21.3: Exercises
( \newcommand{\kernel}{\mathrm{null}\,}\)
Plot the complex numbers in the complex plane.
- 4+2i
- −3−5i
- 6−2i
- −5+i
- −2i
- √2−√2i
- 7
- i
- 0
- 2i−√3
- Answer
-
Add, subtract, multiply, and divide, as indicated.
- (5−2i)+(−2+6i)
- (−9−i)−(5−3i)
- (3+2i)⋅(4+3i)
- (−2−i)⋅(−1+4i)
- 2+3i2+i
- (5+5i)÷(2−4i)
- Answer
-
- 3+4i
- −14+2i
- 6+17i
- 6−7i
- 75+45i
- −12+32i
Find the absolute value |a+bi| of the given complex number, and simplify your answer as much as possible.
- |4+3i|
- |1−2i|
- |−3i|
- |−2−6i|
- |√8−i|
- |−2√3−2i|
- |−5|
- |−√17+4√2i|
- Answer
-
- 5
- √5
- 3
- 2√10
- 3
- 4
- 5
- 7
Convert the complex number into polar form r(cos(θ)+isin(θ)).
- 2+2i
- 4√3+4i
- 3−2i
- −5+5i
- 4−3i
- −4+3i
- −√5−√15i
- √7−√21i
- −5−12i
- 6i
- −10
- −√3+3i
- Answer
-
- 2√2(cos(π4)+isin(π4))
- 8(cos(π6)+isin(π6))
- approximately √13(cos(−.588)+isin(−.588)) or √13(cos(−.187π)+isin(−.187π))
- 5√2(cos(3π4)+isin(3π4))
- approximately 5(cos(−.644)+isin(−.644)) or 5(cos(−.205π)+isin(−.205π))
- approximately 5(cos(2.498)+isin(2.498)) or 5(cos(.795π)+isin(.795π))
- 2√5(cos(4π3)+isin(4π3))
- 2√7(cos(−π3)+isin(−π3))
- approximately 13(cos(4.318)+isin(4.318)) or 13(cos(1.374π)+isin(1.374π))
- 6(cos(π2)+isin(π2))
- 10(cos(π)+isin(π))
- 2√3(cos(2π3)+isin(2π3))
Convert the complex number into the standard form a+bi.
- 6(cos(134∘)+isin(134∘))
- 12(cos(π17)+isin(π17))
- 2(cos(270∘)+isin(270∘))
- cos(π6)+isin(π6)
- 10(cos(7π6)+isin(7π6))
- 6(cos(−5π12)+isin(−5π12))
- Answer
-
- approximately −4.168+4.316i
- approximately .491+0.0919i
- −2i
- √32+12i
- −5√3−5i
- approximately 1.553−5/796i
Multiply the complex numbers and write the answer in standard form a+bi.
- 4(cos(27∘)+isin(27∘))⋅10(cos(33∘)+isin(33∘))
- 7(cos(2π9)+isin(2π9))⋅6(cos(π9)+isin(π9))
- (cos(13π12)+isin(13π12))⋅(cos(−11π12)+isin(−11π12))
- 8(cos(3π7)+isin(3π7))⋅1.5(cos(4π7)+isin(4π7))
- 0.2(cos(196∘)+isin(196∘))⋅0.5(cos(88∘)+isin(88∘))
- 4(cos(7π8)+isin(7π8))⋅0.25(cos(−5π24)+isin(−5π24))
- Answer
-
- 40(cos(60∘)+isin(60∘))=20+20√3i
- 42(cos(π3)+isin(π3))=21+21√3i
- cos(π6)+isin(π6)=√32+12i
- 12(cos(π)+isin(π))=−12
- .1(cos(284∘)+isin(284∘))≈.0242−.0970i
- cos(2π3)+isin(2π3)=−12+√32i
Divide the complex numbers and write the answer in standard form a+bi.
- 18(cos(π2)+isin(π2))3(cos(π6)+isin(π6))
- 10(cos(254∘)+isin(254∘))15(cos(164∘)+isin(164∘))
- √24(cos(11π14)+isin(11π14))√6(cos(2π7)+isin(2π7))
- cos(8π5)+isin(8π5)2(cos(π10)+isin(π10))
- 42(cos(7π4)+isin(7π4))7(cos(5π12)+isin(5π12))
- 30(cos(−175∘)+isin(−175∘))18(cos(144∘)+isin(144∘))
- Answer
-
- 6(cos(π/3)+isin(π/3))=3+3√3i
- 23(cos(90∘)+isin(90∘))=23i
- 2(cos(π/2)+isin(π/2))=2i
- 12(cos(3π2)+isin(3π2))=−12i
- 6(cos(4π3)+isin(4π3))=−3−3√3i
- 53(cos(−319∘)+isin(−319∘))≈1.258+1.093i