MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

As we know that tan θ = 6 5 , we immediately conclude that θ = arctan 6 5 and
⇒ z =
√
61
3 (cos(arctan(6 5 )) + i. sin(arctan(6 5 ))).
8. Let z = 2+3i
5+4i
. This can be simple form as following :
From this equality,
z = 2 + 3i (2 + 3i)(5 − 4i) 22 + 7i
= = == 22
5 + 4i (5 + 4i)(5 − 4i 41 41 + 7 41 i.
|z| =
From above, we can say that
consequently
√
⇒ z =
( 22
41 )2 + ( 7
41 )2 =
√
444
1681 + 49
1681 = √
533
41 .
tan θ = 7 22 ⇒ θ = arctan 7 22 ,
9-15 Determine the principle value of the argument:
√
533
41 (cos(arctan(7 2 )) + i. sin(arctan(7 2 ))).
9. Let z = −1 − i. Then tan α = 1 and since z is in 3. region, we get α = 5π 4 .
10. Let z = −20 + i. For z,
tan α = −1
20 ⇒ α = arctan(−1 ) = 3, 09163
20
On the other hand, for z = −20 − i, tan β = 1
20
and so
⇒ β = arctan( 1 ) = −3, 09163.
20
11. We take z = 4 ∓ 3i. As a result,
tan α = ∓3
4
⇒ α = arctan( ∓3 ) = ∓0, 6435.
4
12. If z = −π 2 , then tan α = 0 and due to this we get α = π.
13. Let z = 7 + 7i. In conclusion
⇒ tan α = 1 ⇒ α = arctan( π ) = 1, 5485.
4
On the other hand, for z = 7 − 7i
tan α = −1 ⇒ α = arctan( −π ) = −1, 5485.
4
14. Take z = (1 + i) 12 . We can this clearly as following :
z = (1 + i) 12 = ((1 + i) 2 ) 6 = (1 + 2i − 1) 6 = 64(i 2 ) 3 = 64(−1) 3 = −64.
5