Mathematics Questions and Answers – Complex Numbers-2

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This set of Mathematics Question Bank focuses on “Complex Numbers-2”.

1. z1=1+2i and z2=2+3i. Find z1z2.
a) 2+6i
b) -4+0i
c) -4+7i
d) 8+7i
View Answer

Answer: c
Explanation: z1z2 = (1+2i) (2+3i)
= 2 + 3i + 4i -6
= -4 + 7i.
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2. i2 =______________________
a) 1
b) -1
c) i
d) -i
View Answer

Answer: b
Explanation: We know, i = \(\sqrt{-1}\)
=> i2 = -1.

3. i7 =______________
a) 1
b) -1
c) i
d) -i
View Answer

Answer: d
Explanation: We know, i = \(\sqrt{-1}\)
=> i2 = -1 => i4 = 1.
So, i7 = i4.i3 = 1*i2*i = (-1)*i = -i.
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4. i241 =________________
a) 1
b) -1
c) i
d) -i
View Answer

Answer: c
Explanation: We know, i = \(\sqrt{-1}\)
=> i2 = -1 => i4 = 1.
So, i241 = (i4)60 * i = 1 * i = i.

5. Square roots of -7 are____________
a) 7i and -7i
b) \(\sqrt{7}\) i
c) –\(\sqrt{7}\) i
d) \(\sqrt{7}\) i and –\(\sqrt{7}\) i
View Answer

Answer: d
Explanation: We know, i2 = -1.
-7 = 7(i2)
Square root of i2 is ±i so, square root of -7 are \(\sqrt{7}\)i and –\(\sqrt{7}\)i.
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6. (-i) (8+5i) =________________
a) 8+5i
b) -8-5i
c) -5-8i
d) 5-8i
View Answer

Answer: d
Explanation: (-i) (8+5i) = -8i – 5 i2
= -8i -5(-1) = 5-8i.

7. (2-i)3 =________________
a) 2-3i
b) 8-i
c) 2-11i
d) 2+11i
View Answer

Answer: c
Explanation: We know, (a-b)3 = a3-b3-3ab(a-b)
So, (2-i)3 = 23-(i)3-3(2)(i) (2-i)
= 8-(-i)-6i(2-i)
= 8+i-12i-6
= 2-11i.
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8. Is z*\(\bar{z}\) = |z|2?
a) True
b) False
View Answer

Answer: a
Explanation: Let z=a+ bi
=>\(\bar{z}\) = a-bi
So, z*\(\bar{z}\) = (a+bi) (a-bi) = a2-(bi)2 = a2-(b2) (-1) = a2+b2
|z|=\(\sqrt{a^2+b^2}\) => |z|2 = a2+b2
Hence, z*\(\bar{z}\) = |z|2.

9. Find multiplicative inverse of 3+5i.
a) 87+145i
b) 87-145i
c) 145-87i
d) 145+87i
View Answer

Answer: b
Explanation: We know, z*\(\bar{z}\) = |z|2.
(1/z) = \(\bar{z}\)|z|2
z-1=(3-5i) (32+52) = (3-5i) (29) = 87-145i.
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10. i-35 =___________________
a) 1
b) -1
c) i
d) -i
View Answer

Answer: c
Explanation: We know, i-35= 1/i35 = i/i36
= i/(i4)9 = i/1 = i.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter