Question

If $z_{1} = 1 -2i ; z_{2} = 1 + i$ and $z_{3 } = 3 + 4i,$ then $\left( \frac{1}{z_{1}} + \frac{3}{z_{2}}\right) \frac{z_{3}}{z_{2}} =$

• A. $13 - 6i$
• B. $13 - 3i$
• C. $6 - \frac{13}{2} i$
• D. $\frac{13}{2} - 3 i$

Answer 1

Answer: (D)

$\frac{13}{2} - 3 i$

Answer 2

Z1 = 1-2i, Z2 = 1+i, and Z3 = 3+4i

(1/Z1 + 3/Z2) x (Z3/Z2) = (1/1-2i + 3/1+i) x (3+4i/1+i)

= [(1+2i)/5 + 3(1-i)/2] [(3+4i)x(1-i)/2]

Solving this equation one gets,

= (17-11i)/10 + (7+i)/2

= (119+17i-77i+11)/20

= 130-60i/20

= (13/2) - 3i

Any integer that may be expressed as a+ib is referred to be a complex number. Complex numbers, such as +3i and 7+8i, are an example. Here i = -1. This allows us to state that i2 = 1. Therefore, we may use i = -1 for any equation that does not have a true solution.

A polynomial with two roots or one of degree two is referred to as a quadratic equation. A quadratic equation has the generic form y=ax2+bx+c. The real numbers a ≠ 0, b, and c are present here.