Learn about calculating the modulus and argument of complex numbers, including using Pythagoras Theorem and trigonometry. Explore examples and practice problems to enhance your understanding.
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The modulus of a complex number is its magnitude you have already seen how to calculate it by using Pythagoras Theorem z y (Imaginary) The argument of a complex number is the angle it makes with the positive real axis x (Real) The argument is usually measured in radians It will be negative if the complex number is plotted below the horizontal axis z
Modulus and argument = + Im z x ( iy The complex number can be represented on an Argand diagram by the coordinates ( ) 2 2 , z ( 1 ) ) 3 3 x, y 1, z 3 z 1 z = + = 2 2 2 z i = + z i 1 3 z i arg 3 3z 2 1 Re 1 3 2 = + 2 2 z x y The modulus of z, 2 z ( ) 1 , 2 z 2 2 1 z = + = 2 1 3 2 Eg = arg z The principal argument 2 z = + = 2 2 2 1 5 Eg is the angle from the positive real axis to in the range ( ) y x z , ( 3 = = tan 1 3 z = + = 2 2 2 2 2 2 Eg ) = 1 arg z 1 3 2 3 3 = = = 1 arg tan z 3 3 4 2 4 1 = = 1 0 463 arg tan . ... 2 2 Remember the definition of arg z = 0 (2dp) 46 . z 2
WB8 a) Find, to two decimal places, the modulus and argument of z = 4 + 5i Use Pythagoras Theorem to find r y (Imaginary) 5i z 42+ 52 ? = = 41 ? = 6.40 (2??) 5 Use Trigonometry to find arg z x (Real) -5 4 5 ???? =? ???? =5 ? 4 ? = 0.90 ??????? (2??) -5i
WB8 b) Find, to two decimal places, the modulus and argument of z = -2 + 4i Use Pythagoras Theorem to find r y (Imaginary) 5i z 22+ 42 ? = = 20 4 ? = 4.47 (2??) 2.03 x (Real) Use Trigonometry to find arg z -5 2 5 ???? =? ???? =4 ? 2 -5i ? = 1.11 ??????? (2??) ? 1.11 = 2.03 ??????? Subtract from to find the required angle (remember radians = 180 ) arg? = 2.03
WB8 c) Find, to two decimal places, the modulus and argument of z = -3 - 3i Use Pythagoras Theorem to find r y (Imaginary) 5i 32+ 32 ? = = 18 ? = 4.24 (2??) Use Trigonometry to find arg z 3 x (Real) -5 5 ?? ? ???? =? ???? =3 3 ? 3 z ? =? 4 ??????? (2??) ? ? 4??????? arg? = 3? -5i 4=3? Subtract from to find the required angle (remember radians = 180 ) As the angle is below the x-axis, its written as negative 4
WB10 z = 2 3i a) Show that z2= 5 12i. b) Find, showing your working, the value of z2 , c) Find the value of arg (z2), giving your answer in radians to 2 decimal places ( )2 = 2 = Im 2 3 4 6 6 9 = 5 z i i i 12 i ( ) ( )2 = 2 2 5 12 = + z i 5 12 Re = 13 ( ) , 5 12 12 = = 1 1 176 tan . ... 5 ( Im ) = = 1. 97 arg z Re z 2 z
WB11 The complex numbers z1 and z2 are ?1= 2 + 8??2= 1 ? Find, showing your working, a) ?1 ?2 in the form a + bi, where a and b are real, b) the value of ?2 c) the value of ??? ?1 ?1 ?2answer in radians to 2 dp + + + 2 2 8 8 + 6 10 i i + 2 8 i 1 z i i = = 3+ = = i 5 1 1 + 1 1 2 + 1 z i i 2 ( ) = + 2 = z x iy = + 2 = 3+ 34 3 5 i 5 b) The modulus of is = arg z The principal argument Im ( ) 3, 5 is the angle from the positive real axis to in the range ( ) y x z , 5 1 . tan = = = arg 2. 11 z = 1 03 3 Re
WB 12 z1= 2 + i a) Find the modulus of z1 b) Find, in radians, the argument of z1 , giving your answer to 2 decimal places. Im ) 1 2, ( ) 2 + = 2 i 5 = 2 + 2 1 ( 1 = = 1 0 463 = tan . ... 2 Re = arg 2. 68 z + = 2 10 ) 5 ) 5 = 28 0 z z Im ( ( 2 z + = 25 28 0 2 Re z = 3 = 5 3 z i 5 i 3 z
modulus and argument Imaginary axis 5 Zx Real axis 5 A complex number has both modulus and argument i z 4 3+ = Then = + = 2 2 3 4 5 z arg = arctan z 4 and 3
The modulus & argument of a product It can also be shown that: It can be shown that: ?2 ??? ?1?2 = arg ?1 + arg ?2 ?1?2 = ?1 The modulus & argument of a quotient It can be shown that: It can also be shown that: ?1 ?2 ?1 ?2 ?1 ?1 ??? = arg ?1 arg ?2 = :Challenge try constructing a proof of either result
WB13a Find the modulus of the product ?1?2 and quotient ?1 ?) ?1= 2 + ? and ?2= 1 3? ?2 if : ( ) 2 ( ) 3 2 .24 1 ( ... + Im = = 1 0 46 tan . ... 1 = = + + 2 2 2 2 z z z z 2 1 1 3 1 2 1 2 1z = = 0 .46 arg ... 1z = 5 2 = 5 10 Re = = 1 0 ) 32 tan = . ... 1 2z ( This is easier than evaluating ?1?2 and then finding the modulus CHECK arg 2z = ... ( ) ) ( )( ) i 3 = = 0 46 1 24 arg . . ... z z = + + = 5 2 6 3 = + i 5 i i 2 1 z z i 1 2 1 2 ( )2 4 = 5 5 = + 2 = z z i = 5 5 50 5 2 1 2
WB13a (cont) Find the modulus of the product ?1?2 and quotient ?1 ?) ?1= 2 + ? and ?2= 1 3? ?2 if : z From previously, z 5 1 1 = 1 = = 1= z z 0 46 . 1 arg arg . ... z z 10 2 2 2 = 24 ( ... 2 This is much easier than evaluating ?1 ?2 and ( ) 2 z ) z = = 0 46 1 24 arg . .71 1 ... ... . ... 1 then finding the modulus CHECK + + 1 3 + + 2 i 2 6 3 z i i i = = 1 = 1+ i 7 + 1 3 + 1 3 10 10 i 1 9 z i 2 ( ) ( ) 10 z = + i 1 7 2 2 = = + = 1 1 1 7 50 10 10 z 10 100 2 2
WB13b Find the modulus of the product ?1?2 and quotient ?1 ?) ?1= 2 + 3? and ?2= 3 2? ?2 if : ?2 ( 2)2+32 32+ ( 2)2= 13 = ?1?2 = ?1 = ? arctan3 + arctan2 =? ??? ?1?2 = arg ?1 + arg ?2 2 3 2 Can you explain why we get this result? ?1 ?2 ?1 ?1 22+ 32 32+ 22= 1 = = ?1 ?2 = 2.159 ??? = arg ?1 arg ?2
KUS objectives BAT find the modulus and argument of complex numbers self-assess One thing learned is One thing to improve is
