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JEE • MATHEMATICS • COMPLEX NUMBERS

Complex numbers practice · Medium

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211 Medium
211 of 337
Application +4 | -1
Consider the following statements.

I. The value of x3+7x2−x+16x^3+7 x^2-x+16, when x=1+2ix=1+2 i is −17+24i-17+24 {i}.

II. If iz3+z2−z+i=0i z^3+z^2-z+i=0 then ∣z∣=1|z|=1

Choose the correct option.
View answer and explanation
Correct option: C
Both are correct
Option C (Both are correct) is correct because both statements are true after evaluating the polynomial at x=1+2i and solving the cubic equation.
212 Medium
212 of 337
Analyzing +4 | -1
If imaginary part of z−5eiα+eiαz−5\frac{z-5}{e^{i \alpha}}+\frac{e^{i \alpha}}{z-5} is zero and if locus of zz is a circle, then its radius is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the radius of the circle representing the locus of z is 1, derived from the given condition.
213 Medium
213 of 337
Application +4 | -1
The value of 4(cos⁡75∘+isin⁡75∘)04(cos⁡30∘+isin⁡30∘)\frac{4\left(\cos 75^{\circ}+i \sin 75^{\circ}\right)}{04\left(\cos 30^{\circ}+i \sin 30^{\circ}\right)} is
View answer and explanation
Correct option: A
102(1+i)\frac{10}{\sqrt{2}}(1+i)
Option A (102(1+i)\frac{{10}}{{\sqrt{{2}}}}(1+i)) is correct because after simplifying using trigonometric identities, the expression equals 102(1+i)\frac{{10}}{{\sqrt{{2}}}}(1+i).
214 Medium
214 of 337
Application +4 | -1
If ∣z−3+2i∣≤4|z-3+2 i| \leq 4 then the difference between the greatest and least value of ∣z∣|z| is -
View answer and explanation
Correct option: B
2132 \sqrt{13}
Option B (2132\sqrt{13}) is correct because the difference between the greatest and least values of ∣z∣|z| is found by using the triangle inequality on ∣(z−3+2i)+(3−2i)∣|(z-3+2i)+(3-2i)|.
215 Medium
215 of 337
Application +4 | -1
If z=cos⁡θ+isin⁡θz=\cos \theta+i \sin \theta, then z2n−1z2n+1\frac{z^{2 n}-1}{z^{2 n}+1}, where nn is an integer, is equal to
View answer and explanation
Correct option: B
itan⁡nθi \tan n \theta
Option B (itan⁡nθi \tan n \theta) is correct because after applying De Moivre's theorem and trigonometric identities, the expression simplifies to itan⁡nθi \tan n \theta.
216 Medium
216 of 337
Application +4 | -1
If ∣z−4z∣=2\left|z-\frac{4}{z}\right|=2, then the difference between the maximum and the minimum values of ∣z∣|z| is ______ (z≠0)(z \neq 0)
View answer and explanation
Correct option: B
2
Option B (2) is correct because by using the properties of complex numbers and inequalities, we find the difference between the maximum and minimum values of |z| is 2.
217 Medium
217 of 337
Understanding +4 | -1
If ∣z1∣=∣z2∣\left| {{z_1}} \right| = \left| {{z_2}} \right| and arg⁡(z1)+arg⁡(z2)=0,\arg \left( {{z_1}} \right) + \arg \left( {{z_2}} \right) = 0, then
View answer and explanation
Correct option: B
z1=zˉ2{z_1} = {\bar z_2}
Option B is correct because the equation represents a circle only if the square of the magnitude of 'a' is greater than 'b'.
218 Medium
218 of 337
Application +4 | -1
The area of the triangle whose vertices are complex numbers z,iz,z+izz, i z, z+i z in the Argand diagram is
View answer and explanation
Correct option: B
1/2∣z∣21 / 2|z|^2
Option B (1/2∣z∣21 / 2|z|^2) is correct. By converting the complex vertices to Cartesian coordinates and calculating the triangle's area, we find it equals 1/2∣z∣21/2 |z|^2.
219 Medium
219 of 337
Application +4 | -1
If zz is any complex number such that ∣z+4∣≤3|z+4| \leq 3, then the least value and greatest value of ∣z+1∣|z+1| are
View answer and explanation
Correct option: B
0,6
Option B is correct because the least value of ∣z+1∣|z+1| is 0 and the greatest value is 6, given the condition ∣z+4∣≤3|z+4| \leq 3.
220 Medium
220 of 337
Application +4 | -1
The value of the determinant ∣1+i1−ii1−ii1+ii1+i1−i∣\left| {\begin{array}{cccccccccccccccccccc} {1 + i}&{1 - i}&i\\ {1 - i}&i&{1 + i}\\ i&{1 + i}&{1 - i} \end{array}} \right|, where i=−1i = \sqrt { - 1} is
View answer and explanation
Correct option: C
4+7i4 + 7i
Option C (4 + 7i) is correct because it is the value obtained after expanding and simplifying the given determinant.
221 Medium
221 of 337
Application +4 | -1
If z1,z2,z3z_1, z_2, z_3 are three complex numbers, then ∣z2+z3∣2+∣z3+z1∣2+∣z1+z2∣2\left|z_2+z_3\right|^2+\left|z_3+z_1\right|^2+\left|z_1+z_2\right|^2 equal
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the given expression does not simplify to any of the provided options. The correct expansion involves terms with individual complex numbers and their conjugates.
222 Medium
222 of 337
Application +4 | -1
For a complex number zz, the minimum value of ∣z∣+∣z−cos⁡α−isin⁡α∣|z|+|z-\cos \alpha-i \sin \alpha| is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the minimum value is the distance between the origin and the point (cos⁡α,sin⁡α)(\cos \alpha, \sin \alpha), which is 1.
223 Medium
223 of 337
Application +4 | -1
The points representing the complex number zz, for which ∣z+3i∣2−∣z−3∣2=6|z+3 i|^2-|z-3|^2=6, lie on the line given by
View answer and explanation
Correct option: A
x+y=1x+y=1
Option A (x+y=1x+y=1) is correct because substituting z=x+iyz=x+iy into the given equation and simplifying leads to the equation of the line x+y=1x+y=1.
224 Medium
224 of 337
Application +4 | -1
(11+2i+31+i)(3+4i2−4i)\left( {\frac{1}{{1 + 2i}} + \frac{3}{{1 + i}}} \right)\left( {\frac{{3 + 4i}}{{2 - 4i}}} \right) is equal to
View answer and explanation
Correct option: D
14+94i\frac{1}{4} + \frac{9}{4}i
Option D (14+94i\frac{1}{4} + \frac{9}{4}i) is correct because it is the result of simplifying the given complex number expression through algebraic manipulation.
225 Medium
225 of 337
Application +4 | -1
If zz satisfies ∣z+1∣<∣z−2∣|z+1| < |z-2|, then w=3z+2+iw=3 z+2+i satisfies
View answer and explanation
Correct option: A
∣w+1∣<∣w−8∣|w+1| < |w-8|
Option A is correct because if ∣z+1∣<∣z−2∣|z+1| < |z-2|, then w=3z+2+iw=3z+2+i satisfies ∣w+1∣<∣w−8∣|w+1| < |w-8|.

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226 Medium
226 of 337
Application +4 | -1
If zr=cos⁡rαn2+isin⁡rαn2z_r=\cos \frac{r \alpha}{n^2}+i \sin \frac{r \alpha}{n^2}, where r=1,2,3,…,nr=1,2,3, \ldots, n, then lim⁡n→∞(z1z2z3…zn)\lim \limits_{{n \rightarrow \infty}}\left(z_1 z_2 z_3 \ldots z_n\right) is equal to
View answer and explanation
Correct option: C
eia/2e^{i a / 2}
Option C (eia/2e^{i a / 2}) is correct because the product of complex numbers in polar form simplifies to eiα/2e^{i \alpha / 2} after applying trigonometric identities and limit properties.
227 Medium
227 of 337
Analyzing +4 | -1
The value of ∑k=1n(n−k)cos⁡2kπn(n≥3)\sum\limits_{{k=1}}^n(n-k) \cos \frac{2 k \pi}{n}(n \geq 3) is
View answer and explanation
Correct option: D
−n-n
Option D (-n) is correct because the summation simplifies to -n using trigonometric identities and series summation formulas.
228 Medium
228 of 337
Analyzing +4 | -1
If a=cos⁡α+isin⁡α,  b=cos⁡β+isin⁡β,  c=cos⁡γ+isin⁡γ  and  bc+ca+ab=1,  a = \cos \mathrm{\alpha } + i\sin \mathrm{\alpha },\;b = \cos \mathrm{\beta } + i\sin \mathrm{\beta },\;c = \cos \mathrm{\gamma } + i\sin \mathrm{\gamma }\;\mathrm{and\;}\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 1,\; then cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)  \cos \left( {\mathrm{\beta } - \mathrm{\gamma }} \right) + \cos \left( {\mathrm{\gamma } - \mathrm{\alpha }} \right) + \cos \left( {\mathrm{\alpha } - \mathrm{\beta }} \right)\mathrm{\;} is equal to
View answer and explanation
Correct option: D
1
Option B (0) is correct because the product of the roots of the given function, as n approaches infinity, is 0.
229 Medium
229 of 337
Application +4 | -1
If ∣z∣=2|z|=2 and locus of 5z−15 z-1 is the circle having radius aa and z12+z22−2z1z2cos⁡θ=0z_1^2+z_2^2-2 z_1 z_2 \cos \theta=0, then ∣z1∣:∣z2∣\left|z_1\right|:\left|z_2\right| is equal to
View answer and explanation
Correct option: C
a:10a: 10
Option C (a:10a: 10) is correct because it correctly relates the radius of the locus of 5z−15z-1 to the ratio of ∣z1∣|z_1| and ∣z2∣|z_2|.
230 Medium
230 of 337
Application +4 | -1
The equation zˉ=aˉ+r2(z−a),   r>0\bar z = \bar a + \frac{{{r^2}}}{{\left( {z - a} \right)}},\;\,r > 0 represents
View answer and explanation
Correct option: C
A circle
Option C (A circle) is correct because the given equation can be rearranged into the general form of a circle equation in the complex plane.
231 Medium
231 of 337
Application +4 | -1
If arg⁡(z−1z+1)=π4\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}, then the locus of zz is
View answer and explanation
Correct option: B
A circle
Option B (A circle) is correct because the given condition leads to the equation of a circle when zz is expressed as x+iyx+iy.
232 Medium
232 of 337
Application +4 | -1
If ∣z+4∣≤3,\left| {z + 4} \right| \le 3, then the greatest and the least value of ∣z+1∣\left| {z + 1} \right| are
View answer and explanation
Correct option: B
6,06,0
Option B (6,0) is correct because the greatest and least values of ∣z+1∣|z+1| are found by interpreting the given inequality geometrically as a circle and finding the distances from the center to the point -1.
233 Medium
233 of 337
Application +4 | -1
If (a1+ib1)(a2+ib2)…(an+ibn)=A+iB\left(a_1+i b_1\right)\left(a_2+i b_2\right) \ldots\left(a_n+i b_n\right)=A+i B, then ∑j=1ntan⁡−1(bjaj)\sum\limits_{{j=1}}^n \tan ^{-1}\left(\frac{b_j}{a_j}\right) is equal to
View answer and explanation
Correct option: C
tan⁡−1(BA)\tan ^{-1}\left(\frac{B}{A}\right)
Option C (tan⁡−1(BA)\tan ^{-1}\left(\frac{{B}}{{A}}\right)) is correct because the argument of a product of complex numbers is the sum of their arguments.
234 Medium
234 of 337
Application +4 | -1
The value of ∑k=16(sin⁡2  π  k7−icos⁡2  π  k7),\sum \limits_{k = 1}^6 \left( {\sin \frac{{2\;\pi \;k}}{7} - i\cos \frac{{2\;\pi \;k}}{7}} \right), is
View answer and explanation
Correct option: D
ii
Option D (ii) is correct because by converting the sum to a geometric series with complex exponentials and simplifying, the result is ii.
235 Medium
235 of 337
Application +4 | -1
cos⁡(ilog⁡a−iba+ib)\cos \left(i \log \frac{a-i b}{a+i b}\right) is equal to
View answer and explanation
Correct option: B
a2−b2a2+b2\frac{a^2-b^2}{a^2+b^2}
Option B (a2−b2a2+b2\frac{a^2-b^2}{a^2+b^2}) is correct because it is the simplified form of cos⁡(ilog⁡a−iba+ib)\cos(i \log \frac{a-ib}{a+ib}) using complex number properties.
236 Medium
236 of 337
Application +4 | -1
(cos⁡2θ+isin⁡2θ)−5(cos⁡3θ−isin⁡3θ)6(\cos 2 \theta+i \sin 2 \theta)^{-5}(\cos 3 \theta-i \sin 3 \theta)^6 (sin⁡θ−icos⁡θ)3(\sin \theta-i \cos \theta)^3 is equal to
View answer and explanation
Correct option: C
sin⁡25θ+icos⁡25θ\sin 25 \theta+i \cos 25 \theta
Option C (sin⁡25θ+icos⁡25θ\sin 25 \theta + i \cos 25 \theta) is correct because after applying De Moivre's theorem and simplifying, the expression equals sin⁡25θ+icos⁡25θ\sin 25 \theta + i \cos 25 \theta.
237 Medium
237 of 337
Analyzing +4 | -1
If for the complex numbers z1z_1 and z2z_2, ∣1−zˉ1z2∣2−∣z1−z2∣2=k(1−∣z1∣2)(1−∣z2∣2)\left|1-\bar{z}_1 z_2\right|^2-\left|z_1-z_2\right|^2=k\left(1-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right), then kk is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because by expanding and simplifying the given expression, it reduces to (1−∣z1∣2)(1−∣z2∣2)(1-|z_1|^2)(1-|z_2|^2), which implies k=1k=1.
238 Medium
238 of 337
Application +4 | -1
If ∣z−i∣=1\left| {z - i} \right| = 1 and arg⁡(z)=θ,\arg \left( z \right) = \mathrm{\theta }, where 0<θ<π20 < \mathrm{\theta } < \frac{\pi }{2}, then cot⁡θ−2z\cot \mathrm{\theta } - \frac{2}{z} equals
View answer and explanation
Correct option: C
ii
Option C (ii) is correct because after substituting z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i\sin \theta ) and simplifying using the given condition ∣z−i∣=1|z - i| = 1, the expression cot⁡θ−2z\cot \mathrm{\theta } - \frac{2}{z} simplifies to ii.
239 Medium
239 of 337
Application +4 | -1
The value of (1+i31−i3)6+(1−i31+i3)6\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right)^6+\left(\frac{1-i \sqrt{3}}{1+i \sqrt{3}}\right)^6 is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the expression simplifies to the sum of two complex numbers which are reciprocals of each other, resulting in 2.
240 Medium
240 of 337
Application +4 | -1
Let α\alpha and β\beta be two fixed non-zero complex numbers and 'zz' a variable complex number. If the lines αzˉ+αˉz+1=0\alpha \bar z + \bar \alpha z + 1 = 0 and β  zˉ+βˉ  z−1=0\mathrm{\beta \;\bar z} + \bar \beta \;z - 1 = 0 are mutually perpendicular, then
View answer and explanation
Correct option: D
α  βˉ+αˉ  β=0\alpha \;\bar \beta + \bar \alpha \;\mathrm{\beta } = 0
Option D (α  βˉ+αˉ  β=0\alpha \;\bar \beta + \bar \alpha \;\mathrm{\beta } = 0) is correct because it represents the condition for the lines to be perpendicular, which involves the sum of the product of the coefficients and their conjugates being zero.
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