Complex Numbers - Test Papers

 CBSE Test Paper 01

CH-05 Complex & Quadratic

---

1. The complex numbers z = x + iy ; x , y ∈$\in$ R which satisfy the equation ∣∣z−3iz+3∣∣=1$\left|\frac{z-3i}{z+3}\right|=1$ lies on

   1. the y axis

   2. the x axis

   3. the line x + y = 0

   4. the line parallel to y axis

2. If (3–√+i)10=a+ib;a,b∈R,${\left(\sqrt{3}+i\right)}^{10}=a+ib;a,b\in R,$ then a and b are respectively :

   1. 64 and - 643–√$\sqrt{3}$

   2. 512 and - 5123–√$\sqrt{3}$

   3. 128 and 1283–√$\sqrt{3}$

   4. none of these

3. z + z¯¯¯≠0$\overline{z}\ne 0$ if and only if

   1. z ≠$\ne$ 0

   2. Re(z) ≠$\ne$ 0

   3. Im ( z ) ≠$\ne$0

   4. | z | ≠ 0$\text{| z | }\ne 0$

4. If α=zz¯¯¯,$\alpha =\frac{z}{\overline{z}},$ then |α|$\text{|}\alpha \text{|}$ is equal to :

   1. -1

   2. 0

   3. 1

   4. none of these

5. 1+i2+i4+i6+i8+.......$1+i^2+i^4+i^6+i^8+.......$ up to 1001 terms is equal to

   1. none of these

   2. 0

   3. 1

   4. -1

6. Fill in the blanks:

   The modulus and argument of z = 1 + i tanα$\alpha$ is ________ and ________ respectively.

7. Fill in the blanks:

   The value of 1i7$\frac{1}{i^7}$ is ________.

8. Solve x2 + 3 = 0

9. Express the complex numbers (1 + i) - (- 1 + i6) in standard form

10. Find the difference of the complex numbers (6 + 5i), (3 +2i).

11. Express the complex numbers (15+25i)−(4+52i)$\left(\frac{1}{5}+\frac{2}{5}i\right)-\left(4+\frac{5}{2}i\right)$ in standard form

12. Solve: ix2 + 4x - 5i = 0.

13. Find the square root of 1 - i.

14. If Re (z2) = 0, |z| = 2, then prove that z = ±2–√±i2–√$\pm \sqrt{2}\pm i\sqrt{2}$.

15. Find all non-zero complex numbers of z satisfying z¯¯¯$\overline{z}$ = iz2 .

CBSE Test Paper 01
CH-05 Complex & Quadratic

---

Solution

1. (c) the line x + y = 0
   Explanation: Let z=x+iy

   Now ∣∣z−3iz+3∣∣=1$\left|\frac{z-3i}{z+3}\right|=1$
   ⇒|z−3i|=|z+3|$\Rightarrow |z-3i|=|z+3|$
   ⇒|(x+iy)−3i|=|x+iy+3|$\Rightarrow |(x+iy)-3i|=|x+iy+3|$
   ⇒|x+i(y−3)|=|(x+3)+iy|$\Rightarrow |x+i(y-3)|=|(x+3)+iy|$
   ⇒(y−3)2+x2−−−−−−−−−−√=(x+3)2+(y)2−−−−−−−−−−−−√$\Rightarrow \sqrt{(y-3{)}^2+x^2}=\sqrt{(x+3{)}^2+(y{)}^2}$
   ⇒(y−3)2+x2=(x+3)2+(y)2$\Rightarrow (y-3{)}^2+x^2=(x+3{)}^2+(y{)}^2$
   ⇒y2−6y+9+x2=x2+6x+9+y2$\Rightarrow y^2-6y+9+x^2=x^2+6x+9+y^2$
   ⇒6x+6y=0$\Rightarrow 6x+6y=0$
   ⇒x+y=0$\Rightarrow x+y=0$

2. (b) 512 and - 5123–√$\sqrt{3}$

   Explanation:

   First we will find the polar representation of the complex number 3–√+i$\sqrt{3}+i$
   Let 3–√+i=r(cosθ+isinθ)⇒rcosθ=3–√andrsinθ=1$\sqrt{3}+i=r(\cos \theta +i\sin \theta )\Rightarrow r\cos \theta =\sqrt{3}andr\sin \theta =1$
   ∴r2(cos2θ+sin2θ)=3+1=4⇒r2=4⇒r=2$\therefore r^2\left({\cos }^2\theta +{\sin }^2\theta \right)=3+1=4\Rightarrow r^2=4\Rightarrow r=2$

   Now cosθ=3√2,sinθ=12$cos\theta =\frac{\sqrt{3}}{2},sin\theta =\frac{1}{2}$ , both are positive .

   So Amplitude =θ=Π6$=\theta =\frac{\mathrm{\Pi }}{6}$

   Hence 3–√+i=2(cosΠ6+isinΠ6)=2eiΠ6$\sqrt{3}+i=2\left(cos\frac{\mathrm{\Pi }}{6}+isin\frac{\mathrm{\Pi }}{6}\right)=2e^{\frac{i\mathrm{\Pi }}{6}}$

   Now,
   (3–√+i)10=(2eiπ6)10=210ei5π3=210(cos(5Π3)+isin(5Π3))$(\sqrt{3}+i{)}^{10}={\left(2e^{\frac{i\pi }{6}}\right)}^{10}=2^{10}{e}^{\frac{i5\pi }{3}}=2^{10}\left(\cos \left(\frac{5\mathrm{\Pi }}{3}\right)+i\sin \left(\frac{5\mathrm{\Pi }}{3}\right)\right)$
   =210(cos(2Π−Π3)+isin(2Π−Π3))=210(cos(−π3)+isin(−π3))$=2^{10}\left(\cos \left(2\mathrm{\Pi }-\frac{\mathrm{\Pi }}{3}\right)+i\sin \left(2\mathrm{\Pi }-\frac{\mathrm{\Pi }}{3}\right)\right)=2^{10}\left(\cos \left(\frac{-\pi }{3}\right)+i\sin \left(\frac{-\pi }{3}\right)\right)$
   =210(cos(Π3)−isin(Π3))=210(12−i3√2)=210(1−3√i2)=29(1−3–√i)$=2^{10}\left(\cos \left(\frac{\mathrm{\Pi }}{3}\right)-i\sin \left(\frac{\mathrm{\Pi }}{3}\right)\right)=2^{10}\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)=2^{10}\left(\frac{1-\sqrt{3}i}{2}\right)=2^9(1-\sqrt{3}i)$

   Hence  (3–√+i)10=a+ib${\left(\sqrt{3}+i\right)}^{10}=a+ib$ ⇒29(1−3–√i)=a+ib$\Rightarrow 2^9\left(1-\sqrt{3}i\right)=a+ib$

   ⇒a=29=512andb=−293–√=−5123–√$\Rightarrow a=2^9=512andb={-2}^9\sqrt{3}=-512\sqrt{3}$

3. (b) Re(z) ≠$\ne$ 0
   Explanation: Let Z=x+iy then we have 

   So Z+Z¯=2x$Z+\overline{Z}=2x$

   Now z + z¯¯¯≠0$\overline{z}\ne 0$
   ⇔2x≠0$\iff 2x\ne 0$
   ⇔x≠0$\iff x\ne 0$
   ⇔Re(z)≠0$\iff Re(z)\ne 0$

4. (c) 1
   Explanation:

   Given α=zz¯¯¯$\alpha =\frac{z}{\overline{z}}$

   Then |α|=∣∣zz¯¯¯∣∣=|z||z¯¯¯|=1[∵∣∣z1z2∣∣=|z1||z2|,|z|=|z¯¯¯|]$|\alpha |=\left|\frac{z}{\overline{z}}\right|=\frac{|z|}{|\overline{z}|}=1\left[\because \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|},|z|=|\overline{z}|\right]$

5. (c) 1
   Explanation:
   1+i2+i4+i6+i8+………. upto 1001 terms $1+i^2+i^4+i^6+i^8+\dots \dots \dots .\text{ upto }1001\text{ terms }$
   =(i2)0+(i2)1+(i2)2+(i2)3+……. upto 1001 terms $={\left(i^2\right)}^0+{\left(i^2\right)}^1+{\left(i^2\right)}^2+{\left(i^2\right)}^3+\dots \dots .\text{ upto }1001\text{ terms }$
   =(i2)0+(i2)1+(i2)2+(i2)3+………+(i2)1000$={\left(i^2\right)}^0+{\left(i^2\right)}^1+{\left(i^2\right)}^2+{\left(i^2\right)}^3+\dots \dots \dots +{\left(i^2\right)}^{1000}$
   =[(i2)0+(i2)1]+[(i2)2+(i2)3]+…………+[(i2)998+(i2)999]+[(i2)1000]$=\left[{\left(i^2\right)}^0+{\left(i^2\right)}^1\right]+\left[{\left(i^2\right)}^2+{\left(i^2\right)}^3\right]+\dots \dots \dots \dots +\left[{\left(i^2\right)}^{998}+{\left(i^2\right)}^{999}\right]+\left[{\left(i^2\right)}^{1000}\right]$
   =[1−1]+[1−1]+………+[1−1]+1$=[1-1]+[1-1]+\dots \dots \dots +[1-1]+1$
   =1

6. secα$\alpha$, α$\alpha$

7. i

8. Here x2 + 3 = 0 ⇒x2=−3⇒x=±−3−−−√=±3–√i$\Rightarrow x^2=-3\Rightarrow x=\pm \sqrt{-3}=\pm \sqrt{3}i$

9. (1 + i) - (-1 + i6)
   1 + i + 1 - 6i = 2 - 5i

10. (6 + 5i) - (3 + 2i) = (6 + 5i) + (- 3 - 2i)
    = (6 - 3) + (5 - 2) i = 3 + 3i

11. (15+25i)−(4+52i)$\left(\frac{1}{5}+\frac{2}{5}i\right)-\left(4+\frac{5}{2}i\right)$
    =15+25i−4−52i$=\frac{1}{5}+\frac{2}{5}i-4-\frac{5}{2}i$
    =(15−4)+(25−52)i$=\left(\frac{1}{5}-4\right)+\left(\frac{2}{5}-\frac{5}{2}\right)i$
    =−195−2110i$=\frac{-19}{5}-\frac{21}{10}i$

12. We have, ix2 + 4x - 5i = 0 ...(i)
    On comparing Eq. (i) with ax2 + bx + c = 0, we get
    a = i, b = 4 and c = - 5i
    ∵$\because$ x = −b±b2−4ac√2a$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
    ∴$\therefore$ x = −4±(4)2−4×i(−5i)√2×i$\frac{-4\pm \sqrt{(4{)}^2-4\times i(-5i)}}{2\times i}$
    = −4±16+20i2√2i$\frac{-4\pm \sqrt{16+20i^2}}{2i}$
    = −4±16−20√2i$\frac{-4\pm \sqrt{16-20}}{2i}$ [∵$\because$ i2 = - 1]
    = −4±−4√2i=−4±2i2i$\frac{-4\pm \sqrt{-4}}{2i}=\frac{-4\pm 2i}{2i}$
    = 4i2±2i2i$\frac{4i^2\pm 2i}{2i}$ = 2i ±$\pm$ 1 [∵$\because$ - 1 = i2]
    ∴$\therefore$ x = 2i + 1 and x = 2i - 1
    Hence, the roots of the given equation are 2i + 1 and 2i - 1.

13. Let x+yi=1−i−−−−√$x+yi=\sqrt{1-i}$
    Squaring both sides, we get
    x2 - y2 + 2xyi = 1 - i
    Equating the real and imaginary parts
    x2 - y2 = 1 and 2xy=-1... . (i)
    ∴xy=−12$\therefore xy=\frac{-1}{2}$
    Using the identity
    (x2 + y2)2 = (x2 - y2)2 + 4x2y2
    =(1)2+44(−12)2$={\left(1\right)}^2+44{\left(-\frac{1}{2}\right)}^2$
    = 1 + 1
    = 2
    ∴x2+y2=2–√$\therefore x^2+y^2=\sqrt{2}$ . . . . (ii) [Neglecting (-) sign as x2 + y2 > 0]
    Solving (i) and (ii) we get
    x2=2√+12$x^2=\frac{\sqrt{2}+1}{2}$ and y=2√−12$y=\frac{\sqrt{2}-1}{2}$
    ∴x=±2√+12−−−−√$\therefore x=\pm \sqrt{\frac{\sqrt{2}+1}{2}}$ and y=±2√−12−−−−√$y=\pm \sqrt{\frac{\sqrt{2}-1}{2}}$
    Since the sign of xy is negative.
    ∴$\therefore$ if x=2√+12−−−−√$x=\sqrt{\frac{\sqrt{2}+1}{2}}$ then y=−2√−12−−−−√$y=-\sqrt{\frac{\sqrt{2}-1}{2}}$
    and if x=−2√+12−−−−√$x=-\sqrt{\frac{\sqrt{2}+1}{2}}$ then y=2√−12−−−−√$y=\sqrt{\frac{\sqrt{2}-1}{2}}$
    ∴1−i−−−−√=±(2√+12−−−−√−2√−12−−−−√i)$\therefore \sqrt{1-i}=\pm \left(\sqrt{\frac{\sqrt{2}+1}{2}}-\sqrt{\frac{\sqrt{2}-1}{2}}i\right)$

14. Let z = x + iy
    ⇒$\Rightarrow$ z2 = (x + iy)2 [squaring both sides]
    ⇒$\Rightarrow$ z2 = x2 + i2 y2 + 2 ixy
    ⇒$\Rightarrow$ z2 = (x2 - y2) + i (2xy)
    ∴$\therefore$ Re (z2) = 0
    ⇒$\Rightarrow$ x2 - y2 = 0 ⇒$\Rightarrow$ x = ±$\pm$ y
    ⇒$\Rightarrow$ x = y ...(i)
    and x = - y ...(ii)
    Again, |z| = 2 [given]
    ⇒$\Rightarrow$ |z|2 = 4
    ⇒$\Rightarrow$ x2 + y2 = 4 ...(iii)
    From Eqs. (i) and (iii), we get
    y2 + y2 = 4 ⇒$\Rightarrow$ 2y2 = 4
    ⇒$\Rightarrow$ y2 = 2 ⇒$\Rightarrow$ y = ±$\pm$2–√$\sqrt{2}$
    Therefore, from Eq. (i), we get
    x =  ±$\pm$2–√$\sqrt{2}$ 
    ∴$\therefore$ z = ±$\pm$2–√$\sqrt{2}$ ±$\pm$ i2–√$\sqrt{2}$
    On putting the value of x from Eq. (ii) in Eq. (iii), we get
    (- y)2 + y2 = 4 ⇒$\Rightarrow$ 2 y2 = 4
    ⇒$\Rightarrow$ y2 = 2 ⇒$\Rightarrow$ y = ±$\pm$2–√$\sqrt{2}$
    From Eq. (ii), x = ±$\pm$ 2–√$\sqrt{2}$
    ∴$\therefore$ z = x + iy ⇒$\Rightarrow$ z =  ±$\pm$ 2–√$\sqrt{2}$  ±$\pm$ i2–√$\sqrt{2}$
    Hence proved.

15. Let z = x + iy
    Given: z¯¯¯$\overline{z}$ = iz2
    ⇒$\Rightarrow$ x - iy = i(x2 - y2 + 2i xy)
    ⇒$\Rightarrow$ x - iy = i(x2 - y2) - 2xy 
    ⇒$\Rightarrow$ (x + 2 xy) - i (x2 - y2 + y) = 0
    ⇒$\Rightarrow$ x + 2xy = 0 ...(i) and x2 - y2 + y = 0 ...(ii)
    Now,
    x + 2 xy = 0 ⇒$\Rightarrow$ x (1 + 2y) = 0 ⇒$\Rightarrow$ x = 0 or 1 + 2y = 0 ⇒$\Rightarrow$ x = 0 or y = -12$\frac{1}{2}$
    CASE I: When x = 0
    Putting x = 0 in (ii), we have
    ⇒$\Rightarrow$ - y2 + y = 0 ⇒$\Rightarrow$ y(y - 1) = 0 ⇒$\Rightarrow$ y = 0, y = 1
    Thus, we have the following pairs of values of x and y :
    x = 0, y = 0; x = 0, y = 1
    ∴$\therefore$ z = 0 + i 0 = 0 and z = 0 + 1i = i
    CASE II: When y = -12$\frac{1}{2}$
    Putting y = −12$\frac{-1}{2}$ in (ii), we get
    x2 - y2 + y = 0 ⇒$\Rightarrow$ x2 - 14$\frac{1}{4}$ - 12$\frac{1}{2}$ = 0 ⇒$\Rightarrow$ x2 - 34$\frac{3}{4}$ = 0 ⇒$\Rightarrow$ x = ±$\pm$ 3√2$\frac{\sqrt{3}}{2}$
    Thus, we have the following pairs of values of x and y:
    x = 3√2$\frac{\sqrt{3}}{2}$, y = −12$\frac{-1}{2}$ and x = −3√2$\frac{-\sqrt{3}}{2}$, y = −12$\frac{-1}{2}$ 
    ∴$\therefore$ z = 3√2$\frac{\sqrt{3}}{2}$ - 12$\frac{1}{2}$i and z = −3√2$\frac{-\sqrt{3}}{2}$ - 12$\frac{1}{2}$i
    Hence, all non-zero complex numbers of z are i, 3√2$\frac{\sqrt{3}}{2}$ - 12$\frac{1}{2}$i, -3√2$\frac{\sqrt{3}}{2}$ - 12$\frac{1}{2}$i