}}+{{i}^{574}}}-1=\]
A)
\[-1\] done
clear
B)
- 2 done
clear
C)
\[-3\] done
clear
D)
- 4 done
clear
View Solution play_arrow
-
question_answer9)
\[1+{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+.....+{{i}^{2n}}\]is [EAMCET 1980]
A)
Positive done
clear
B)
Negative done
clear
C)
Zero done
clear
D)
Cannot be determined done
clear
View Solution play_arrow
-
question_answer10)
\[{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+......\]upto \[(2n+1)\] terms = [EAMCET 1980; Kerala (Engg.) 2005]
A)
\[i\] done
clear
B)
\[-i\] done
clear
C)
1 done
clear
D)
\[-1\] done
clear
View Solution play_arrow
-
question_answer11)
If \[i=\sqrt{-1}\], then \[1+{{i}^{2}}+{{i}^{3}}-{{i}^{6}}+{{i}^{8}}\] is equal to [RPET 1995]
A)
\[2-i\] done
clear
B)
1 done
clear
C)
3 done
clear
D)
\[-1\] done
clear
View Solution play_arrow
-
question_answer12)
If \[{{i}^{2}}=-1\], then the value of \[\sum\limits_{n=1}^{200}{{{i}^{n}}}\]is [MP PET 1996]
A)
\[50\] done
clear
B)
- 50 done
clear
C)
0 done
clear
D)
100 done
clear
View Solution play_arrow
-
question_answer13)
The value of the sum \[\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}\], where \[i=\sqrt{-1}\], equals [IIT 1998]
A)
\[i\] done
clear
B)
\[i-1\] done
clear
C)
\[-i\] done
clear
D)
0 done
clear
View Solution play_arrow
-
question_answer14)
The least positive integer \[n\] which will reduce \[{{\left( \frac{i-1}{i+1} \right)}^{n}}\] to a real number, is [Roorkee 1998]
A)
2 done
clear
B)
3 done
clear
C)
4 done
clear
D)
5 done
clear
View Solution play_arrow
-
question_answer15)
The value of \[{{i}^{1+3+5+...+(2n+1)}}\] is [AMU 1999]
A)
i if n is even, - i if n is odd done
clear
B)
1 if n is even, - 1 if n is odd done
clear
C)
1 if n is odd, - 1 if n is even done
clear
D)
i if n is even, - 1 if n is odd done
clear
View Solution play_arrow
-
question_answer16)
If \[\]\[x+\frac{1}{x}=2\cos \theta ,\] then x is equal to [RPET 2001]
A)
\[\cos \theta +i\,\sin \theta \] done
clear
B)
\[\cos \theta -i\,\sin \theta \] done
clear
C)
\[\cos \theta \pm i\,\sin \theta \] done
clear
D)
\[\sin \theta \pm i\,\cos \theta \] done
clear
View Solution play_arrow
-
question_answer17)
The value of \[{{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}\,,\,(n\in N)\] is [RPET 2001]
A)
0 done
clear
B)
1 done
clear
C)
2 done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer18)
The value of \[{{(1+i)}^{8}}+{{(1-i)}^{8}}\] is [RPET 2001; KCET 2001]
A)
16 done
clear
B)
- 16 done
clear
C)
32 done
clear
D)
- 32 done
clear
View Solution play_arrow
-
question_answer19)
\[{{(1+i)}^{10}}\], where \[{{i}^{2}}=-1,\] is equal to [AMU 2001]
A)
32 i done
clear
B)
64 + i done
clear
C)
24 i - 32 done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer20)
The value of \[{{(1+i)}^{6}}+{{(1-i)}^{6}}\] is [RPET 2002]
A)
0 done
clear
B)
27 done
clear
C)
26 done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer21)
If \[{{i}^{2}}=-1\], then sum \[i+{{i}^{2}}+{{i}^{3}}+...\]to 1000 terms is equal to [Kerala (Engg.) 2002]
A)
1 done
clear
B)
- 1 done
clear
C)
i done
clear
D)
0 done
clear
View Solution play_arrow
-
question_answer22)
If \[x=3+i\], then \[{{x}^{3}}-3{{x}^{2}}-8x+15=\] [UPSEAT 2003]
A)
6 done
clear
B)
10 done
clear
C)
- 18 done
clear
D)
- 15 done
clear
View Solution play_arrow
-
question_answer23)
The smallest positive integer \[n\]for which \[{{(1+i)}^{2n}}={{(1-i)}^{2n}}\]is [Karnataka CET 2004]
A)
1 done
clear
B)
2 done
clear
C)
3 done
clear
D)
4 done
clear
View Solution play_arrow
-
question_answer24)
The values of \[x\] and \[y\] satisfying the equation \[\frac{(1+i)x-2i}{3+i}\] \[+\frac{(2-3i)\,y+i}{3-i}=i\] are [IIT 1980; MNR 1987]
A)
\[x=-1,\,y=3\] done
clear
B)
\[x=3,\,y=-1\] done
clear
C)
\[x=0,\,y=1\] done
clear
D)
\[x=1,y=0\] done
clear
View Solution play_arrow
-
question_answer25)
If \[{{z}_{1}}\] and \[{{z}_{2}}\] be two complex number, then Re\[({{z}_{1}}{{z}_{2}})=\]
A)
Re \[({{z}_{1}}).\operatorname{Re}({{z}_{2}})\] done
clear
B)
Re \[({{z}_{1}})\]. Im \[({{z}_{2}})\] done
clear
C)
Im \[({{z}_{1}}).\operatorname{Re}\,({{z}_{2}})\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer26)
\[\left( \frac{1}{1-2i}+\frac{3}{1+i} \right)\,\,\left( \frac{3+4i}{2-4i} \right)=\] [Roorkee 1979; RPET 1999; Pb. CET 2003]
A)
\[\frac{1}{2}+\frac{9}{2}i\] done
clear
B)
\[\frac{1}{2}-\frac{9}{2}i\] done
clear
C)
\[\frac{1}{4}-\frac{9}{4}i\] done
clear
D)
\[\frac{1}{4}+\frac{9}{4}i\] done
clear
View Solution play_arrow
-
question_answer27)
Additive inverse of \[1-i\]is
A)
\[0+0i\] done
clear
B)
\[-1-i\] done
clear
C)
\[-1+i\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer28)
\[\operatorname{Re}\frac{{{(1+i)}^{2}}}{3-i}\] =
A)
\[-1/5\] done
clear
B)
1/5 done
clear
C)
1/10 done
clear
D)
-1/10 done
clear
View Solution play_arrow
-
question_answer29)
If \[(1-i)x+(1+i)y=1-3i,\] then \[(x,y)=\]
A)
\[(2,-1)\] done
clear
B)
\[(-2,\,1)\] done
clear
C)
\[(-2,-1)\] done
clear
D)
(2, 1) done
clear
View Solution play_arrow
-
question_answer30)
\[\frac{3+2i\sin \theta }{1-2i\sin \theta }\]will be real, if \[\theta \] = [IIT 1976; EAMCET 2002]
A)
\[2n\pi \] done
clear
B)
\[n\pi +\frac{\pi }{2}\] done
clear
C)
\[n\pi \] done
clear
D)
None of these [Where \[n\] is an integer] done
clear
View Solution play_arrow
-
question_answer31)
\[\frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}=\]
A)
\[-\frac{3}{2}i\] done
clear
B)
\[\frac{3}{2}i\] done
clear
C)
\[-\frac{3}{2}\] done
clear
D)
\[\frac{3}{2}\] done
clear
View Solution play_arrow
-
question_answer32)
If z and\[{z}'\] are complex numbers such that \[z.z'=z\], then \[z'=\]
A)
\[0+i\,0\] done
clear
B)
\[1+0i\] done
clear
C)
\[0+i\] done
clear
D)
\[1+i\] done
clear
View Solution play_arrow
-
question_answer33)
If \[{{a}^{2}}+{{b}^{2}}=1,\] then \[\frac{1+b+ia}{1+b-ia}=\]
A)
1 done
clear
B)
2 done
clear
C)
\[b+ia\] done
clear
D)
\[a+ib\] done
clear
View Solution play_arrow
-
question_answer34)
\[\frac{3+2i\sin \theta }{1-2i\sin \theta }\] will be purely imaginary, if \[\theta =\] [IIT 1976; Pb. CET 2003]
A)
\[2n\pi \pm \frac{\pi }{3}\] done
clear
B)
\[n\pi +\frac{\pi }{3}\] done
clear
C)
\[n\pi \pm \frac{\pi }{3}\] done
clear
D)
None of these [Where \[n\] is an integer] done
clear
View Solution play_arrow
-
question_answer35)
The real part of \[{{(1-\cos \theta +2i\sin \theta )}^{-1}}\]is [IIT 1978, 86]
A)
\[\frac{1}{3+5\cos \theta }\] done
clear
B)
\[\frac{1}{5-3\cos \theta }\] done
clear
C)
\[\frac{1}{3-5\cos \theta }\] done
clear
D)
\[\frac{1}{5+3\cos \theta }\] done
clear
View Solution play_arrow
-
question_answer36)
If \[{{(x+iy)}^{1/3}}=a+ib,\]then \[\frac{x}{a}+\frac{y}{b}\]is equal to [IT 1982; Karnataka CET 2000]
A)
\[4({{a}^{2}}+{{b}^{2}})\] done
clear
B)
\[4({{a}^{2}}-{{b}^{2}})\] done
clear
C)
\[4({{b}^{2}}-{{a}^{2}})\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer37)
\[{{\left\{ \frac{2i}{1+i} \right\}}^{2}}=\] [BIT Ranchi 1992]
A)
1 done
clear
B)
\[2i\] done
clear
C)
\[\alpha -i\beta \,(\alpha ,\beta \,\text{real),}\] done
clear
D)
\[\left( \frac{3-4ix}{3+4ix} \right)=\] done
clear
View Solution play_arrow
-
question_answer38)
The real values of \[x\]and\[y\]for which the equation is \[(x+iy)\] \[(2-3i)\]= \[4+i\] is satisfied, are [Roorkee 1978]
A)
\[x=\frac{5}{13},y=\frac{8}{13}\] done
clear
B)
\[x=\frac{8}{13},y=\frac{5}{13}\] done
clear
C)
\[x=\frac{5}{13},y=\frac{14}{13}\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer39)
The real values of \[x\] and \[y\] for which the equation \[({{x}^{4}}+2xi)-(3{{x}^{2}}+yi)=\]\[(3-5i)+(1+2yi)\] is satisfied, are [Roorkee 1984]
A)
\[x=2,y=3\] done
clear
B)
\[x=-2,y=\frac{1}{3}\] done
clear
C)
Both (a) and (b) done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer40)
The imaginary part of \[\frac{{{(1+i)}^{2}}}{(2-i)}\]is
A)
\[\frac{1}{5}\] done
clear
B)
\[\frac{3}{5}\] done
clear
C)
\[\frac{4}{5}\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer41)
If \[z\ne 0\] is a complex number, then
A)
\[\operatorname{Re}(z)=0\Rightarrow \operatorname{Im}({{z}^{2}})=0\] done
clear
B)
\[\operatorname{Re}({{z}^{2}})=0\Rightarrow \operatorname{Im}({{z}^{2}})=0\] done
clear
C)
\[\operatorname{Re}(z)=0\Rightarrow \operatorname{Re}({{z}^{2}})=0\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer42)
If \[\frac{5(-8+6i)}{{{(1+i)}^{2}}}=a+ib\], then\[(a,\,b)\] equals [RPET 1986]
A)
(15, 20) done
clear
B)
(20, 15) done
clear
C)
\[(-15,\,20)\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer43)
The true statement is [Roorkee 1989]
A)
\[1-i<1+i\] done
clear
B)
\[2i+1>-2i+1\] done
clear
C)
\[2i>1\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer44)
\[\frac{1-2i}{2+i}+\frac{4-i}{3+2i}=\] [RPET 1987]
A)
\[\frac{24}{13}+\frac{10}{13}i\] done
clear
B)
\[\frac{24}{13}-\frac{10}{13}i\] done
clear
C)
\[\frac{10}{13}+\frac{24}{13}i\] done
clear
D)
\[\frac{10}{13}-\frac{24}{13}i\] done
clear
View Solution play_arrow
-
question_answer45)
\[a+ib>c+id\]can be explained only when
A)
\[b=0,c=0\] done
clear
B)
\[b=0,d=0\] done
clear
C)
\[a=0,c=0\] done
clear
D)
\[a=0,d=0\] done
clear
View Solution play_arrow
-
question_answer46)
If \[x+iy=\frac{3}{2+\cos \theta +i\sin \theta },\]then \[{{x}^{2}}+{{y}^{2}}\] is equal to
A)
\[3x-4\] done
clear
B)
\[4x-3\] done
clear
C)
\[4x+3\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer47)
If \[\frac{{{(p+i)}^{2}}}{2p-i}=\mu +i\lambda ,\]then \[{{\mu }^{2}}+{{\lambda }^{2}}\] is equal to
A)
\[\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}-1}\] done
clear
B)
\[\frac{{{({{p}^{2}}-1)}^{2}}}{4{{p}^{2}}-1}\] done
clear
C)
\[\frac{{{({{p}^{2}}-1)}^{2}}}{4{{p}^{2}}+1}\] done
clear
D)
\[\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}+1}\] done
clear
View Solution play_arrow
-
question_answer48)
If \[z=3-4i\], then \[{{z}^{4}}-3{{z}^{3}}+3{{z}^{2}}+99z-95\]is equal to
A)
5 done
clear
B)
6 done
clear
C)
- 5 done
clear
D)
- 4 done
clear
View Solution play_arrow
-
question_answer49)
If \[{{z}_{1}}=1-i\] and \[{{z}_{2}}=-2+4i\], then \[\operatorname{Im}\left( \frac{{{z}_{1}}{{z}_{2}}}{{{z}_{1}}} \right)=\]
A)
1 done
clear
B)
2 done
clear
C)
3 done
clear
D)
4 done
clear
View Solution play_arrow
-
question_answer50)
If \[\frac{3x+2iy}{5i-2}=\frac{15}{8x+3iy}\], then
A)
\[x=1,y=-3\] done
clear
B)
\[x=-1,y=3\] done
clear
C)
\[x=1,y=3\] done
clear
D)
\[x=-1,y=-3\]or \[x=1,\]\[y=3\] done
clear
View Solution play_arrow
-
question_answer51)
If \[\sum\limits_{k=0}^{100}{{{i}^{k}}}=x+iy\], then the values of \[x\] and \[y\]are
A)
\[x=-1,y=0\] done
clear
B)
\[x=1,y=1\] done
clear
C)
\[x=1,y=0\] done
clear
D)
\[x=0,y=1\] done
clear
View Solution play_arrow
-
question_answer52)
If \[z(1+a)=b+ic\] and \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1\], then \[\frac{1+iz}{1-iz}=\]
A)
\[\frac{a+ib}{1+c}\] done
clear
B)
\[\frac{b-ic}{1+a}\] done
clear
C)
\[\frac{a+ic}{1+b}\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer53)
Let \[{{z}_{1}},{{z}_{2}}\] be two complex numbers such that \[{{z}_{1}}+{{z}_{2}}\] and \[{{z}_{1}}{{z}_{2}}\] both are real, then [RPET 1996]
A)
\[{{z}_{1}}=-{{z}_{2}}\] done
clear
B)
\[{{z}_{1}}={{\bar{z}}_{2}}\] done
clear
C)
\[{{z}_{1}}=-{{\bar{z}}_{2}}\] done
clear
D)
\[{{z}_{1}}={{z}_{2}}\] done
clear
View Solution play_arrow
-
question_answer54)
If \[(x+iy)(p+iq)=({{x}^{2}}+{{y}^{2}})i\], then
A)
\[p=x,q=y\] done
clear
B)
\[p={{x}^{2}},\,\,q={{y}^{2}}\] done
clear
C)
\[x=q,y=p\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer55)
\[A+iB\] form of \[\frac{(\cos x+i\sin x)(\cos y+i\sin y)}{(\cot u+i)(1+i\tan v)}\] is [Roorkee 1980]
A)
\[\sin u\cos v\,[\cos (x+y-u-v)+i\sin (x+y-u-v)]\] done
clear
B)
\[\sin u\cos v\,[\cos (x+y+u+v)+i\sin (x+y+u+v)]\] done
clear
C)
\[\sin u\cos v\,[\cos (x+y+u+v)-i\sin (x+y+u+v)]\] done
clear
D)
None of these done
clear
View Solution play_arrow
-
question_answer56)
If \[x,y\in R\]and \[(x+iy)(3+2i)=1+i\], then \[(x,\,y)\] is
A)
\[\left( 1,\frac{1}{5} \right)\] done
clear
B)
\[\left( \frac{1}{13},\frac{1}{13} \right)\] done
clear
C)
\[\left( \frac{5}{13},\frac{1}{13} \right)\] done
clear
D)
\[\left( \frac{1}{5},\frac{1}{5} \right)\] done
clear
View Solution play_arrow
-
question_answer57)
If \[{{\left( \frac{1-i}{1+i} \right)}^{100}}=a+ib\], then [MP PET 1998]
A)
\[a=2,b=-1\] done
clear
B)
\[a=1,b=0\] done
clear
C)
\[a=0,b=1\] done
clear
D)
\[a=-1,b=2\] done
clear
View Solution play_arrow
-
question_answer58)
If \[{{z}_{1}}=(4,5)\] and \[{{z}_{2}}=(-3,2)\]then \[\frac{{{z}_{1}}}{{{z}_{2}}}\] equals [RPET 1996]
A)
\[\left( \frac{-23}{12},\frac{-2}{13} \right)\] done
clear
B)
\[\left( \frac{2}{13},\frac{-23}{13} \right)\] done
clear
C)
\[\left( \frac{-2}{13},\frac{-23}{13} \right)\] done
clear
D)
\[\left( \frac{-2}{13},\frac{23}{13} \right)\] done
clear
View Solution play_arrow
-
question_answer59)
If \[z=1+i,\] then the multiplicative inverse of z2 is (where i = \[\sqrt{-1}\]) [Karnataka CET 1999]
A)
2 si done
clear
B)
1 - i done
clear
C)
- i/2 done
clear
D)
i/2 done
clear
View Solution play_arrow
-
question_answer60)
If \[\,\left| \begin{align} & \,6i\,\,\,\,\,-3i\,\,\,\,\,\,\,\,\,1 \\ & \,\,4\,\,\,\,\,\,\,\,\,3i\,\,\,\,\,\,-1 \\ & \,20\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,i \\ \end{align} \right|\,\]=\[x+iy\], then (x, y) is [MP PET 2000]
A)
(3, 1) done
clear
B)
(1, 3) done
clear
C)
(0, 3) done
clear
D)
(0, 0) done
clear
View Solution play_arrow
-
question_answer61)
If \[a=\cos \,\theta +i\,\sin \,\theta ,\] then \[\frac{1+a}{1-a}=\] [Karnataka CET 2000]
A)
\[\cot \theta \] done
clear
B)
\[\cot \frac{\theta }{2}\] done
clear
C)
\[i\,\cot \frac{\theta }{2}\] done
clear
D)
\[i\,\tan \frac{\theta }{2}\] done
clear
View Solution play_arrow
-
question_answer62)
Solving \[3-2yi={{9}^{x}}-7i\], where \[{{i}^{2}}=-1,\] for x and y real, we get [AMU 2000]
A)
\[x=0.5\,\,,\,\,y=3.5\] done
clear
B)
\[x=5\,\,,\,\,y=3\] done
clear
C)
\[x=\frac{1}{2}\,\,,\,\,y=7\] done
clear
D)
\[x=0,\,y=\frac{3+7i}{2i}\] done
clear
View Solution play_arrow
-
question_answer63)
The complex number \[\frac{1+2i}{1-i}\] lies in which quadrant of the complex plane [MP PET 2001]
A)
First done
clear
B)
Second done
clear
C)
Third done
clear
D)
Fourth done
clear
View Solution play_arrow
-
question_answer64)
The real part of \[\frac{1}{1-\cos \theta +i\,\sin \theta }\] is equal to [Karnataka CET 2001, 05]
A)
1/4 done
clear
B)
1/2 done
clear
C)
tan q/2 done
clear
D)
1/1- cos q done
clear
View Solution play_arrow
-
question_answer65)
The statement \[(a+ib)<(c+id)\] is true for [RPET 2002]
A)
\[{{a}^{2}}+{{b}^{2}}=0\] done
clear
B)
\[{{b}^{2}}+{{c}^{2}}=0\] done
clear
C)
\[{{a}^{2}}+{{c}^{2}}=0\] done
clear
D)
\[{{b}^{2}}+{{d}^{2}}=0\] done
clear
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-
question_answer66)
The multiplication inverse of a number is the number itself, then its initial value is [RPET 2003]
A)
i done
clear
B)
- 1 done
clear
C)
2 done
clear
D)
- i done
clear
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-
question_answer67)
If \[z=x+iy,\,{{z}^{1/3}}=a-ib\] and \[\frac{x}{a}-\frac{y}{b}=k\,({{a}^{2}}-{{b}^{2}})\] then value of k equals [DCE 2005]
A)
2 done
clear
B)
4 done
clear
C)
6 done
clear
D)
1 done
clear
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