4^x-4^x-1=24,then find (2x)^x
1 answer:
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Answer:
1) w₁=4 - i w₂= -4 + i
2) w₁= 3 - i w₂= -3 + i
3) w₁= 1 + 2i w₂= - 1 - 2i
4) w₁= 2- 3i w₂= -2 + 3i
5) w₁= 5 - 2i w₂= -5 + 2i
6) w₁= 5 - 3i w₂= -5 + 3i
Step-by-step explanation:
The root of a complex number is given by:
where:
r: is the module of the complex number
θ: is the angle of the complex number to the positive axis x
n: index of the root
1) z = 15 − 8i ⇒ r=17 θ= -0.4899 rad
w₁==4-i
w₂==-1+i
2) z = 8 − 6i ⇒ r=10 θ= -0.6435 rad
w₁== 3 - i
w₂== -3 + i
3) z = −3 + 4i ⇒ r=5 θ= -0.9316 rad
w₁== 1 + 2i
w₂== -1 - 2i
4) z = −5 − 12i ⇒ r=13 θ= 0.4426 rad
w₁== 2- 3i
w₂== -2 + 3i
5) z = 21 − 20i ⇒ r=29 θ= -0.8098 rad
w₁== 5 - 2i
w₂== -5 + 2i
6) z = 16 − 30i ⇒ r=34 θ= -1.0808 rad
w₁== 5 - 3i
w₂== -5 + 3i