Solve z? – 2z+4 = 0. Let u and v be the solutions of z? – 2z +4 = 0, then: A. u = -1 – iv3 and v = –1+iv3. B. u = 1 – 3i and v = 1+3i. C. u = 1 – iv3 and v = 1+iv3. D. u = V3 +i and v = V3 – i. - Then the standard form of " is: 1 V3

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Question

Solve z^2−2z+4=0. Let u and v be the solutions of z^2−2z+4=0, then:

Solve z2 – 2z + 4 = 0.
Let u and v be the solutions
of z2 – 2z + 4 = 0, then:
A. u = -1 – iv3 and
v = –1+iv3.
B. u = 1 – 3i and
v = 1+ 3i.
C. u = 1– iv3 and
v = 1+iv3.
D. u = v3 + i and
- i.
v = V3
Then the standard form of
* is:
V3
E.글 + i(을).
F. - i().
G. – + i().
H. - - i().
1
2
Then
is in the:
I. First quadrant.
J. Second quadrant.
K. Third quadrant.
Transcribed Image Text:Solve z2 – 2z + 4 = 0. Let u and v be the solutions of z2 – 2z + 4 = 0, then: A. u = -1 – iv3 and v = –1+iv3. B. u = 1 – 3i and v = 1+ 3i. C. u = 1– iv3 and v = 1+iv3. D. u = v3 + i and - i. v = V3 Then the standard form of * is: V3 E.글 + i(을). F. - i(). G. – + i(). H. - - i(). 1 2 Then is in the: I. First quadrant. J. Second quadrant. K. Third quadrant.
L. Fourth quadrant.
Therefore Arg(-) is:
M. .
3
N. – .
3
4т
O.
3
With that, then the Euler
form of - is:
Р. 4е
3
Q. e'3
R. e-i.
Transcribed Image Text:L. Fourth quadrant. Therefore Arg(-) is: M. . 3 N. – . 3 4т O. 3 With that, then the Euler form of - is: Р. 4е 3 Q. e'3 R. e-i.
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