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Complex Numbers: Problems with Solutions

Theory

Addition and subtraction of complex numbers:

Let (a + bi) and (c + di) be two complex numbers, then:
(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) - (c + di) = (a - c) + (b - d)i

Reals are added with reals and imaginary with imaginary.

Complex numbers multiplication:

Complex numbers division:

Problems with Solutions

Problem 1

Let [tex]z=(-1,2)[/tex] and [tex]w=(3,2)[/tex] be complex numbers written as ordered pairs [tex](a,b)=a+bi[/tex].
Calculate [tex]5z-3w[/tex] and plot it in the complex plane.

$(-7,2)$

$(14,-4)$

$(-14,-4)$

$(-14,4)$

Solution:
[tex]5z-3w=5(-1,2)-3(3,2)=(-5,10)-(9,6)=(-14,4)[/tex]
Answer: [tex]5z-3w=(-14,4)=-14+4i[/tex]
Graph:

Problem 2

Let [tex]z=(2,-1)[/tex] and [tex]w=(3,2)[/tex] be complex numbers written as ordered pairs [tex](a,b)=a+bi[/tex].
Find [tex]x=z\overline{w}[/tex].

$(2,-7)$

$(4,-14)$

$(-4,7)$

$(4,-7)$

Solution:
[tex]\overline{w}=(3,-2)[/tex]
Multiplication rule:
[tex](a,b)\cdot(c,d)=(ac-bd,\;ad+bc)[/tex]
[tex]x=(2,-1)\cdot(3,-2)=\left(6-2,\;-4-3\right)=(4,-7)[/tex]
Answer: [tex]x=(4,-7)[/tex]

Problem 3

Let [tex]z=(3,-2)[/tex] and [tex]w=(-3,1)[/tex] be complex numbers written as ordered pairs [tex](a,b)=a+bi[/tex].
Find [tex]x=\frac{z}{w}[/tex].

$\left( -\frac{7}{10},\frac{3}{10}\right)$

$\left( -\frac{11}{10},\frac{3}{10}\right)$

$\left( -70, 30\right)$

$\left( -7, 3\right)$

Solution:
Division rule:
[tex]\frac{(a,b)}{(c,d)}=\frac{1}{c^{2}+d^{2}}\,(a,b)\cdot(c,-d)[/tex]
[tex]x=\frac{1}{9+1}\,(3,-2)\cdot(-3,-1)=\frac{1}{10}\left(-9-2,\;-3+6\right)=\frac{1}{10}(-11,3)[/tex]
Answer: [tex]x=\left(-\frac{11}{10},\frac{3}{10}\right)[/tex]

Problem 4

Let [tex]z=(-2,3)[/tex] and [tex]w=(4,-1)[/tex] be complex numbers written as ordered pairs [tex](a,b)=a+bi[/tex].
Find [tex]x=z^{2}-w^{2}[/tex].

$(20,-4)$

$(-20,-4)$

$(10,-4)$

$(-10,-4)$

Solution:
[tex]z^{2}=(-2,3)\cdot(-2,3)=(4-9,\;-6-6)=(-5,-12)[/tex]
[tex]w^{2}=(4,-1)\cdot(4,-1)=(16-1,\;-4-4)=(15,-8)[/tex]
[tex]x=(-5,-12)-(15,-8)=(-20,-4)[/tex]
Answer: [tex]x=(-20,-4)[/tex]

Problem 5

Let [tex]z=(2,4)[/tex] and [tex]w=(4,-1)[/tex] be complex numbers written as ordered pairs [tex](a,b)=a+bi[/tex].
Calculate [tex]\overline{z-w}-\overline{z+w}[/tex].

$(8,-2)$

$(-8,-2)$

$(-8,2)$

$(-2,-5)$

Solution:
[tex]z-w=(2,4)-(4,-1)=(-2,5)[/tex]
[tex]z+w=(2,4)+(4,-1)=(6,3)[/tex]
[tex]\overline{z-w}-\overline{z+w}=(-2,-5)-(6,-3)=(-2-6,\;-5+3)=(-8,-2)[/tex]
Answer: [tex](-8,-2)[/tex]

Problem 6

Let [tex]z=(3,-4)[/tex] and let [tex]\rho=|z|[/tex] be its modulus.
Calculate [tex]x=\rho\overline{z}+\frac{10}{\rho}z[/tex].

$(7,6)$

$(23,2)$

$(21,12)$

$(19,20)$

Solution:
[tex]\rho=\sqrt{3^{2}+(-4)^{2}}=\sqrt{25}=5[/tex]
[tex]\overline{z}=(3,4)[/tex]
[tex]x=5(3,4)+\frac{10}{5}(3,-4)=(15,20)+(6,-8)=(21,12)[/tex]
Answer: [tex]x=(21,12)[/tex]

Problem 7

Find the square roots of [tex]z=(3,4)[/tex].

$\left( 2,1\right)$

$\left(-2,-1\right)$

$\left(1,1\right)$, $\left(-1,-1\right)$

$\left( 2,1\right)$, $\left(-2,-1\right)$

Solution:
For [tex]z=(a,b)[/tex] with [tex]b>0[/tex] the square roots are
[tex]\pm\left(\sqrt{\frac{\rho+a}{2}},\;\sqrt{\frac{\rho-a}{2}}\right),\qquad \rho=|z|=\sqrt{a^{2}+b^{2}}[/tex]
[tex]\rho=\sqrt{3^{2}+4^{2}}=5[/tex]
[tex]\pm\left(\sqrt{\frac{5+3}{2}},\;\sqrt{\frac{5-3}{2}}\right)=\pm\left(\sqrt{4},\sqrt{1}\right)=\pm(2,1)[/tex]
Check: [tex](2+i)^{2}=4+4i+i^{2}=3+4i[/tex].
Answer: [tex](2,1)\text{ and }(-2,-1)[/tex]

Problem 8

Let [tex]z=(4,-3)[/tex]. Is the following statement true?
[tex]\operatorname{Re}\left(z\overline{z}\right)=|z|^{2}[/tex]

True

False

Solution:
[tex]\overline{z}=(4,3)[/tex]
[tex]z\overline{z}=(4,-3)\cdot(4,3)=\left(16+9,\;12-12\right)=(25,0)[/tex]
[tex]\operatorname{Re}\left(z\overline{z}\right)=25[/tex]
[tex]|z|^{2}=4^{2}+(-3)^{2}=25[/tex]
The statement is true.

Problem 9

Let [tex]z=(4,-3)[/tex]. Is the following statement true?
[tex]z+\overline{z}=2\operatorname{Re}(z)[/tex]

True

False

Solution:
[tex]\overline{z}=(4,3)[/tex]
[tex]z+\overline{z}=(4,-3)+(4,3)=(8,0)=8[/tex]
[tex]2\operatorname{Re}(z)=2\times 4=8[/tex]
The statement is true.

Problem 10

Calculate [tex]i^{761}[/tex].

1

i

-1

-i

Solution:
The powers of [tex]i[/tex] repeat every four steps:
[tex]i^{1}=i,\quad i^{2}=-1,\quad i^{3}=-i,\quad i^{4}=1[/tex]
Divide the exponent by [tex]4[/tex]: the quotient is [tex]190[/tex] and the remainder is [tex]1[/tex].
[tex]761=4\times190+1[/tex]
[tex]i^{761}=\left(i^{4}\right)^{190}\cdot i=1^{190}\cdot i=i[/tex]
Answer: [tex]i^{761}=i[/tex]

Problem 11

Calculate [tex]\frac{(2-3i)-(3+2i)}{(3+2i)-(2+i)}[/tex].

$3-2i$

$3+2i$

$-3+2i$

$-3-2i$

Solution:
Numerator and denominator:
[tex](2-3i)-(3+2i)=(2-3)+(-3-2)i=-1-5i[/tex]
[tex](3+2i)-(2+i)=(3-2)+(2-1)i=1+i[/tex]
We multiply the numerator and the denominator by the conjugate of the denominator.
[tex]\frac{-1-5i}{1+i}=\frac{(-1-5i)(1-i)}{(1+i)(1-i)}=\frac{-1+i-5i+5i^{2}}{1-i^{2}}=\frac{-6-4i}{2}=-3-2i[/tex]
Answer: [tex]-3-2i[/tex]

Problem 12

Calculate [tex]\frac{i^{326}-1}{i^{545}+1}[/tex].

$1-i$

$1+i$

$-1+i$

$-1-i$

Solution:
[tex]326=4\times81+2\;\Rightarrow\; i^{326}=i^{2}=-1[/tex]
[tex]545=4\times136+1\;\Rightarrow\; i^{545}=i^{1}=i[/tex]
[tex]\frac{i^{326}-1}{i^{545}+1}=\frac{-1-1}{i+1}=\frac{-2}{1+i}[/tex]
We multiply the numerator and the denominator by the conjugate of the denominator.
[tex]\frac{-2(1-i)}{(1+i)(1-i)}=\frac{-2+2i}{2}=-1+i[/tex]
Answer: [tex]-1+i[/tex]

Problem 13

Write the conjugate of [tex]z=4+4i[/tex] in rectangular (algebraic) form and in polar (trigonometric) form.

$\overline{z}=4\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)$

$\overline{z}=4\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)$

$\overline{z}=4\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)$

$\overline{z}=4\sqrt{2}\left(\cos\left(\frac{3\pi}{4}\right)+i\sin\left(\frac{3\pi}{4}\right)\right)$

Solution:
Rectangular form:
[tex]\overline{z}=4-4i[/tex]
Modulus:
[tex]r=|\overline{z}|=\sqrt{4^{2}+(-4)^{2}}=\sqrt{32}=4\sqrt{2}[/tex]
The point [tex](4,-4)[/tex] lies in the fourth quadrant and [tex]\tan\varphi=\frac{-4}{4}=-1[/tex], so [tex]\varphi=-\frac{\pi}{4}[/tex] (or [tex]\frac{7\pi}{4}[/tex]).
Answer: [tex]\overline{z}=4-4i=4\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)[/tex]

Problem 14

Multiply the complex numbers [tex](5+2i)(2-3i)[/tex].

$10-6i$

$11-15i$

$16-11i$

$12-12i$

Solution:
We expand and use [tex]i^{2}=-1[/tex]:
[tex](5+2i)(2-3i)=10-15i+4i-6i^{2}=10-11i+6=16-11i[/tex]
Answer: [tex]16-11i[/tex]

Problem 15

Divide the complex numbers [tex]\frac{3-2i}{5+2i}[/tex].

$\frac{16}{29}i$

$\frac{3}{5}-i$

$\frac{11}{29}-\frac{16}{29}i$

$\frac{11}{29}i$

Solution:
We multiply the numerator and the denominator by the conjugate of the denominator.
[tex]\frac{3-2i}{5+2i}=\frac{(3-2i)(5-2i)}{(5+2i)(5-2i)}=\frac{15-6i-10i+4i^{2}}{25+4}=\frac{11-16i}{29}[/tex]
Answer: [tex]\frac{11}{29}-\frac{16}{29}i[/tex]

Problem 16

Let [tex]z=2+5i[/tex] and [tex]w=3+2i[/tex]. Calculate [tex]\left(z\overline{w}\right)^{2}[/tex] and find its conjugate.

$135+352i$

$135-352i$

$16-11i$

$377-352i$

Solution:
[tex]\overline{w}=3-2i[/tex]
[tex]z\overline{w}=(2+5i)(3-2i)=6-4i+15i-10i^{2}=16+11i[/tex]
[tex]\left(z\overline{w}\right)^{2}=(16+11i)^{2}=256+352i+121i^{2}=135+352i[/tex]
The conjugate changes the sign of the imaginary part:
Answer: [tex]\overline{\left(z\overline{w}\right)^{2}}=135-352i[/tex]

Problem 17

Let [tex]z=3+i[/tex], [tex]t=1+i[/tex], [tex]w=2-3i[/tex] and [tex]s=-1+2i[/tex]. Calculate [tex]|3zt-2ws|[/tex].

0

$\sqrt{2}$

2

$2\sqrt{2}$

Solution:
[tex]3zt=(3+i)(3+3i)=9+9i+3i+3i^{2}=6+12i[/tex]
[tex]2ws=(4-6i)(-1+2i)=-4+8i+6i-12i^{2}=8+14i[/tex]
[tex]3zt-2ws=(6+12i)-(8+14i)=-2-2i[/tex]
[tex]|-2-2i|=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{8}=2\sqrt{2}[/tex]
Answer: [tex]2\sqrt{2}[/tex]

Problem 18

What is the polar form of the complex number [tex]z=-1+\sqrt{3}i[/tex]?

$z=\sqrt{3} \angle 120^{\circ}$

$z=-1\angle 120^{\circ}$

$z=2\angle 120^{\circ}$

$z=\frac{-1}{\sqrt{3}}\angle 120^{\circ}$

Solution:
[tex]|z|=\sqrt{(-1)^{2}+\left(\sqrt{3}\right)^{2}}=\sqrt{4}=2[/tex]
[tex]\tan\varphi=\frac{\sqrt{3}}{-1}=-\sqrt{3}[/tex]. The point [tex]\left(-1,\sqrt{3}\right)[/tex] lies in the second quadrant, so [tex]\varphi[/tex] is not [tex]\arctan\left(-\sqrt{3}\right)=-60^{\circ}[/tex] (that angle is in the fourth quadrant) but [tex]\varphi=180^{\circ}-60^{\circ}=120^{\circ}[/tex].
Answer: [tex]z=2\angle 120^{\circ}=2\left(\cos 120^{\circ}+i\sin 120^{\circ}\right)[/tex]

Problem 19

Calculate [tex](5+2i)+(-8+3i)-(4-2i)[/tex] and write the result in polar (trigonometric) form.

$7\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right)$

$7\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)$

$7\sqrt{2}\left(\cos\left(\frac{3\pi}{4}\right)+i\sin\left(\frac{3\pi}{4}\right)\right)$

$\sqrt{2}\left(\cos\left(\frac{3\pi}{4}\right)+i\sin\left(\frac{3\pi}{4}\right)\right)$

Solution:
We add the real parts and the imaginary parts separately:
[tex](5-8-4)+(2+3+2)i=-7+7i[/tex]
[tex]r=|-7+7i|=\sqrt{(-7)^{2}+7^{2}}=\sqrt{98}=7\sqrt{2}[/tex]
The point [tex](-7,7)[/tex] lies in the second quadrant and [tex]\tan\varphi=\frac{7}{-7}=-1[/tex], so [tex]\varphi=\frac{3\pi}{4}[/tex].
Answer: [tex]-7+7i=7\sqrt{2}\left(\cos\left(\frac{3\pi}{4}\right)+i\sin\left(\frac{3\pi}{4}\right)\right)[/tex]

Problem 20

Find the distance between the complex numbers [tex]z=2-3i[/tex] and [tex]w=-3+2i[/tex].

$5\sqrt{2}$

$\sqrt{2}$

$\sqrt{5}$

$3\sqrt{2}$

Solution:
The distance is the modulus of the difference:
[tex]d(z,w)=|z-w|=|(2-3i)-(-3+2i)|=|5-5i|[/tex]
[tex]|5-5i|=\sqrt{5^{2}+(-5)^{2}}=\sqrt{50}=5\sqrt{2}[/tex]
Answer: [tex]5\sqrt{2}[/tex]

Problem 21

What is the midpoint of the segment joining [tex]z=6-3i[/tex] and [tex]w=2+5i[/tex]?

8+8i

4+i

4+2i

4

Solution:
The midpoint of the segment between [tex]a[/tex] and [tex]b[/tex] is [tex]\frac{a+b}{2}[/tex]:
[tex]\frac{z+w}{2}=\frac{(6-3i)+(2+5i)}{2}=\frac{8+2i}{2}=4+i[/tex]
Answer: [tex]4+i[/tex]

Problem 22

Let [tex]s[/tex] be the sum and [tex]r[/tex] the difference [tex]z-w[/tex] of the complex numbers [tex]z=2+3i[/tex] and [tex]w=1-4i[/tex]. Find the midpoint of the segment joining [tex]s[/tex] and [tex]r[/tex].

$4+6i$

$2-3i$

$2+3i$

$3+2i$

Solution:
[tex]s=(2+3i)+(1-4i)=3-i[/tex]
[tex]r=(2+3i)-(1-4i)=1+7i[/tex]
The midpoint of the segment between [tex]a[/tex] and [tex]b[/tex] is [tex]\frac{a+b}{2}[/tex]:
[tex]\frac{s+r}{2}=\frac{4+6i}{2}=2+3i[/tex]
Answer: [tex]2+3i[/tex]

Problem 23

Let [tex]z=2-i[/tex] and [tex]w=-3+2i[/tex]. What is the midpoint of the segment joining [tex]2z[/tex] and [tex]\overline{w}[/tex]?

$-\frac{1}{2}-\frac{3}{2}i$

$\frac{1}{2}+2i$

$-\frac{1}{2}-2i$

$\frac{1}{2}-2i$

Solution:
[tex]2z=2(2-i)=4-2i[/tex]
[tex]\overline{w}=-3-2i[/tex]
The midpoint of the segment between [tex]a[/tex] and [tex]b[/tex] is [tex]\frac{a+b}{2}[/tex]:
[tex]\frac{2z+\overline{w}}{2}=\frac{(4-2i)+(-3-2i)}{2}=\frac{1-4i}{2}=\frac{1}{2}-2i[/tex]
Answer: [tex]\frac{1}{2}-2i[/tex]

Problem 24

Find the distance between [tex]z=-1+i[/tex] and [tex]w=2+3i[/tex].

$\sqrt{5}$

$\sqrt{13}$

$13$

$5$

Solution:
[tex]d(z,w)=|z-w|=|(-1+i)-(2+3i)|=|-3-2i|[/tex]
[tex]|-3-2i|=\sqrt{(-3)^{2}+(-2)^{2}}=\sqrt{9+4}=\sqrt{13}[/tex]
Answer: [tex]\sqrt{13}[/tex]

Problem 25

Let [tex]z=2-i[/tex], [tex]w=5+i[/tex] and [tex]t=-3+2i[/tex]. Calculate [tex]\frac{2z}{w-t}[/tex].

$34-12i$

$\frac{34}{65}-\frac{12}{65}i$

$\frac{34}{65}+\frac{12}{65}i$

$\frac{12}{65}-\frac{34}{65}i$

Solution:
[tex]2z=4-2i[/tex]
[tex]w-t=(5+i)-(-3+2i)=8-i[/tex]
We multiply the numerator and the denominator by the conjugate of the denominator.
[tex]\frac{4-2i}{8-i}=\frac{(4-2i)(8+i)}{(8-i)(8+i)}=\frac{32+4i-16i-2i^{2}}{64+1}=\frac{34-12i}{65}[/tex]
Answer: [tex]\frac{34}{65}-\frac{12}{65}i[/tex]

Problem 26

Let [tex]z=2-i[/tex], [tex]w=5+i[/tex] and [tex]t=-3+2i[/tex]. Calculate [tex]\overline{wt-3z}[/tex].

$23+10i$

$-23+10i$

$23-10i$

$-23-10i$

Solution:
[tex]wt=(5+i)(-3+2i)=-15+10i-3i+2i^{2}=-17+7i[/tex]
[tex]wt-3z=(-17+7i)-(6-3i)=-23+10i[/tex]
Answer: [tex]\overline{wt-3z}=-23-10i[/tex]