12th Grade > Mathematics
COMPLEX NUMBERS MCQs
Complex Numbers
Total Questions : 60
| Page 6 of 6 pages
Answer: Option D. -> x2=1−2y
:
D
z=x+iyiz=−y+ix
Given,|z|=Re(iz)+1⇒√x2+y2=−y+1⇒x2+y2=y2−2y+1⇒x2=1−2y
:
D
z=x+iyiz=−y+ix
Given,|z|=Re(iz)+1⇒√x2+y2=−y+1⇒x2+y2=y2−2y+1⇒x2=1−2y
Answer: Option C. -> 6+8i, 6+17i
:
C
We have ∣∣z−12z−8i∣∣=53, ∣∣z−4z−8∣∣=1
Let z=x+iy, then
∣∣z−12z−8i∣∣=53 ⇒ 3|z-12| = 5|z-8i|
⇒ 9(x−12)2+9y2 = 25x2+25(y−8)2 ⋯⋯(i)
∣∣z−4z−8∣∣=1 ⇒ |z-4| = |z-8|
⇒ (x−4)2+y2 = y2+(x−8)2⇒ x=6
Putting x=6 in (i), we get y2-25y+136=0
y=17, 8
Hence z=6+17i or z=6+8i
Trick: Check it with options.
:
C
We have ∣∣z−12z−8i∣∣=53, ∣∣z−4z−8∣∣=1
Let z=x+iy, then
∣∣z−12z−8i∣∣=53 ⇒ 3|z-12| = 5|z-8i|
⇒ 9(x−12)2+9y2 = 25x2+25(y−8)2 ⋯⋯(i)
∣∣z−4z−8∣∣=1 ⇒ |z-4| = |z-8|
⇒ (x−4)2+y2 = y2+(x−8)2⇒ x=6
Putting x=6 in (i), we get y2-25y+136=0
y=17, 8
Hence z=6+17i or z=6+8i
Trick: Check it with options.
Answer: Option A. -> (-2,-1) or (2,-1)
:
A
According to condition, 3-ix2 y =x2 + y + 4i
⇒x2 + y = 3 Andx2y = -4⇒ x =±2, y = -1
⇒ (x,y) = (2,-1) or (-2,-1)
:
A
According to condition, 3-ix2 y =x2 + y + 4i
⇒x2 + y = 3 Andx2y = -4⇒ x =±2, y = -1
⇒ (x,y) = (2,-1) or (-2,-1)
Answer: Option D. -> 4
:
D
Let z = x+iy, so that ¯z = x - iy, therefore
z2+¯z=0⇔(x2−y2+x)+i(2xy−y) = 0
Equating real and imaginary parts , we get
x2−y2+x = 0 .......(i)
And 2xy - y = 0⇒ y = 0 or x = 12
if y = 0 , then (i) gives x2 + x = 0⇒ x = 0 or
x = -1
If x = 12,
Then x2−y2+x=0⇒y2=14+12=34⇒y=±√32
Hence, there are four solutions in all.
:
D
Let z = x+iy, so that ¯z = x - iy, therefore
z2+¯z=0⇔(x2−y2+x)+i(2xy−y) = 0
Equating real and imaginary parts , we get
x2−y2+x = 0 .......(i)
And 2xy - y = 0⇒ y = 0 or x = 12
if y = 0 , then (i) gives x2 + x = 0⇒ x = 0 or
x = -1
If x = 12,
Then x2−y2+x=0⇒y2=14+12=34⇒y=±√32
Hence, there are four solutions in all.
Answer: Option B. -> cosπ2-isinπ2
:
B
1−i1+i=(1−i)(1−i)(1+i)(1−i)=1+(i)2−2i1+1=-i
Which can be written as cosπ2-isinπ2
:
B
1−i1+i=(1−i)(1−i)(1+i)(1−i)=1+(i)2−2i1+1=-i
Which can be written as cosπ2-isinπ2
Answer: Option C. -> Both z+¯z and z¯z are real.
:
C
Let z=x+iy
Here, z+¯z=(x+iy)+(x−iy)=2x (Real)
And z¯z=(x+iy)(x−iy)=x2+y2 (Real).
:
C
Let z=x+iy
Here, z+¯z=(x+iy)+(x−iy)=2x (Real)
And z¯z=(x+iy)(x−iy)=x2+y2 (Real).
Answer: Option C. -> Either z is purely real or purely imaginary
:
C
Let z = x+iy, then its coonjucate ¯z = x-iy
Given that z2=(¯z)2
⇒ x2 -y2 + 2ixy =x2 -y2 - 2ixy⇒ 4ixy = 0
if x ≠ 0 then y = 0 and if y≠ 0 then x = 0
:
C
Let z = x+iy, then its coonjucate ¯z = x-iy
Given that z2=(¯z)2
⇒ x2 -y2 + 2ixy =x2 -y2 - 2ixy⇒ 4ixy = 0
if x ≠ 0 then y = 0 and if y≠ 0 then x = 0
Answer: Option A. -> z=0
:
A
Let z = x+iy,¯z = x-iy
∴ z¯z = 0⇒ (x+iy)(x-iy) = 0 ⇒ x2+y2 = 0
It is possible onle when x and y oth simultaneously zero
i.e, z = 0 +0i = 0
:
A
Let z = x+iy,¯z = x-iy
∴ z¯z = 0⇒ (x+iy)(x-iy) = 0 ⇒ x2+y2 = 0
It is possible onle when x and y oth simultaneously zero
i.e, z = 0 +0i = 0
Answer: Option A. -> Equal to 1
:
A
1=∣∣1z1+1z2+1z3∣∣=∣∣∣z1∗¯z1z1+z2∗¯z2z2+z3∗¯z3z3∣∣∣
(hence, |z1|2=1=z1¯¯¯¯¯z1, etc)
⇒ |z1+z2+z3|=|¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯z1+z2+z3|= |z1+z2+z3|
(hence,|¯¯¯¯¯z1|=|z1|)
:
A
1=∣∣1z1+1z2+1z3∣∣=∣∣∣z1∗¯z1z1+z2∗¯z2z2+z3∗¯z3z3∣∣∣
(hence, |z1|2=1=z1¯¯¯¯¯z1, etc)
⇒ |z1+z2+z3|=|¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯z1+z2+z3|= |z1+z2+z3|
(hence,|¯¯¯¯¯z1|=|z1|)
Answer: Option B. -> -160
:
B
x+5 =4i ⇒ x2+10x+25=-16
Now, x4+9x3+35x2-x+4
=(x2+10x+41)(x2-x+4)-160=-160
:
B
x+5 =4i ⇒ x2+10x+25=-16
Now, x4+9x3+35x2-x+4
=(x2+10x+41)(x2-x+4)-160=-160
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