PYQs Learn Notes Mocks Board Questions MHT-CET Maths Complex Numbers 67 MHT-CET Maths previous year questions (PYQs) from 65 papers, 2021 to 2026, with answers and worked solutions.
About this chapter Difficulty: 6 easy · 45 moderate · 16 hard. Most-asked subtopics: Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity (30), Modulus and Argument — Polar Form, De Moivre and Square Roots (26), Locus in the Argand Plane — Circles, Lines and Greatest/Least Modulus (11). Updated 13 October 2026. #1 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If n ∈ Z n \in Z n ∈ Z , then the value of 2 n ( 1 + i ) 2 n + 2 n ( 1 − i ) 2 n + ( 1 + i ) 2 n + ( 1 − i ) 2 n 2 n \dfrac{2^n}{(1+i)^{2n}} + \dfrac{2^n}{(1-i)^{2n}} + \dfrac{(1+i)^{2n} + (1-i)^{2n}}{2^n} ( 1 + i ) 2 n 2 n + ( 1 − i ) 2 n 2 n + 2 n ( 1 + i ) 2 n + ( 1 − i ) 2 n is equal to A 4 cos ( n π 2 ) 4\cos\left(\dfrac{n\pi}{2}\right) 4 cos ( 2 nπ ) B 2 ( n + 1 ) cos ( n π 2 ) 2(n+1)\cos\left(\dfrac{n\pi}{2}\right) 2 ( n + 1 ) cos ( 2 nπ ) C 2 [ 1 + ( − 1 ) n ] 2[1 + (-1)^n] 2 [ 1 + ( − 1 ) n ] D #2 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If ∣ z − 5 i ∣ = ∣ z + 3 ∣ |z - 5i| = |z + 3| ∣ z − 5 i ∣ = ∣ z + 3∣ , where z z z is a complex number, then the equation of the locus of z z z in Cartesian coordinates is ... A 3 x + 5 y − 8 = 0 3x + 5y - 8 = 0 3 x + 5 y − 8 = 0 B 3 x − 5 y − 8 = 0 3x - 5y - 8 = 0 3 x − 5 y − 8 = 0 C − 3 x + 5 y − 8 = 0 -3x + 5y - 8 = 0 − 3 x + 5 y − 8 = 0 D 3 x + 5 y + 8 = 0 3x + 5y + 8 = 0 3 x + 5 y + 8 = 0 #3 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If ω \omega ω is a complex cube root of unity, then the value of the expression ( 1 − ω + ω 2 ) 7 + ( 1 + ω − ω 2 ) 7 (1 - \omega + \omega^2)^7 + (1 + \omega - \omega^2)^7 ( 1 − ω + ω 2 ) 7 + ( 1 + ω − ω 2 ) 7 is equal to... #4 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
The value of ∑ n = 1 4 ( i n + i n + 1 ) \displaystyle\sum_{n=1}^{4}\left(i^n + i^{n+1}\right) n = 1 ∑ 4 ( i n + i n + 1 ) is #5 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If x = 25 3 − 4 i x = \dfrac{25}{3 - 4i} x = 3 − 4 i 25 then 2 x 3 − 11 x 2 + 44 x + 27 = 2x^3 - 11x^2 + 44x + 27 = 2 x 3 − 11 x 2 + 44 x + 27 = Done a few? Here’s your next move
MHT-CET · 2026 · 12 May Shift I · Q56 Moderate Add #6 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If n ∈ Z n \in Z n ∈ Z , then the expression 2 n ( 1 − i ) 2 n + ( 1 + i ) 2 n 2 n \dfrac{2^n}{(1 - i)^{2n}} + \dfrac{(1 + i)^{2n}}{2^n} ( 1 − i ) 2 n 2 n + 2 n ( 1 + i ) 2 n is equal to ... A B C [ 1 + ( − 1 ) n ] i n [1 + (-1)^n]i^n [ 1 + ( − 1 ) n ] i n D #7 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If i = − 1 i = \sqrt{-1} i = − 1 then [ i 18 + ( 1 i ) 25 ] 3 = \left[i^{18} + \left(\dfrac{1}{i}\right)^{25}\right]^3 = [ i 18 + ( i 1 ) 25 ] 3 = #8 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If w w w is a complex cube root of unity , then the value of w 10 − w 7 + w 5 − w 2 + 1 w^{10} - w^7 + w^5 - w^2 + 1 w 10 − w 7 + w 5 − w 2 + 1 is. #9 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
The smallest positive integer n n n for which ( 1 + i ) n ( 1 − i ) n − 2 \dfrac{(1 + i)^n}{(1 - i)^{n-2}} ( 1 − i ) n − 2 ( 1 + i ) n is a real number, is ... MHT-CET · 2026 · 18 April Shift I · Q51 Easy Add #10 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
The value of ( − 1 + i 3 2 ) 18 + ( − 1 − i 3 2 ) 18 \left(\dfrac{-1 + i\sqrt{3}}{2}\right)^{18} + \left(\dfrac{-1 - i\sqrt{3}}{2}\right)^{18} ( 2 − 1 + i 3 ) 18 + ( 2 − 1 − i 3 ) 18 is A B C D 3 2 \dfrac{\sqrt{3}}{2} 2 3 #11 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If α \alpha α and β \beta β are the distinct roots of the equation x 2 − x + 1 = 0 x^2 - x + 1 = 0 x 2 − x + 1 = 0 , then the value of α 200 + β 206 + 2 \alpha^{200} + \beta^{206} + 2 α 200 + β 206 + 2 is equal to #12 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If ω \omega ω is a complex cube root of unity, then the value of sin [ π ( ω 10 + ω 23 ) − π 4 ] = \sin\left[\pi(\omega^{10} + \omega^{23}) - \dfrac{\pi}{4}\right] = sin [ π ( ω 10 + ω 23 ) − 4 π ] = A − 3 2 -\dfrac{\sqrt{3}}{2} − 2 3 B − 1 2 -\dfrac{1}{\sqrt{2}} − 2 1 C 1 2 \dfrac{1}{\sqrt{2}} 2 1 D 3 2 \dfrac{\sqrt{3}}{2} 2 3 #13 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If x = − 1 − − 1 − − 1 − … ∞ x = \sqrt{-1 - \sqrt{-1 - \sqrt{-1 - \ldots\infty}}} x = − 1 − − 1 − − 1 − … ∞ , where ω \omega ω is a non-real complex cube root of unity, then the value of x x x is... #14 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If α \alpha α and β \beta β are the roots of the equation x 2 + x + 1 = 0 x^2 + x + 1 = 0 x 2 + x + 1 = 0 then α 2026 + β 2026 = \alpha^{2026} + \beta^{2026} = α 2026 + β 2026 = #15 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If ω \omega ω is a complex cube root of unity, then the value of the expression 2 ( 1 + 1 ω ) ( 1 + 1 ω 2 ) + 3 ( 2 + 1 ω ) ( 2 + 1 ω 2 ) + … + ( n + 1 ) ( n + 1 ω ) ( n + 1 ω 2 ) 2\left(1+\dfrac{1}{\omega}\right)\left(1+\dfrac{1}{\omega^2}\right)+3\left(2+\dfrac{1}{\omega}\right)\left(2+\dfrac{1}{\omega^2}\right)+\ldots+(n+1)\left(n+\dfrac{1}{\omega}\right)\left(n+\dfrac{1}{\omega^2}\right) 2 ( 1 + ω 1 ) ( 1 + ω 2 1 ) + 3 ( 2 + ω 1 ) ( 2 + ω 2 1 ) + … + ( n + 1 ) ( n + ω 1 ) ( n + ω 2 1 ) is... A [ n ( n + 1 ) 2 ] 2 + n \left[\dfrac{n(n+1)}{2}\right]^2+n [ 2 n ( n + 1 ) ] 2 + n B [ n ( n + 1 ) 2 ] 2 − n \left[\dfrac{n(n+1)}{2}\right]^2-n [ 2 n ( n + 1 ) ] 2 − n C [ n ( n + 1 ) 2 ] 2 \left[\dfrac{n(n+1)}{2}\right]^2 [ 2 n ( n + 1 ) ] 2 D [ n ( n − 1 ) 2 ] 2 \left[\dfrac{n(n-1)}{2}\right]^2 [ 2 n ( n − 1 ) ] 2 MHT-CET · 2025 · 25 April Shift I · Q129 Hard Add #16 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
Let z be the complex number with I m ( z ) = 10 Im(z) = 10 I m ( z ) = 10 and satisfying 2 z − n 2 z + n = 2 i − 1 \frac{2z-n}{2z+n}= 2i- 1 2 z + n 2 z − n = 2 i − 1 , where i = − 1 i=\sqrt{- 1} i = − 1 , for some natural number ' n n n ' then A n = 20 n = 20 n = 20 and R e ( z ) = 10 Re(z) = 10 R e ( z ) = 10 B n = 20 n = 20 n = 20 and R e ( z ) = − 10 Re(z) = - 10 R e ( z ) = − 10 C n = 40 n = 40 n = 40 and R e ( z ) = 10 Re(z) = 10 R e ( z ) = 10 D n = 40 n = 40 n = 40 and R e ( z ) = − 10 Re(z) = - 10 R e ( z ) = − 10 #17 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If x = − 2 + − 3 x= - 2 +\sqrt{- 3} x = − 2 + − 3 , then the value of 2 x 4 + 5 x 3 + 7 x 2 − x + 38 2x^{4}+ 5x^{3}+ 7x^{2}-x+ 38 2 x 4 + 5 x 3 + 7 x 2 − x + 38 is equal to #18 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
z = 3 + 2 i s i n θ 1 − 2 i s i n θ , ( i = − 1 ) z =\frac{3 + 2isin\theta}{1 - 2isin\theta},(i =\sqrt{- 1}) z = 1 − 2 i s in θ 3 + 2 i s in θ , ( i = − 1 ) will be purely imaginary if θ = \theta = θ = A 2 n π ± π 8 2n\pi \pm \frac{\pi}{8} 2 nπ ± 8 π , where n ∈ Z n \in \mathbb{Z} n ∈ Z B n π + π 8 n\pi + \frac{\pi}{8} nπ + 8 π , where n ∈ Z n \in \mathbb{Z} n ∈ Z C n π ± π 3 n\pi \pm \frac{\pi}{3} nπ ± 3 π , where n ∈ Z n \in \mathbb{Z} n ∈ Z D n π n\pi nπ , where n ∈ Z n \in \mathbb{Z} n ∈ Z #19 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
The complex numbers sin x + i c o s 2 x \sin x+icos2x sin x + i cos 2 x and cos x − i s i n 2 x , ( i = − 1 ) \cos x- isin2x,(i =\sqrt{- 1}) cos x − i s in 2 x , ( i = − 1 ) are conjugate to each other for, A x = n π , n ∈ Z \ x = n\pi,n \in \mathbb{Z} x = nπ , n ∈ Z B x = ( n + 1 2 ) π , n ∈ Z \ x = \left( n + \frac{1}{2} \right)\pi,n \in \mathbb{Z} x = ( n + 2 1 ) π , n ∈ Z C x = ( 3 n − 1 ) π , n ∈ Z x = (3n - 1)\pi,n \in \mathbb{Z} x = ( 3 n − 1 ) π , n ∈ Z D #20 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
Let ω = − 1 2 + i 3 2 , i = − 1 \omega = -\frac{1}{2}+i\frac{\sqrt{3}}{2},\ i=\sqrt{-1} ω = − 2 1 + i 2 3 , i = − 1 , then the value of ∣ 1 1 1 1 − 1 − ω 2 ω 2 1 ω 2 ω 4 ∣ \begin{vmatrix}1&1&1\\1&-1-\omega^2&\omega^2\\1&\omega^2&\omega^4\end{vmatrix} 1 1 1 1 − 1 − ω 2 ω 2 1 ω 2 ω 4 is A B 3 ω ( ω − 1 ) 3\omega(\omega-1) 3 ω ( ω − 1 ) C D 3 ω ( 1 − ω ) 3\omega(1-\omega) 3 ω ( 1 − ω ) MHT-CET · 2023 · 10th May Shift 2 · Q120 Moderate Add #21 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If ω = − 1 − i 3 2 \omega = \frac{-1-i\sqrt{3}}{2} ω = 2 − 1 − i 3 where i = − 1 i = \sqrt{-1} i = − 1 , then the value of ∣ 1 ω ω 2 ω ω 2 1 ω 2 1 ω ∣ \begin{vmatrix}1&\omega&\omega^2\\\omega&\omega^2&1\\\omega^2&1&\omega\end{vmatrix} 1 ω ω 2 ω ω 2 1 ω 2 1 ω is #22 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If w = − 1 + i 3 2 w=\frac{-1+i\sqrt{3}}{2} w = 2 − 1 + i 3 , where i = − 1 i=\sqrt{-1} i = − 1 , then the value of ( 3 + w + 3 w 2 ) 4 (3+w+3w^2)^4 ( 3 + w + 3 w 2 ) 4 is #23 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
Let ( − 2 − 1 3 i ) 3 = x + i y 27 \left(-2-\dfrac{1}{3}i\right)^{3}=\dfrac{x+iy}{27} ( − 2 − 3 1 i ) 3 = 27 x + i y , i = − 1 i=\sqrt{-1} i = − 1 , where x x x and y y y are real numbers, then ( y − x ) (y-x) ( y − x ) has the value MHT-CET · 2023 · 3rd May Shift 2 · Q149 Moderate Add #24 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If z 2 + z + 1 = 0 z^{2}+z+1=0 z 2 + z + 1 = 0 then ( z 3 + 1 z 3 ) 2 + ( z 4 + 1 z 4 ) 2 = \left(z^{3}+\frac{1}{z^{3}}\right)^{2}+\left(z^{4}+\frac{1}{z^{4}}\right)^{2}= ( z 3 + z 3 1 ) 2 + ( z 4 + z 4 1 ) 2 = where z = ω = z=\omega= z = ω = complex cube root of unity #25 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
Let z ∈ C z \in \mathbb{C} z ∈ C with Im ( z ) = 10 \text{Im}(z) = 10 Im ( z ) = 10 and it satisfies 2 z − n 2 z + n = 2 i − 1 \frac{2z-n}{2z+n} = 2i-1 2 z + n 2 z − n = 2 i − 1 for some natural number n, then A n = 20 n=20 n = 20 and Re ( z ) = − 10 \text{Re}(z) = -10 Re ( z ) = − 10 B n = 40 n=40 n = 40 and Re ( z ) = − 10 \text{Re}(z) = -10 Re ( z ) = − 10 C n = 40 n=40 n = 40 and Re ( z ) = 10 \text{Re}(z) = 10 Re ( z ) = 10 D n = 20 n=20 n = 20 and Re ( z ) = 10 \text{Re}(z) = 10 Re ( z ) = 10 #26 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If ( 3 x + 2 ) − ( 5 y − 3 ) i (3x+2)-(5y-3)i ( 3 x + 2 ) − ( 5 y − 3 ) i and ( 6 x + 3 ) + ( 2 y − 4 ) i (6x+3)+(2y-4)i ( 6 x + 3 ) + ( 2 y − 4 ) i are conjugates of each other, then the value of x − y x + y \frac{x-y}{x+y} x + y x − y is #27 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If x = 5 1 − 2 i x = \frac{5}{1-2i} x = 1 − 2 i 5 , i = − 1 i=\sqrt{-1} i = − 1 , then the value of x 3 + x 2 − x + 22 x^3+x^2-x+22 x 3 + x 2 − x + 22 is #28 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If z = x + i y z=x+iy z = x + i y and z 1 / 3 = p + i q z^{1/3}=p+iq z 1/3 = p + i q , where x , y , p , q ∈ R x,y,p,q\in\mathbb{R} x , y , p , q ∈ R and i = − 1 i=\sqrt{-1} i = − 1 , then value of x p + y q \frac{x}{p}+\frac{y}{q} p x + q y is A B 4 ( p 2 − q 2 ) 4(p^2-q^2) 4 ( p 2 − q 2 ) C D 4 ( p 2 + q 2 ) 4(p^2+q^2) 4 ( p 2 + q 2 ) #29 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If Z 1 = 2 + i Z_1 = 2+i Z 1 = 2 + i and Z 2 = 3 − 4 i Z_2 = 3-4i Z 2 = 3 − 4 i and Z 1 ‾ Z 2 ‾ = a + b i \dfrac{\overline{Z_1}}{\overline{Z_2}} = a+bi Z 2 Z 1 = a + bi , then the value of − 7 a + b -7a + b − 7 a + b is (where i = − 1 i = \sqrt{-1} i = − 1 and a , b ∈ R a,b \in \mathbb{R} a , b ∈ R ) MHT-CET · 2024 · 9th May Shift 1 · Q110 Moderate Add #30 · Algebra of Complex Numbers — Conjugates, Powers of i and Cube Roots of Unity
If Z 1 = 4 i 40 − 5 i 35 + 6 i 17 + 2 Z_1=4i^{40}-5i^{35}+6i^{17}+2 Z 1 = 4 i 40 − 5 i 35 + 6 i 17 + 2 , Z 2 = − 1 + i Z_2=-1+i Z 2 = − 1 + i , where i = − 1 i=\sqrt{-1} i = − 1 , then ∣ Z 1 + Z 2 ∣ = |Z_1+Z_2|= ∣ Z 1 + Z 2 ∣ = #31 · Modulus and Argument — Polar Form, De Moivre and Square Roots
If z 1 = 3 ⋅ e 5 π 12 i z_1 = 3 \cdot e^{\frac{5\pi}{12}i} z 1 = 3 ⋅ e 12 5 π i and z 2 = 4 ⋅ e π 12 i z_2 = 4 \cdot e^{\frac{\pi}{12}i} z 2 = 4 ⋅ e 12 π i then z 1 ⋅ z 2 = ⋅ s ⋅ s z_1 \cdot z_2 = \cdot s\cdot s z 1 ⋅ z 2 = ⋅ s ⋅ s A B C e π 2 i e^{\frac{\pi}{2}i} e 2 π i D #32 · Modulus and Argument — Polar Form, De Moivre and Square Roots
If z = 3 + i 2 z = \dfrac{\sqrt{3} + i}{2} z = 2 3 + i , then the value of z 69 z^{69} z 69 is #33 · Modulus and Argument — Polar Form, De Moivre and Square Roots
The modulus and argument of ( 1 + i 3 ) 8 (1 + i\sqrt{3})^8 ( 1 + i 3 ) 8 are respectively... A B 256 256 256 and 7 π / 3 7\pi/3 7 π /3 C 256 256 256 and 2 π / 3 2\pi/3 2 π /3 D 64 64 64 and 4 π / 3 4\pi/3 4 π /3 #34 · Modulus and Argument — Polar Form, De Moivre and Square Roots
If z = ∑ n = 0 2026 i n z = \sum_{n=0}^{2026} i^n z = ∑ n = 0 2026 i n , where i = − 1 i = \sqrt{-1} i = − 1 , then one of the values of z \sqrt{z} z is... A e i π 4 e^{i\frac{\pi}{4}} e i 4 π B 1 2 e i π 4 \dfrac{1}{\sqrt{2}}e^{i\frac{\pi}{4}} 2 1 e i 4 π C e i π 2 e^{i\frac{\pi}{2}} e i 2 π D 1 2 e i π 2 \dfrac{1}{\sqrt{2}}e^{i\frac{\pi}{2}} 2 1 e i 2 π #35 · Modulus and Argument — Polar Form, De Moivre and Square Roots
( cos 2 θ + i sin 2 θ ) 7 ( cos 4 θ + i sin 4 θ ) 3 = \dfrac{(\cos 2\theta + i\sin 2\theta)^7}{(\cos 4\theta + i\sin 4\theta)^3} = ( cos 4 θ + i sin 4 θ ) 3 ( cos 2 θ + i sin 2 θ ) 7 = A cos 2 θ + i sin 2 θ \cos 2\theta + i\sin 2\theta cos 2 θ + i sin 2 θ B ( cos 2 θ + i sin 2 θ ) 4 (\cos 2\theta + i\sin 2\theta)^4 ( cos 2 θ + i sin 2 θ ) 4 C cos 4 θ + i sin 4 θ \cos 4\theta + i\sin 4\theta cos 4 θ + i sin 4 θ D ( cos 4 θ + i sin 4 θ ) − 2 (\cos 4\theta + i\sin 4\theta)^{-2} ( cos 4 θ + i sin 4 θ ) − 2 #36 · Modulus and Argument — Polar Form, De Moivre and Square Roots
If n n n is a positive integer, then ( 1 + i 3 ) 2 n + ( 1 − i 3 ) 2 n (1 + i\sqrt{3})^{2n} + (1 - i\sqrt{3})^{2n} ( 1 + i 3 ) 2 n + ( 1 − i 3 ) 2 n is equal to ......... A 2 2 n + 1 cos ( 2 n π 3 ) 2^{2n+1}\cos\left(\dfrac{2n\pi}{3}\right) 2 2 n + 1 cos ( 3 2 nπ ) B 2 2 n cos ( n π 3 ) 2^{2n}\cos\left(\dfrac{n\pi}{3}\right) 2 2 n cos ( 3 nπ ) C 2 2 n + 1 cos ( n π 3 ) 2^{2n+1}\cos\left(\dfrac{n\pi}{3}\right) 2 2 n + 1 cos ( 3 nπ ) D 2 n cos ( 2 n π 3 ) 2^n\cos\left(\dfrac{2n\pi}{3}\right) 2 n cos ( 3 2 nπ ) #37 · Modulus and Argument — Polar Form, De Moivre and Square Roots
If ( 1 + i ) ⋅ ( 1 + 2 i ) … … … ( 1 + n i ) = x + i y (1 + i)\cdot(1 + 2i)\ldots\ldots\ldots(1 + ni) = x + iy ( 1 + i ) ⋅ ( 1 + 2 i ) ……… ( 1 + ni ) = x + i y (Where i = − 1 i = \sqrt{-1} i = − 1 ), then the value of ( 2 ) ⋅ ( 5 ) ⋅ ( 10 ) … … … ( 1 + n 2 ) (2)\cdot(5)\cdot(10)\ldots\ldots\ldots(1 + n^2) ( 2 ) ⋅ ( 5 ) ⋅ ( 10 ) ……… ( 1 + n 2 ) A x + y 2 \dfrac{\sqrt{x} + \sqrt{y}}{2} 2 x + y B C x + y x − y \dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} - \sqrt{y}} x − y x + y D #38 · Modulus and Argument — Polar Form, De Moivre and Square Roots
The polar form of the complex number z = 1 1 + i z = \dfrac{1}{1 + i} z = 1 + i 1 , (where i = − 1 i = \sqrt{-1} i = − 1 ) is A 1 2 ( cos 7 π 4 + i sin 7 π 4 ) \dfrac{1}{\sqrt{2}}\left(\cos\dfrac{7\pi}{4} + i\sin\dfrac{7\pi}{4}\right) 2 1 ( cos 4 7 π + i sin 4 7 π ) B 1 2 ( cos π 4 + i sin π 4 ) \dfrac{1}{\sqrt{2}}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) 2 1 ( cos 4 π + i sin 4 π ) C 1 2 ( cos π 4 + i sin π 4 ) \dfrac{1}{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) 2 1 ( cos 4 π + i sin 4 π ) D 1 2 ( cos 7 π 4 + i sin 7 π 4 ) \dfrac{1}{2}\left(\cos\dfrac{7\pi}{4} + i\sin\dfrac{7\pi}{4}\right) 2 1 ( cos 4 7 π + i sin 4 7 π ) #39 · Modulus and Argument — Polar Form, De Moivre and Square Roots
If − 1 + − 3 = r e i θ -1 + \sqrt{-3} = re^{i\theta} − 1 + − 3 = r e i θ , then the value of θ \theta θ is A − 2 π 3 -\dfrac{2\pi}{3} − 3 2 π B C D #40 · Modulus and Argument — Polar Form, De Moivre and Square Roots
Let z z z be the complex number such that ∣ z ∣ + z = 3 + i |z| +z= 3 +i ∣ z ∣ + z = 3 + i where i = − 1 i=\sqrt{- 1} i = − 1 , then ∣ z ∣ = |z| = ∣ z ∣ = A 34 3 \frac{\sqrt{34}}{3} 3 34 B C 41 4 \frac{\sqrt{41}}{4} 4 41 D #41 · Modulus and Argument — Polar Form, De Moivre and Square Roots
The modulus of the square root of the complex number 6 + 8 i 6 + 8i 6 + 8 i (where i = − 1 i=\sqrt{- 1} i = − 1 ) is A B C 2 ⋅ 5 \sqrt{2} \cdot \sqrt{5} 2 ⋅ 5 D #42 · Modulus and Argument — Polar Form, De Moivre and Square Roots
A particle P P P starts from the point Z 0 = 1 + 2 i Z_{0}= 1 + 2i Z 0 = 1 + 2 i where i = − 1 i =\sqrt{- 1} i = − 1 . It moves first horizontally away from the origin by 5 units and then vertically upwards parallel to positive Y -axis by 3 units to reach a point Z 1 Z_{1} Z 1 . From Z 1 Z_{1} Z 1 the particle moves 2 \sqrt{2} 2 units in the direction of vector i ^ + j ^ \widehat{i}+\widehat{j} i + j and then it moves through an angle π 2 \frac{\pi}{2} 2 π in anticlockwise direction on a circle with centre at origin to reach at point Z 2 Z_{2} Z 2 , then Z 2 = Z_{2}= Z 2 = MHT-CET · 2025 · 20 April Shift II · Q140 Hard Add #43 · Modulus and Argument — Polar Form, De Moivre and Square Roots
The value of ( cos θ + i s i n θ ) 4 ( sin θ + i c o s θ ) 5 = \frac{(\cos\theta+isin\theta)^{4}}{(\sin\theta+icos\theta)^{5}}= ( s i n θ + i cos θ ) 5 ( c o s θ + i s in θ ) 4 = where i = − 1 i =\sqrt{- 1} i = − 1 A cos θ − i s i n θ \ \cos\theta - isin\theta cos θ − i s in θ B cos 9 θ − i s i n 9 θ \cos9\theta - isin9\theta cos 9 θ − i s in 9 θ C sin θ − i c o s θ \sin\theta - icos\theta sin θ − i cos θ D sin 9 θ − \sin9\theta - sin 9 θ − i cos 9 θ \cos9\theta cos 9 θ MHT-CET · 2025 · 19 April Shift I · Q144 Moderate Add #44 · Modulus and Argument — Polar Form, De Moivre and Square Roots
The modulus of the square root of the conjugate of − 7 + 24 − 1 - 7 + 24\sqrt{- 1} − 7 + 24 − 1 is _ _ _ _ \_\_\_\_ ____ #45 · Modulus and Argument — Polar Form, De Moivre and Square Roots
If z 1 = 5 − 2 i z_1 = 5-2i z 1 = 5 − 2 i and z 2 = 3 + i z_2 = 3+i z 2 = 3 + i , where i = − 1 i = \sqrt{-1} i = − 1 , then arg ( z 1 + z 2 z 1 − z 2 ) \arg\left(\frac{z_1+z_2}{z_1-z_2}\right) arg ( z 1 − z 2 z 1 + z 2 ) is A tan − 1 22 19 \tan^{-1}\frac{22}{19} tan − 1 19 22 B tan − 1 22 13 \tan^{-1}\frac{22}{13} tan − 1 13 22 C tan − 1 21 19 \tan^{-1}\frac{21}{19} tan − 1 19 21 D tan − 1 19 22 \tan^{-1}\frac{19}{22} tan − 1 22 19 #46 · Modulus and Argument — Polar Form, De Moivre and Square Roots
Let z z z be a complex number such that ∣ z ∣ + z = 2 + i |z|+z = 2+i ∣ z ∣ + z = 2 + i , where i = − 1 i=\sqrt{-1} i = − 1 , then ∣ z ∣ |z| ∣ z ∣ is equal to #47 · Modulus and Argument — Polar Form, De Moivre and Square Roots
Let Z be a complex number such that ∣ Z ∣ + Z = 2 + i |Z| + Z = 2 + i ∣ Z ∣ + Z = 2 + i (where i = − 1 i = \sqrt{-1} i = − 1 ), then ∣ Z ∣ |Z| ∣ Z ∣ is equal to #48 · Modulus and Argument — Polar Form, De Moivre and Square Roots
If Z = − 2 1 + 3 i Z=\frac{-2}{1+\sqrt{3}i} Z = 1 + 3 i − 2 , i = − 1 i=\sqrt{-1} i = − 1 , then the value of arg ( Z ) \arg(Z) arg ( Z ) is #49 · Modulus and Argument — Polar Form, De Moivre and Square Roots
If a > 0 a>0 a > 0 and z = ( 1 + i ) 2 a − i z=\dfrac{(1+i)^2}{a-i} z = a − i ( 1 + i ) 2 , i = − 1 i=\sqrt{-1} i = − 1 , has magnitude 2 5 \sqrt{\dfrac{2}{5}} 5 2 , then z ˉ \bar{z} z ˉ is equal to A 1 5 − 3 5 i \dfrac{1}{5} - \dfrac{3}{5}i 5 1 − 5 3 i B − 1 5 − 3 5 i -\dfrac{1}{5} - \dfrac{3}{5}i − 5 1 − 5 3 i C − 1 5 + 3 5 i -\dfrac{1}{5} + \dfrac{3}{5}i − 5 1 + 5 3 i D − 3 5 − 1 5 i -\dfrac{3}{5} - \dfrac{1}{5}i − 5 3 − 5 1 i #50 · Modulus and Argument — Polar Form, De Moivre and Square Roots
If a > 0 a>0 a > 0 and z = ( 1 + i ) 2 a − i z=\dfrac{(1+i)^2}{a-i} z = a − i ( 1 + i ) 2 , i = − 1 i=\sqrt{-1} i = − 1 , has magnitude 2 5 \dfrac{2}{\sqrt{5}} 5 2 , then z z z is A − 2 5 − 4 5 i -\dfrac{2}{5}-\dfrac{4}{5}i − 5 2 − 5 4 i B − 2 5 + 4 5 i -\dfrac{2}{5}+\dfrac{4}{5}i − 5 2 + 5 4 i C 2 5 − 4 5 i \dfrac{2}{5}-\dfrac{4}{5}i 5 2 − 5 4 i D 2 5 + 4 5 i \dfrac{2}{5}+\dfrac{4}{5}i 5 2 + 5 4 i Showing the first 50 of 67.
See all 67 questions Studying this with friends? Send them this chapter's questions. It's free for them too.
More chapters in this subject 