InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Show that `{i^(23)+((1)/(i))^(29)}^(2) = -4`. |
|
Answer» We have `i^(23)=i^((4xx5+3))=(i^(4))^(5)xx i^(3) = 1 xx (-i) = -i." "[because i^(4) = 1 and i^(3) = -i]` `((1)/(i))^(29) = (1)/(i^(29))=(1)/(i^(29))xx(i^(3))/(i^(3))=(i^(3))/(i^(32))=(-i)/(1) =-i." "[because i^(3) =-i and i^(32) =1]` `therefore" "{i^(23)+((1)/(i))^(29)}^(2)=(-i-i)^(2)=(-2i)^(2)=4i^(2)=4xx(-1)=-4`. |
|
| 152. |
Evaluate `sqrt(4+3sqrt(-20))+sqrt(4-3sqrt(-20))`. |
|
Answer» `sqrt(4+3sqrt(-20))=sqrt(4+6 sqrt(5)i)and sqrt(4-3 sqrt(-20))=sqrt(4-6sqrt(5)i)`. Let `sqrt(4+6 sqrt(5)i)=(x+iy)." "...(i)` On squaring both sides of (i), we get `(4+6 sqrt(5)i)=(x+iy)^(2)rArr (4+6 sqrt(5)i)=(x^(2)-y^(2))+i(2xy)." "...(ii)` On comparing real parts and imaginary parts on both sides of (ii), we get `(x^(2)-y^(2))=4 and 2xy=6 sqrt(5)`. `therefore" "(x^(2)+y^(2))=sqrt((x^(2)-y^(2))^(2)+4x^(2)y^(2))=sqrt((4)^(2)+(6 sqrt(5))^(2))=sqrt(16+180)=sqrt(196)=14`. Thus, `(x^(2)-y^(2))=4" "...(iii) and (x^(2)+y^(2))=14" "...(iv)`. On solving (iii) and (iv), we get `x^(2)=9 and y^(2)=5`. `therefore" "x = +- 3 and y = +- sqrt(5)`. Since `xy gt 0`, so x and y are of the same sign. `therefore" "(x = 3, y = sqrt(5))or(x = -3, y = -sqrt(5))`. `therefore" "sqrt(4+3 sqrt(-20))=sqrt(4+6 sqrt(5)i)=(3-sqrt(5)i)or(-3+sqrt(5)i)`. Similarly, `sqrt(4-3 sqrt(-20))=sqrt(4-6 sqrt(5)i)=(3-sqrt(5)i)or(-3+sqrt(5)i)`. `therefore" "sqrt(4+3 sqrt(-20))+sqrt(4-3 sqrt(-20))={{:({(3+sqrt(5)i)+(-3+sqrt(5)i)}=6),(" "or),({-3-sqrt(5)i}+{-3+sqrt(5)i}=-6):}` Hence, `sqrt(4+3sqrt(-20))+sqrt(4-3 sqrt(-20))=6 or -6`. |
|
| 153. |
Find the multiplicative inverse of the complex number.`sqrt(5)+3i` |
|
Answer» let `sqrt 5 + 3i=z` so`z^-1 = 1/z` `= 1/(sqrt5+3i)` rationalizing it , we get `1/(sqrt5 + 3i)*(sqrt5 - 3i)/(sqrt5- 3i) ` `= (sqrt5 - 3i)/(5+9)` `= (sqrt5 -3i)/14` answer |
|
| 154. |
Solve the equation: `x^2+3=0` |
|
Answer» (i) `x^2= -3` `x= +- sqrt3` `= +-isqrt3` (ii)` 2x^2+x+1=0` `D= b^2-4ac = 1-4*2*1 = -7` `x= (-b+-sqrt D)/(2a)` `=(-1+- sqrt7)/4` |
|
| 155. |
Evaluate: `sqrt(1+4sqrt(-3))` |
| Answer» Correct Answer - `(2+sqrt(3)i)or(-2-sqrt(3)i)` | |
| 156. |
Evaluate : `(i) i^(23)" "(ii) i^(998)" "(iii)i^(-998)" "(iv) i^(-71)` `(v) (sqrt(-1))^(91)" "(vi) (i^(37)xx i^(-61))" "(vii) i^(-1)` |
|
Answer» We have `(i)" "i^(23) = i^((4xx5)+3)=(i^(4))^(5)xx i^(3) = i^(3) = -i." "[because i^(4)=1]` `(ii)" "i^(998) = i^(4xx249+2)=(i^(4))^(249)xx i^(2) = (1 xx i^(2)) = i^(2) = -1." "[because i^(4)=1]` `(iii)" "i^(-998) = (1)/(i^(998))xx(i^(2))/(i^(2))=(i^(2))/(i^(1000))=(-1)/(1)= -1." "[because i^(1000)=(i^(4))^(250)=1]` `(iv)" "i^(-71) = (1)/(i^(71))xx(i)/(i)=(i)/(i^(72))=(i)/((i^(4))^(18))=(i)/(1)= i." "[because i^(4)=1]` `(v)" "(sqrt(-1))^(91)= i^(91) = i^(4 xx 22 + 3) = (i^(4))^(22)xx i^(3) = 1 xx(-i) = -i." "[because i^(3)=i]` `(vi)" "i^(37) = i^(4 xx 9 + 1) = (i^(4))^(9) xx i = 1 xx i = i`. `" "i^(-61)=(1)/(i^(61))xx(i^(3))/(i^(3))=(i^(3))/(i^(64))=(-i)/((i^(4))^(16))=(-i)/(1)=-i`. `" "therefore" "(i^(37)+i^(-61))=i+(-i)=0`. `(vii)" "i^(-1)=(1)/(i)=(1)/(i)xx(i^(3))/(i^(3))=(-i)/(i^(4))=(-i)/(1) = -i`. |
|
| 157. |
Prove that : `(i)" "i^(n) + i^(n+1) + i^(n+2) + i^(n+3) = 0` `(ii) (1+i)^(4) xx (1 + (1)/(i))^(4) = 16`. |
|
Answer» We have `(i)" "i^(n) + i^(n+1) + i^(n+2) + i^(n+3)` `= i^(n)(1 + i + i^(2) + i^(3))` `= i^(n)(1+i-1-i)=(i^(n)xx0)=0." "[because i^(2) = -1 and i^(3) = -i]` `(ii)" "i^(107)+i^(112)+i^(117)+i^(122)` `= i^(107)(1+i^(5)+i^(10)+i^(15))=i^(107)(1+i^(4)xx i + i^(8) xx i^(2) + i^(12) xx i^(3))` `=i^(107)(1 + i + i^(2) + i^(3))" "[because i^(4) = 1, i^(8) = 1 and i^(12) = 1]` `=i^(107)(1+i-1-i)=(i^(107)xx0)=0." "[because i^(2) = -1, i^(3) = -i]` `(1+i)^(4) xx (1 + (1)/(i))^(4)` `=(1+i)^(4) xx (1+(1)/(i)xx(i)/(i))^(4)(1+i)^(4)(1-i)^(4)" "[because i^(2) = -1]` `={(1+i)(1-i)}^(4) = (1-i^(2))^(4) = {1-(-1)}^(4) = 2^(4) = 16`. |
|
| 158. |
If `za n dw`are two complex number such that `|z w|=1a n da rg(z)-a rg(w)=pi/2`, then show that ` z w=-idot` |
|
Answer» Let `z = r_(1) (cos theta_(1) + isin theta_(1)) "and" w = r_(2) (costheta_(2) + isin theta _(2)) ` Also, `|zw| = |z||w| = r_(1) r_(2) = 1 ` `:. r_(1) r_(2) = 1 ` Futher, `arg (z) = theta _(1) and arg (w) = theta_(2)` But ` arg (z) - arg (w) = (pi)/(2)` `rarr theta_(1) - theta_(2) = (pi)/(2)` `rarr arg((z)/(w)) = (pi)/(2)` Now. to prove `barzw = - i` LHS = `barzw` ` = r_(1) (costheta _(1) - isintheta _(1)) r_(2) (costheta_(2) + isin theta_(2))` ` = r_(1)r_(2) [cos(theta _(1) - theta_(2))+isin (theta_(2)-theta_(1) ) ]` ` = r_(1)r_(2) [cos(-pi//2)+isin (-pi//2 ) ]` `1 [0 - i]` `-i = RHS " "Hence proved.` |
|
| 159. |
The real value of `theta` for which the expression `(1 + icos theta)/(1 - 2i cos theta)` is real number isA. `npi+(pi)/(4)`B. `npi+(-1)^(2)(pi)/(4)`C. `2npipm(pi)/(2)`D. None of these |
|
Answer» Correct Answer - C Given expression = `(1+icostheta)/(1-2icostheta)=((1+icostheta)(1+2icostheta))/((1-2icostheta)(1+2icostheta))` `=(1+icostheta+2icostheta+2i^(2)cos^(2)theta)/(1-4i^(2) cos^(2)theta)` `=(1+3icostheta-2cos^(2)theta)/(1-4 cos^(2)theta)` For real value of `theta, (3costheta)/(1+cos^(2)theta)=0` `rArr 3costheta= 0` `rArr costheta = cos(pi)/(2)` `rArr theta= 2npipm(pi)/(2)` |
|
| 160. |
Convert of the complex number in the polar form: `- 3` |
|
Answer» `rcos theta = -3` `rsin theta = 0` `r^2(sin^2 theta + cos^2 theta) = 9` `|r| = 3` `rcos theta = -3` `3 cos theta = -3` `cos theta= -1` `theta = pi` `-3 = 3(cos pi + i sin pi)` answer |
|
| 161. |
Evaluate `sqrt(-i)`. |
|
Answer» Let `sqrt(-i)=(x+iy)." "...(i)` On squaring both sides of (i), we get `(-i)=(x-iy)^(2) rArr -i = (x^(2)-y^(2))-i(2xy)." "...(ii)` On comparing real parts and imaginary parts on both sides of (ii), we get `(x^(2)-y^(2))=0 and 2xy=1` `therefore" "(x^(2)+y^(2))=sqrt((x^(2)-y^(2))^(2)+4x^(2)y^(2))=sqrt(0^(2)+ 1^(2))=sqrt(0+1)=sqrt(1)=1` Thus, `(x^(2)-y^(2))=0" "...(iii) and (x^(2)+y^(2))=1" "...(iv)`. On solving (iii) and (iv), we get `x^(2)=(1)/(2) and y^(2)=(1)/(2)`. `therefore" "x = +-(1)/(sqrt(2))and y = +-(1)/(sqrt(2))`. Since `xy gt 0`, so x and y are of the same sign. `therefore" "(x=(1)/(sqrt(2)),y=(1)/(sqrt(2)))or(x=(-1)/(sqrt(2)),y=(-1)/(sqrt(2)))`. Hence, `sqrt(-i)=((1)/(sqrt(2))-i.(1)/(sqrt(2)))or((-1)/(sqrt(2))+i.(1)/(sqrt(2)))`. |
|
| 162. |
Evaluate: `sqrt(-2+2sqrt(3)i)` |
| Answer» Correct Answer - `(1+sqrt(3)i)or(-1-sqrt(3)i)` | |
| 163. |
Evaluate : `(i)" "sqrt(-25)xx sqrt(-49)" "(ii)" "sqrt(-36)xx sqrt(16)` |
|
Answer» We have `(i)" "sqrt(-25)xxsqrt(-49)=(5i)xx(7i)=(35xxi^(2))=35 xx (-1) = -35`. `(ii)" "sqrt(-36) xx sqrt(16) = (6i) xx 4 = 24 i`. |
|
| 164. |
`|z_(1)+z_(2)|=|z_(1)|+|z_(2)|` is possible, ifA. `z_(2) = barz_(1)`B. `z_(2)=(1)/(z_(1))`C. arg`(z_(1))=arg(z_(2))`D. `|z_(1)|+|z_(2)|` |
|
Answer» Correct Answer - C Given that, `rArr |r_(1)(costheta_(1)+isintheta_(1))+r_(2)(costheta_(2)+isintheta_(2))|=r_(1)(costheta_(1)+isintheta_(2))|+r_(2)(costheta_(2) +isintheta_(2))|` `rArr |r_(1)(costheta_(1)+r_(2)costheta)+i(r_(1)sintheta_(1)+r_(2)sintheta_(2)|=r_(1)+r_(2)` `rArr sqrt(r_(1)^(2)cos^(2)theta_(1)+r_(2)^(2)costheta+2r_(1)r_(2)costheta_(1)costheta_(2) + r_(1)^(2)sin^(2)theta_(1)+r_(2)^(2)sin^(2)theta_(2))` `rArr sqrt(+2r_(1)r_(2)sintheta_(1)+sintheta_(2))=r_(1)+r_(2)` `rArr sqrt(r_(1)^(2)+r_(2)^(2)+ 2r_(1)r_(2)[ cos(theta_(1)-theta_(2))]=r_(1)+r_(2)` On squaring both sides, we get ` r_(1)^(2)+r_(2)^(2)+ 2r_(1)r_(2) cos(theta_(1)-theta_(2))=r_(1)+r_(2)+2r_(1)r_(2)` `rArr 2r_(1)r_(2)[1 -cos(theta_(1)-theta_(2))]=0` `rArr 1-cos(theta_(1)-theta_(2))=1` `rArr cos(theta _(1)-theta_(2)) = cos 0^(@)` `rArr theta_(1)= theta_(2)` arg`(z_(1)) = arg (z_(2)) |
|
| 165. |
Convert the complex number `z=(i-1)/(cospi/3+isinpi/3)`in the polar form. |
|
Answer» Given that, `z =(1-i)/("cos"(pi)/(3)+i " sin" (pi)/(3))=(-sqrt(2)[(-1)/(sqrt(2))+i(1)/sqrt(2)])/("cos"(pi)/(3)+i " sin" (pi)/(3)) ` `= (-sqrt(2)[cos(pi = pi//4) + i sin (pi - pi //4))]/(cos pi//3 + i sin pi//3)` ` = (-sqrt(2) [cos3pi//4 + i sin pi//4]) /(cos pi//3 + i sin pi//3)` `=sqrt(2)[cos((3pi)/(4) - (pi)/(3)) + i sin ((3pi)/(4) - (pi)/(3))]` `=sqrt(2)["cos"(5pi)/(12) + "i sin" (5pi)/(12)]` |
|
| 166. |
If `Arg((z-1)/(z+1))= pi/2` then the locus of z is `) circle with radius 2 2) circle with radius 1 3) straight line 4) pair of lines |
|
Answer» `arg ( z_1/z_2) = theta` `arg z_1 - arg z_2 = theta` `ar((z-1)/(z+1)) = pi/2` `arg(z +1) = theta` `arg(z-1) = pi/2 + theta` in `/_ ZAB` `/_ AZB = 180^@ - /_A - /_B` `= 180^@ - theta - pi/2 + theta` `= pi/2` Answer |
|
| 167. |
Find the square root of the following complex number: `-5+12 i` |
|
Answer» Let `sqrt(-5+12i)=(x+iy)." "...(i)` On squaring both sides of (i), we get `-5+12i=(x+iy)^(2) rArr -5+12i=(x^(2)-y^(2))+i(2xy)." "...(ii)` On comparing real parts and imaginary parts on both sides of (ii), we get `x^(2)-y^(2)=-5 and 2xy = 12` `rArr" "x^(2)-y^(2)=-5 and xy = 6` `rArr" "(x^(2)+y^(2))=sqrt((x^(2)-y^(2))^(2)+4x^(2)y^(2))=sqrt((-5)^(2)+4xx36)=sqrt(169)=13` `rArr" "x^(2)-y^(2)=-5 and x^(2)+y^(2)=13` `rArr" "2x^(2)=8 and 2y^(2)=18` `x^(2)=4 and y^(2)=9` `rArr" "x = +- 2 and y = +- 3`. Since `xy gt 0`, so x and y are of the same sign. `therefore" "(x=2 and y=3) or (x =-2 and y =-3)`. Hence, `sqrt(-5+12i) = (2 + 3i) or (-2 -3i)`. |
|
| 168. |
Convert of the complex number in the polar form: `sqrt(3)+i` |
|
Answer» Here, `z = -sqrt(3) + i` Comparing it with polar form, we get, `rcostheta = sqrt3 and rsintheta = 1` Squaring and adding these two terms, we get, `r^2cos^2theta+r^2sin^2theta = 4 ` `r^2(cos^2theta+sin^2theta) = 4` `r^2 = 4=> r = 2` So, modulus `|r|` is `2`. Now, `rsintheta = 1=>sintheta = 1/2` `theta = pi/6` As `costheta` is positive and `sintheta` is positive, theta lies in first quadrant. So, `theta = pi/6` Polar form, `z = 2(cospi/6+isinpi/6)` |
|
| 169. |
Evaluate: `sqrt(-7+24i)` |
| Answer» Correct Answer - `(3+4i) or (-3-4i)` | |
| 170. |
Evaluate : `sqrt(-16) + 3 sqrt(-25) + sqrt(-36) - sqrt(-625)`. |
|
Answer» We have `sqrt(-16) + sqrt(-25) + sqrt(-36) - sqrt(-625)` `=(4i + 3 xx 5i + 6i - 25i)=(4i + 15i + 6i - 25i) = 0`. |
|
| 171. |
If z is a complex number, thenA. `|z^(2)|gt|z|`B. `|z^(2)| = |z^(2)|`C. `|z^(2)|lt|z|^(2)`D. `|z^(2)|ge|z|^(2)` |
|
Answer» Correct Answer - B If z is a complex number, then `z = x + iy ` `|z| = |x + iy| "and" |z|^(2) = |x +iy|^(2)` `rArr |z|^(2) = x^(2) +y^(2)" "...(i)` and ` z^(2) = (x + iy)^(2) = x^(2) + i^(2) y^(2) + i2xy` `rArr |z^(2)|=sqrt((x^(2) - y^(2) )^(2) + (2xy)^(2))` `rArr |z^(2)| = sqrt(x^(4) + y^(4) - 2x^(2)y^(2) + 4x^(2) y^(2))` `rArr |z^(2)| = sqrt(x^(4) + y^(4) - 2x^(2)y^(2)) =sqrt((x^(2) +y^(2)))^(2)` `rArr |z^(2)| = x^(2) + y^(2)` From Eqs (i) and (ii), `|z|^(2)=|z^(2)|` |
|
| 172. |
Let complex numbers `alpha and 1/alpha` lies on circle `(x-x_0)^2(y-y_0)^2=r^2 and (x-x_0)^2+(y-y_0)^2=4r^2` respectively. If `z_0=x_0+iy_0` satisfies the equation `2|z_0|^2=r^2+2` then `|alpha|` is equal to (a) `1/sqrt2` (b) `1/2` (c) `1/sqrt7` (d) `1/3` |
|
Answer» `|alpha-Z_0|=r-(1)` `|1/alpha-Z_0|=2r` `z|z_0|^2=r^2+2-(3)` `|alpha/|alpha|^2-Z_0|=2r-(2)` `alpha*overlinealpha=|alpha|^2` `1/overlinealpha=alpha/|alpha|^2` squaring equation 1 `|alpha|^2+|Z_0|^2-(alphaoverlineZ_0+overlinealphaZ_0)=r^2` `|alpha|^2+r^2/2+1-(alphaoverlineZ_0+overlinealphaZ_0)=r^2-(4)` `|alpha|^2/|alpha|^4+|Z_0|^2-((alphaoverlineZ_0+overlinealphaZ_0)/|alpha|^2)=4r^2` `1/|alpha|^2(1-(alphaoverlineZ_0+overlinealphaZ_0))=4r^2-|Z_0|^2` `=4r^2-((r^2+2)/2)` `=(7r^2)/2-1` `(1-(alphaoverlineZ_0+overlinealphaZ_o))=|alpha|^2[(7r^2)/2-1]` From equation 4 `|alpha|^2+r^2/2+1-(1+|alpha|^2(1-(7r^2)/2))=r^2` `r^2/2+|alpha|^2*(7r^2)/2=r^2` `|alpha|^2*(7r^2/2)=r^2/2` `|alpha|^2=1/7` `|alpha|=1/sqrt7` Option C is correct. |
|
| 173. |
Locate the complex number z such that `log_[cospi/6][ [|z-2| +5] /[4| z-2 | -4]] |
|
Answer» `(t+5)/(4(t-1))>(sqrt3/2)^2` `(t+5)/(4(t-1)-3/4)>0` `(t+5-3t+3)/(4t-4)>0` `(8-2t)/(4(t-1))>0` `8-2t>0` `2t<8` `t<4` `|z-2|>1,|z-2|<4`. |
|
| 174. |
Find square root of `8-15 i` |
|
Answer» Let `sqrt(8-15i)=(x+iy)." "...(i)` On squaring both sides of (i), we get `(8-15i)=(x-iy)^(2) rArr (8-15i)=(x^(2)-y^(2))-i(2xy)." "...(ii)` On comparing real parts and imaginary parts on both sides of (ii), we get `x^(2)-y^(2)=8 and 2xy = 15` `rArr" "x^(2)-y^(2)=8 and xy = (15)/(2)` `rArr" "(x^(2)+y^(2))=sqrt((x^(2)-y^(2))^(2)+4x^(2)y^(2))=sqrt(64+225)=sqrt(289)=17` `rArr" "x^(2)-y^(2)=8 and x^(2)+y^(2)=17` `rArr" "2x^(2)=25 and 2y^(2)=9` `x^(2)=(25)/(2) and y^(2)=(9)/(2)` `rArr" "x = +- (5)/(sqrt(2)) and y = +- (3)/(sqrt(2))`. Since `xy gt 0`, so x and y are of the same sign. Hence, `sqrt(8-15i) = ((5)/(sqrt(2))-(3)/(2)i)or((-5)/(sqrt(2))+(3)/(sqrt(2))i)`. |
|
| 175. |
If `z=(lambda+3)+isqrt(5-lambda^2)` then the locus of Z is |
|
Answer» `z = (lambda+3) +isqrt(5-lambda^2)` `=>x+iy = (lambda+3) +isqrt(5-lambda^2)` `=> x = lambda+3=> lambda = x-3` `=>lambda^2 = (x-3)^2->(1)` `=>y = sqrt(5-lambda^2) => y^2 = (5-lambda^2)` `=>lambda^2 = 5-y^2->(2)` From (1) and (2), `=> (x-3)^2 = 5-y^2` `=>(x-3)^2 +y^2 = 5` This is an equation of a circle with center `(3,0)` and radius `sqrt5`. So, option -`(b)` is the correct option. |
|
| 176. |
Evaluate `sqrt(24-10i)`. |
|
Answer» Let `sqrt(-24-10i)=(x+iy)." "...(i)` On squaring both sides of (i), we get `(-24-10i)=(x^(2)-y^(2))-i(2xy)." "...(ii)` On comparing real parts and imaginary parts on both sides of (ii), we get `x^(2)-y^(2)=-24 and 2xy = 10` `rArr" "x^(2)-y^(2)=-24 and xy = 5` `rArr" "(x^(2)+y^(2))=sqrt((x^(2)-y^(2))^(2)+4x^(2)y^(2))=sqrt((-24)^(2)+4 xx 24)=sqrt(676)=26` `rArr" "x^(2)-y^(2)=-24 and x^(2)+y^(2)=26` `rArr" "2x^(2)=2 and 2y^(2)=50` `x^(2)=1 and y^(2)=25 rArr x = +-1 and y = +-5`. Since `xy gt 0`, so x and y are of the same sign. `therefore" "sqrt(-24-10i)=(1-5i)or(-1+5i)`. |
|
| 177. |
Evaluate: `sqrt(5+12i)` |
| Answer» Correct Answer - `(3+2i)or(-3-2i)` | |
| 178. |
Simplify : `{:((i),3(6+6i)+i(6+6i),(ii),(1-i)-(-3+6i)),((iii),((1)/(3)-(2)/(3)i)-(4+(3)/(2)i),(iv),{((1)/(5)+(7)/(5)i)-(6+(1)/(5)i)}-((-4)/(5)+i)):}` |
|
Answer» We have `(i)" "3(6+6i)+i(6+6i)=18+18i +6i + 6i^(2)=18 + 24i - 6 = 12 + 24i`. `(ii)" "(1-i)-(-3+6i)=1 - i + 3 - - 6i = 4 - 7i`. `(iii)" "((1)/(3)-(2)/(3)i)-(4+(3)/(3)i)=(1)/(3)-(2)/(3)i - 4 - (3)/(2)i` `" "((1)/(3)-4)-((2)/(3)+(3)/(2))i = (-11)/(3) -(13)/(6)i`. `(iv)" "{((1)/(5)+(7)/(5)i)-(6+(1)/(5)i)}-((-4)/(5)+i)` `" "=((1)/(5)+(7)/(5)i-6-(1)/(5)i)-((-4)/(5)+i)={(1)/(5)-6)+((7)/(5)-(1)/(5))i}-((-4)/(5)+i)` `" "=(-29)/(5)+(6)/(5)i+(4)/(5)-i=((-29)/(5)+(4)/(5))+((6)/(5)i-i)=(-5+(1)/(5)i)`. |
|
| 179. |
If `alpha` and `beta` are the roots of the equation `x^2-2x+4=0`, prove that `alpha^n+beta^n=2^(n+1)cos(npi/3)` |
|
Answer» `x^2-2x+4=0` `(x-1)^2+3=0` `(x-1)^2=-3` `(x-1)=pmsqrt3i` `x=pmsqrt3i+1` `x=2(1/2pmsqrt3/2i)` `x-2(cospi/3pmisinpi/3)` `x=2e^(pmipi/3)` `alpha^n=2e^(ipi/3)`,`beta^n=2^n e^(-i npi/3)` `alpha^n+beta^n=2^n(cosnpi/3+isinnpi/3+cosnpi/3-isinnpi/3)` `=2^n*2*cos(npi/3)` `=2^(n+1) cos(npi/3)`. |
|
| 180. |
The complex numberwhich satisfies the condition `|(i+z)/(i-z)|=1 `lies on`c i r c l e x^2+y^2=1`b. `t h e x-a xi s`c. `t h e y-a xi s`d. `t h e l in e x+y=1`A. Circle `x^(2) + y^(2) = 1`B. the X-axisC. the Y-axisD. the line `x + y = 1` |
|
Answer» Correct Answer - B Given that, `|(i+x)/(i-z)| = 1` Let z = x + iy `:. |(x + i(y+1))/(-x -i(y-1))| = 1 rArr (x^(2) + (y+1)^(2))/(x^(2) + (y-1)^(2)) =1` `rArr x^(2) + (y+1)^(2)= x^(2)(y-1)^(2)` `rArr 4y= 0 rArr y = 0` So, z lies on X-axis (real axis). |
|
| 181. |
Convert the complex number `(-16)/(1+isqrt(3))`into polar form. |
|
Answer» `z = (-16)/(1+isqrt3) = (-16)/(1+isqrt3)**(1-isqrt3)/(1-isqrt3)` `=(-16+16sqrt3i)/(1+3) = -4+4sqrt3i` `:.-4+4sqrt3i = r(costheta+isintheta) ` `rcostheta = -4 and rsintheta = 4sqrt3` `r^2(cos^2theta+sin^2theta) =16+48` `r^2 = 64=> r = 8` `:.costheta = -4/8 = -1/2 and sintheta = 4sqrt3/8 = sqrt3/2` So, `theta = (2pi)/3` so, polar form will be, `z = 8(cos((2pi)/3)+isin((2pi)/3))` |
|
| 182. |
`{sqrt(5+12i)+sqrt(5-12i)}/(sqrt(5+12i)-sqrt(5-12i)``=` |
|
Answer» We have `z=(sqrt(5+12i)+sqrt(5-12i))/(sqrt(5+12i)-sqrt(5-12i))` `=((sqrt(5+12i)+sqrt(5-12i)))/((sqrt(5+12i)-sqrt(5-12i)))xx((sqrt(5+12i)+sqrt(5-12i)))/((sqrt(5+12i)+sqrt(5-12i)))` `=((sqrt(5+12i)+sqrt(5-12i))^(2))/((5+12i)-(5-12i))=((5+12i)+(5-12i)+2sqrt((5)^(2)-(12i)^(2)))/(24i)` `=(10+2 sqrt(25+144))/(24i)=(10+2 sqrt(169))/(24i)=((10+2xx13))/(24i)` `=(36)/(24i)=(3)/(2i)=(3)/(2i)xx(i)/(i)=(3i)/(2i^(2))=-(3)/(2)i`. Hence, `z =(0-(3)/(2)i) and bar(z) = bar((0-(3)/(2)i))=(0+(3)/(2)i)`. |
|
| 183. |
Simplify each of the following and express it in the form a + ib : `{:((i),2(3+4i)+i(5-6i),(ii),(3+sqrt(-16))-(4-sqrt(-9))),((iii),(-5+6i)-(2+i),(iv),(8-4i)-(-3+5i)),((v),(1-i)^(2)(1+i)-(3-4i)^(2),(vi),(5+sqrt(-3))(5-sqrt(-3))),((vii),(3+4i)(2-3i),(viii),(-2+sqrt(-3))(-3+2 sqrt(-3))):}` |
| Answer» Correct Answer - `{:((i),(12+13i),(ii),(-1+7i),(iii),(-3+5i),(iv),"(11-9i)"),((v),(9+22i),(vi),28,(vii),(18-i),(vii),-7sqrt(3)i):}` | |
| 184. |
If `alpha` is non real and `alpha=1^[1/5]` then value of `2^|1+alpha+alpha^2+alpha^-2-alpha^-1|` is equal to |
|
Answer» `|1+alpha+alpha^2+1/alpha^2-1/alpha|` `|(alpha^2+alpha^3+alpha^4+1+alpha-alpha-alpha)/alpha^2|` `|(-2alpha)/alpha^2|` `|-2/alpha|` `2^|(-2/alpha)|` `2^(2/|alpha|)` `2^2` `4`. |
|
| 185. |
The condition for `ax^2 +2cxy+by^2 +2bx+2ay+c` is resolvable into two linear factors is |
|
Answer» `ax^2+2hxy+by^2+2yx+2fy+c=0` `/_=abc+2fgh-af^2-bg^2-ch^2=0` `=abc+2abc-a^3-b^3-c^3=0` `=a^3+b3+c^3=3abc`. |
|
| 186. |
If P and Q are two complex numbers then the modulus of the quotient of P and Q is (a) greater than the quotient of their moduli (b) less than the quotient of their moduli (c) less than or equal to the quotient of their moduli (d) equal to the quotient of their moduli |
|
Answer» `P=x_1+iy_1` `Q=x_2+iy_2` `P/Q=(x_1+iy_1)/(x_2+iy_2)` `|P/Q|=|P|/|Q|=sqrt(x_1^2+y_1^2)/sqrt(x_2^2+y_2^2` Option D is correct. |
|
| 187. |
Separate `((3+sqrt(-1))/(2-sqrt(-1)))` into real and imaginary parts and hence find it modulus. |
|
Answer» Let `z=((3+sqrt(-1))/(2-sqrt(-1)))=((3+i)/(2-i))=((3+i)/(2-i))=((3+i))/((2-i))xx((2+i))/((2+i))` `((3+i)(2+i))/((2-i)(2+i))=((6+i^(2))+5i)/((4-i^(2)))=(5+5i)/({4-(-1)})=(5(1+i))/(5)=(1+i)`. `therefore" "|z|=sqrt(1^(2)+1^(2))=sqrt(2)`. Hence, `z = (1 + i) and |z| = sqrt(2)`. |
|
| 188. |
If `a`, `b` are real and `a^2+b^2=1`, then show that the equation `(sqrt(1+x)-isqrt(1-x))/(sqrt(1+x)+isqrt(1-x))=a-ib` is satisfied y a real value of `x`. |
|
Answer» let x e real `(sqrt(1+x)-isqrt(1-x))/(sqrt(1+x)+isqrt(1-x))` `{(sqrt(1+x)-isqrt(1-x))/((a+x)+(1-x))}` `(1+x-1_x-2isqrt(1-x^2))/2` `(2x-2isqrt(1-x^2))/2` `x-isqrt(1-x^2)` `a=x` `b=sqrt(1-x^2)` `a^2+b^2=x^2+1-x^2=1`. |
|
| 189. |
If `sqrt((1+i)/(1-i))=(a+ib)` then show that `(a^(2)+b^(2))=1`. |
|
Answer» We have `(a+ib)=sqrt((1+i)/(1-i))=(sqrt(1+i))/(sqrt(1-i))xx(sqrt(1+i))/(sqrt(1+i))=((1+i))/(sqrt(1-i^(2)))` `=((1+i))/(sqrt(2))=((1)/(sqrt(2))+(1)/(sqrt(2))i)` `rArr" "|a+ib|^(2)=((1)/(sqrt(2)))^(2)+((1)/(sqrt(2)))^(2)=((1)/(2)+(1)/(2))=1`. `rArr" "(a^(2)+b^(2)) = 1`. Hence, `(a^(2)+b^(2))=1`. |
|
| 190. |
`z_1 and z_2`, lie on a circle with centre at origin. The point of intersection of the tangents at`z_1 and z_2` is given by |
|
Answer» `|Z_2|^2+|Z_3-Z_2|^2=|Z_3|^2` `|Z_1|^2+|Z_3-Z_1|^2=|Z_3|^2` `Z_2Z_2+(Z_3-Z_2)*(Z_3-Z_2)=Z_2Z_3` `Z_2Z_2+Z_3Z_3-Z_2Z_3-Z_3Z_2=Z_3Z_3` `2Z_2-Z_3-Z_3Z_2/Z_3=0-(1)` `2Z_1-Z_3-Z_3Z_1/Z_2=0-(2)` subtracting equation 2 from equation 1 `2(Z_2-Z_1)-Z_3(Z_2/Z_2-Z_1/Z_1)=0` `Z_3=(2(Z_2-Z_1)Z_1Z_2)/(Z_1Z_2-Z_1Z_2)` option b is correct. |
|
| 191. |
If zis a complex number, then `|3z-1|=3|z-2|` represents |
|
Answer» Let `z = x+iy` Then, `|3(x+iy)-1| = 3|x+iy-2|` `=>|(3x-1)+3iy| = 3|(x-2)+iy|` `=>sqrt((3x-1)^2+(3y)^2) = 3sqrt((x-2)^2+y^2)` Now, squaring both sides, `=>(3x-1)^2+(3y)^2 = 9((x-2)^2+y^2)` `=>9x^2+1- 6x +9y^2 = 9x^2+36-36x+9y^2` `=> 30x = 35` `=> x = 7/6` This is an equation of a line parallel to `y`-axis. So, option - `(d)` is the correct option. |
|
| 192. |
If `alpha, beta , gamma, delta` are the roots of the equation `x^4+x^2+1=0` then the equation whose roots are `alpha^2, beta^2, gamma^2, delta^2` is |
|
Answer» let `x^2=t` `t^2+t+1` `1+w+w^2=0` `x=e^(ipi/3),-e^(ipi/3),e^(i2.3pi)...` `alpha^2beta^2gamma^2rho^2=e^(12/3pi)...` `w^6=1` `alpha^2+beta^2+gamma^2+rho^2=e^(i2/3pi),e^(i2/3pi)+e^(i4/3pi+e^(i4/3pi)` `=2(w+w^2)=-2` option b is correct. |
|
| 193. |
What is the argument of the complex number `(-1-i)` where `i=sqrt(-1)`? |
|
Answer» `Z=-1-hati` `-x<=avg(Z)<=pi` `tan(pi/4)=1` From diagram`avg(Z)=x+pi/4` `avg(Z)=-3/4pi`. |
|
| 194. |
If `ax^2 + bx + c = 0` is satisfied by every real value of x, then |
|
Answer» Let `y = ax^2+bx+c` So, with the given equation,`ax^2+bx+c = 0` `y = 0` If we draw this, it will be a straight line along `X`-axis.It means, ` c = 0` For, any real value of `x`, `ax^2+bx` will be `0`, only when `a = b = 0` So, `a = b = c = 0` So, option `(c)` is the correct option. |
|
| 195. |
If `((1+i)/(1-i))^m=1,`then find the least positive integral value of `mdot` |
|
Answer» We have `((1+i)/(1-i))=((1+i))/((1-i))xx((1+i))/((1+i))=((1+i)^(2))/((1-i^(2)))=((1+2i+i^(2)))/(2)=(2i)/(2)=i`. `therefore" "((1+i)/(1-i))^(m)=1 rArr i^(m) = 1`. And, we know that 4 is the least positive integer such that `i^(4) = 1` and therefore m = 4. |
|
| 196. |
If the ratio of the roots of `x^2 + bx + c =0` and ` x^2 + qx + r = 0` is same then |
|
Answer» Let `alpha,beta` are the roots of `x^2+bx+c = 0` Then, `alpha+beta =-b` and `alphabeta = c` Let `A,B` are the roots of `x^2+qx+r = 0` Then, `A+B =-q` and `AB = r` We are given, `alpha/beta = A/B->(1)` `=>beta/alpha = B/A->(2)` Adding (1) and (2), `alpha/beta+beta/alpha = A/B+B/A` `=>(alpha^2+beta^2)/(alphabeta) = (A^2+B^2)/(AB)` `=>((alpha+beta)^2-2alphabeta)/(alphabeta) = ((A+B)^2-2AB)/(AB)` `=>(b^2-2c)/c = (q^2-2r)/r` `=>b^2r-2cr = q^2c-2cr` `=>b^2r = q^2c` So, option `C` is the correct option. |
|
| 197. |
If `(5z_1)/(7z_2)` is purely imaginary then `|(2z_1+3z_2)/(2z_1-3z_2)|=` |
|
Answer» `z_1=r_1e^(itheta_1)` `z_2=r_1e^(itheta_2)` `5z_1/7z_2->(5r_1e^(itheta_1))/(7r_2e^(itheta_2)` `theta_1-theta_2=2npipmpi/2` `|(2z_1+3z_2)/(2z_1-3z_2)|` `|(1+3/2z_2/z_1)/(1-3/2z_2/z_1)|` `|(1+3/2x i)/(1-3/2x i)|` `sqrt(1^2+(3/2x)^2)/sqrt(1^2+(-3/2x)^2)` `1` |
|
| 198. |
Let `alpha` and `beta(alpha |
|
Answer» `x^2+bx+c=0` `x=(-bpmsqrt(b^2-4c))/2` `b^2-4c>b^2` `sqrt(b^2-4c)>|b|` `|alpha|>beta` `-alpha>beta` `alpha+beta=alphabeta-b+C<0` `alpha^2beta+alphabeta^2=alphabeta(alpha+beta)<0` option 2 is correct. |
|
| 199. |
Statement-I Given equation `x^6-12x^5+ax^4+bx^3+cx^2+dx+64=0` has six positive roots `a_i(i=1,2,......6)` then the value of `|a+b|=100`Statement-II For the equation `x^6-12x^5+ax^4+bx^3+cx^2+dx+64=0` with six positive roots `a_i(i=1,2...6). `a=sum a_i a_2 and b=- suma_1 a_2 a_3`. |
|
Answer» `alpha_!+alpha_2+alpha_3+alpha_4+alpha_5+alpha_6=12` `alpha_1alpha_2alpha_3alpha_4alpha_5alpha_6=64` `AM=12/6=2` `GM=(64)^(1/6)=2` `AM=GM` `alpha_1=alpha+2=alpha_3=alpha_4=alpha_5=alpha+6=2` `a=6C_2*4=60` `b=-6C_3*6=-160` `|a+b|=100` option 1 is correct. |
|
| 200. |
If z = 2 + i, prove that `z^(3) + 3z^(2) - 9z + 8 = (1 + 14i)`. |
|
Answer» We have `z = 2 + i rArr z - 2 = i rArr (z-2)^(2) = i^(2) rArr z^(2) - 4z + 5 = 0." "...(i)` `therefore" "z^(3) + 3z^(2) - 9z + 8` `= z(z^(2)-4z+5) + 7(z^(2) - 4z + 5) + 14z - 27` `=(z xx 0) + (7 xx 0) + 14z - 27 = (14z - 27)" "["using (i)"]` `= 14 (2 + i) - 27 = (1 + 14i)`. Hence, `z^(3) + 3z^(2) - 9z + 8 = (1 + 14i)`. |
|